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Q. Draw a motion diagram and a free-body diagram for a roller coaster car at the  top of a "loop-de-loop" (i.e., when the car is upside down).  Do not ignore  air resistance, and assume that the car is moving rapidly enough to remain in  firm contact with the rails.  In particular, explain  why there is no outward force on the car.

A. The forces on the roller coaster are ( in the absence of air friction and friction between rails and the car)

a. Weight mg downwards

b. Normal force N by the rails on the car, also downwards

From Newton's second law N + mg = (mv² )/ r = centripetal force to execute the circle with velocity v

Min speed at the top = (rg)^-2 where r is the radius of the loop

This speed is derived from conservation of energy. If air friction is taken into account, then v must be greater than (rg)^-2 as part of the energy will be used to overcome friction.

Outward force or cetrifugal force is a pseudo force which needs to be taken into account if we describe the motion of the car on the roller coaster from a rotating frame which is non inertial and still use Newton's laws. As we are observing the roller coaster from the ground frame, which is inertial, there is no pseudo force.

Q The ac generator in Fig below supplies 120 V (rms) at 60 Hz. With the switch open as in the diagram, the resulting current leads the generator emf by 20 degs. With the switch in position 1 the current lags the generator emf by 10 deg. When the switch is in position 2 the rms current is 2.0 A. find the values of R,L, and c.

E_v = 120V , f=60 Hz , w=2pif = 360 radians

Current I leads by 20 degs

Phase is negative, X_l < X_c

Tan (-20deg)= (X_l - X_c)/R = -0.36 ---- (1)

Or (X_c -X_l)/R =0.36

I lags by 10 degs and phase is positive

X_l > X_c1

Tan 10 degs = 0.18 = (X_l - X_c1)/R ------(2)

Z=X_c - X_l = E_v /I_v = 120/2 =60 ---------(3)

Substituting 3 in 1

60/R=0.36

R=60/0.36 = 500/3 ohms

Equation 2 implies wL-1/2wC = 0.06 x 500/3 = 30

Equation 3 implies 1/wC -wL=60 ----------(a)

wL = 30 + 1/2wC ------(b)

substituting in a

1/wC -30 - 1/2wC =60

1/2wC = 90 -------------(c)

C=0.15 µF

Substituting c in b

wL=30 + 90

L= 1/3 H

Q Find the magnetic field at the center of the loop shown.

B = (µ_o/4pi)(IL/r²)

B = 0

Since I₁R₁ = I₂R₂ or I₁L₁ = I₂L₂

Or B = (µ_o/4pir²)( I₁L₁ - I₂L₂) = 0

 

Force Q. Two people pull a box resting on a frictionless surface. One person pulls with a force of 10 N to the north. The other person pulls with a force of 15 N to the west. Find the magnitude and direction of the resultant using a graphical and algebraic approach.

A. Algebraic approach

The resultant force will be found by squaring the two given forces, adding them and then finding the square root.

Resultant force= square root of (100 + 225) = 18 N

Here 10 ² = 100 and

15 ² = 225

The graphical method would be to plot the forces on a graph and then to find the resultant as shown in the figure below.

Latest Answers

Q. An automatic rifle fires 100 gram bullets at a speed of 571 m/s at a rate of 16 bullets per second. What is the average recoil force against the gun's mount?

A.

Mass of bullet m = 100 grams = 0.1 kg

Velocity of bullet v = 571 m/s

Rate = 16 bullets / s

there fore Mass / s = 16 ´ 0.1 kg/s = 1.6 kg/s

Average force F_av = 1.6 ´ 571 = D p / D t = D mv / D t = v. D m / D t = 91.36 N

where D p = small change in force and D t = small change in time

The average recoil against the gun's mount = 91.36 N

Q. Suppose a comet of mass 5.71e+22kg moving at 1.41e+4m/s collides with the earth, initially at rest, and sticks to the earth. The mass of the earth is 5.98e+24 kg. Find the final speed of the earth.

A.

Mass of comet M_c = 5.71 x 10²² kg

Initial velocity of comet u_c = 1.41 x 10⁴ m/s

Mass of earth M_e = 5.98 x 10²⁴ kg

Initial velocity of earth u_e = 0 m/s

After hitting the earth the comet sticks to it. The collision is in-elastic. This implies that linear momentum is conserved and kinetic energy is not conserved.

therefore M_c u_c =( M_e + M_c) v

where v is the final velocity of both earth and comet.

hence

v = M_c u_c / ( M_e + M_c) = 5.71 x 10²² x 1.41 x 10⁴ / (5.98 x 10²⁴ + 5.71 x 10²²)

= 1.33 x 10² m/s

Q. A constant torque of 1.56 N-m is applied to a rotating solid disk of radius 15 cm and mass 5.71 kg. What is the angular velocity of the disk after 10 seconds?

A.

Torque T = 1.56 Nm

Radius r = 15 cm = 0.15 m

Mass m = 5.71 kg

Time t = 10 s

Initial angular velocity w ₀ of the disc is not given. We assume it to be zero.

Final angular velocity of the disc = w = w ₀ + a t

where a is the angular acceleration

and torque T = I a

where I is the moment of inertia, which for a solid disc, about an axis passing through its center and perpendicular to it is = ½ mr²

hence I = ½ x 0.15 x 0.15 x 5.71 = 0.064

hence a = 1.56 / 0.064 = 24.28

and w = 24.28 x 10 rad/s

Q. Find the moment of inertia of a drum shaped like a right circular cylinder of radius 0.156 m and mass 5.71 kg.

A. Moment of inertia I of a hollow cylinder about an axis passing through the center and perpendicular to its cross section is mr².

hence I = 5.71 x (0.156)² = 0.14 kg m²

Q. A solid 1.56 kg sphere is rolling along a level surface, without slipping at 1.41 m/s. What is its kinetic energy?

A.

Moment of inertia of a solid sphere = ²/₅ mr² = I

Kinetic energy = kinetic energy of translation + kinetic energy of rotation

=½ mv² + ½ I?² = ½ mv² + ½(²/₅ mr²)( v²/ r²) as v = ?r

=½ mv² + ¹/₅ mv²= ⁷/₁₀ mv² =⁷/₁₀ x 1.56 x (1.41)²

= 2.17 J

Q. A cord is 5.71e+1 meters long and 1.56 mm in diameter. When it supports a 1.41 kg load it stretches 3.50 cm. What is the Young's modulus of the cord's material?

A.

length l = 5.71 x 10 m = 57.1 m

d = 1.56 mm = 1.56 x 10^-3 m

r = 0.78 x 10^-3m

m = 1.41 kg

Increase in length ?l = 3.5 x 10^-2 m

Area of cross section A = p (0.78 x 10 ^-3)² m²

=1.91 x 10^-6 m²

Young's modulus Y = (F/A)/(?l/l) = stress / strain = F l /A ?l

where F=mg=1.41 x 9.8 = 13.82

Y = (13.82 x 57.1)/(1.91 x 10^-6 x 3.5 x ^-2) = 11.8 x 10⁹ N/m²

 

Q. In a spring-loaded gun the spring constant is 1.41e+02 n/m. The spring is compressed 5.71e-02 m and the gun is fired straight up, how high will a 1.56e+02 g projectile rise?

A.

k = 1.41 x 10² N/m

m = 1.56 x 10² g = 1.56 x 10^-1 kg

x = 5.71 x 10^-2 m

Elastic PE = KE = gravitational PE

½ kx² =mgh

h = ½ kx² / mg = {½ x 1.41 x 10² x (5.71 x 10^-2)²} / 1.56 x 10^-1 x 9.8

= 0.15 m = 15 cm

Q. When a mass of 5.71e+1 grams is hung on a spring, it stretches the spring by 1.56 cm. What is the frequency if the mass is then set to vibrating?

A.

m = 5.71 x 10¹ = 57.1 g

x = 1.56 cm

F = kx

hence k = F/x = 5.71 x 980 / 1.56

= 3.6 x 10⁴ dyne / cm

f = ¹/_2p (k/m)^-2 = ¹/_2p (3.6x10⁴)/57.1

=0.04 x 10² = 4 vibrations/s

Q. A simple pendulum has a length of 5.71e-01 meters and vibrates with a period of 1.56 s. What is the acceleration of gravity there?

A.

L= 5.71 x 10^-1 m

T = 1.56 s

T = 2p(L/g)^-2 hence T² = 4 p² L / g

or g = 4 p² L / T² = (4 p² x 0.571)/(1.56)²

=9.25 m/s²

Q. A block moves with simple harmonic motion with amplitude 5.71 cm and period 1.41 sec. Find the speed of the block when its displacement is 1.56 cm.

A.

a = 5.71 cm

T = 1.41 s

y = 1.56 cm

y = a Sin ?t

v=a?Cos?t = a?(1-Sin²?t) =a?(1-y²/a²)² = ?(a² - y²)

?=2p/T = 2p/1.41 = 4.45

v=4.45{(5.71)² - (1.56)²}^-2

= 24.4 cm/s

 

ATOMS

Atoms Q. How do we know atoms exist if we can't see them?
A. Even the smallest speck which can be seen through an ordinary microscope contains billions of atoms. However, scientists use very powerful instruments and techniques to study atoms. When X rays are passed through a crystal, the atoms in the crystal diffracts the X-rays in a certain way. These diffracted rays produce patterns on photographic film. In this way scientists are able to find out how the atoms are arranged and how far apart they are. Extremely powerful Scanning Electron Microscopes allow scientists to observe the positions of individual atoms.

Auto Q. I understand that car engines lose power when they gain altitude because of reduction of oxygen and other factors. How can I figure out how much reduction takes place, if I only know sea level horsepower and my current altitude?
A. The density of the atmosphere decreases as the altitude increases. If one goes up to an altitude of 6000m, the density will reduce to half. If we go up by a further 6000m, the density reduces to one fourth of the density at sea level. The reduction in density can be taken as a reduction of 1/120 every 100 meters The oxygen content will also reduce by the same ratio. However, this is not a simple variation and thus it will not be possible to work out a mathematical formula for calculation of reduction in horsepower. But one can safely say that the horsepower will also reduce approximately by the same ratio as the decrease in the density. This means that the power will reduce by 1/120 every 100 meters increase in height.

By fitting a turbocharger or a supercharger we are able to counter this effect, as compressing air is the same as making it denser.

AUTOS

Autos  Q. How does a car's design affect the temperature inside the passenger compartment on a hot sunny day? In the answer consider all sources of heat.

A. The following will affect the temperature inside the car:-

The color of the car. The darker the color the more the heat absorbed

The size of the windows and the windshields of the car. The bigger these are the more will be the increase of temperature due to the green house effect.

The engine heat. The amount of heat transferred to the passenger compartment will depend upon the effectiveness of the insulation.

The road surface heat.

The other sources of heat will be. Friction heat from air friction, tire heat, heat of suspension, heat of brake system and heat of drive line components.

ELECTRICITY

Q. A horizontal rod 2 meters long is released in a region where there is a uniform horizontal magnetic induction of 0.6 Tesla perpendicular to the rod. How far has the rod fallen when the induced emf is 120V.

E=Blu

120 = 0.6 x 2x u

u = 100 m/s

initially u = o finally u =100 m/s

v² - u² =2as = 2gh

h= (100² -0)/(2x10) =500 m

Q What is the back emf in a motor that draws 7.1 A from a 220 volt line if the field coils have a resistance of 200 ohms and are wound in parallel with an armature with a resistance of 3 ohms?

I=7.1 A , v = 220 V , R₁ = 200 ohms, R₂ = 3 ohms

Let E = back emf

I =(V-E)/R

1/R =1/3 +1/200 ie R = 600/203

E = V - IR = 220 - 21 = 199 V

Q. A coil of diameter 3 m having 120 turns and resistance 8.0 m ohms is flat on a table, and a magnetic induction perpendicular to its plane changes at a steady rate from 0 to 0.8T. How much charge flowed through the coil during the time required for the field to change?

A Flux Ø = nAB

dØ = -nAdB

dØ/dt = -nAdB/dt = -e

this implies that e = nAdB/dt =iR

which implies that nAdB = iRdt

nAdB = Rdq

Integrating both sides of the above equation with limits

q = nAB/R = nAx0.8/R

= (120 x pi x (1.5)² x 0.8)/8 = 84.78 Coulomb

Electricity Q. Explain Ohm's Law and resistance?

A. Ohm's law gives the relationship between current and the resistance to it. The unit of measuring resistance is called ohm. This law states that the voltage equals the current multiplied by the resistance, through which the current flows. For example if a current of 5 amperes passes through a resistance of 4 ohms, the voltage which will be needed will be 20 volts ( 5 amperes X 4 Ohms)

Electricity Q. How much work is needed to transfer 20 coulombs of charge between two points having a potential difference of 100 volts?
A. Work done = charge ? potential difference = 20 ? 100 = 2000J.

Electricity Q. Is it possible, under special conditions, for the energy in an electric circuit to be 100% conserved?
This meaning that none of it is ever lost through heat energy caused by friction through the conduit or load, or any other type of transformation of the energy that eventually leaves the circuit.
When would this happen, if possible, or why not, if impossible?

A. The electrical resistivity of a metal arises from the interactions of the conduction electrons with impurities, defects and the vibrating ions of the crystal lattice. As the temperature is lowered, the amplitudes of the lattice vibrations diminish, so one would expect the resistivity also to decrease gradually toward a small, but finite, value determined by impurities and defects. Many materials manifest this behavior. However, in 1911, H Kammerlingh Ownes discovered that as the temperature of a specimen of mercury was reduced its resistance suddenly dropped to extremely small values at 4.15 K. The metal had made transition to a new superconducting state. The resistivity of a superconductor is at least a factor of 10^-12 less than that for an ordinary conductor. We can usually take it to be zero. An electric current induced in a superconductor has been observed to flow for several years without any applied potential difference-provided the temperature is maintained below the critical temperature.

Electricity Q. What are the factors involved in the functioning of a light bulb?

A. The light bulb glows due to the conversion of electric energy into heat energy. As the electric current passes through the filament of a light bulb, the atoms of the filament try of prevent the flow of this current. In other words the filament act as a resistant. This leads to the release of heat energy, which in turn increases the temperature of the filament. The filament becomes white hot and starts to glow. If the light bulb did not have a vacuum in side, the filament would fuse or burn due to this heat.

ELECTRONICS

Electronics Q. In the circuit below... a) what is the potential difference between points A and B in Fig1. When switch S is open? b) Which point A or B is at the higher potential? c) What is the final potential of point B when switch S is closed? d) How much charge flows through switch S when it is closed?

A. i = 18/(6+3) = 2 A

potential at A = 18-(2x6) = 6 V

At steady state, capacitor will be fully charged

Let C_eff be the effective capacitance of the 6µF and 3µF capacitors in series

C_eff = 1/6 + 1/3 = 1/2

C_eff = 2 µF

Some charge will flow through 6 µF and 3 µF

Q=2x10^-6 x 18 = 36 x 10^-6

Potential at B = 18 - (36 x 10^-6 )/(6 x 10^-6 ) = 12V

Therefore B is at an higher potential by 6V

once the switch is closed the potential at B will become 6V

potential difference across 6µF=6V

charge on 6 µF= (6 x 10^-6 x 6) µC

this is the charge which will flow from B to A

Electronics Q. A capacitor of 6µf is connected in parallel with the combination of 4µF capacitor and a 3µF capacitor that are in series. Find (a) the net capacitance of the entire combination. and (b) The Potential Difference across the 3µF capacitor when 20V is maintained across the 6µF capacitor.

A. Capacitors 3 µF and 4 µF are in series

let C_s = capacitance of a single capacitor which replaces the above two capacitors in series. Then

1/C_s = 1/4 + 1/3 =(4+3)/12=7/12

therefore C_s = 12/7 (see Fig a above)

let C_p = capacitance of the capacitor which replaces the capacitors in parallel on Fig b above

C_p = (12/7+6) µF = 54/7 µF = net capacitance

20 V potential difference is across 6 µF as well as 12/7 µF as they are in parallel

Charge through 12/7 µF capacitor = capacitance x voltage

=12/7 µF x 20 = 240/7 µC

Charge on 4 µF and 3 µF capacitors will be the same as they are in series.

Potential difference across 3 µF = (240 µC) / (7 x 3 µF) = 80/7 V = 11.4 V

Electronics Q. One electron has a mass of 9.1 ? 10^-31 kg. How many coulombs of charge would there be in 1 kg of electrons? How much force would this charge exert on another 1 kg of electrons 1.0 km away?

A. The number of electrons in one kg of electrons will be 1 / (9.1 ? 10^-31 ). The charge on one electron is 1.6 ????^????coulombs. Thus the charge on kg of electrons will be (1.6 ????^???????1 / (9.1 ? 10^-31 ) coulombs. Force can be calculated using the Coulombs law, F = (1/4? ? ₀) q1q2/r² where q1 and q2 will be the charge on one kg of electrons each and r is 1000m, the distance by which q1 and q2 are separated.

Electronics Q. What is the difference between analog and digital devices?
A. Electronic devices work with two basic types of signals, analog and digital. Digital signals represent all information with a limited number of voltage signals. Each signal has a distinct value. Analog signals vary continuously in voltage or current, corresponding to input information. Many digital circuits can process information much faster than analog circuits. Digital circuits do the majority of processing.

ENERGY

 

Q. Determine the kinetic energy of an electron that has a wavelength of 1m.

A. l = h/p

h = 6.63 x 10 ^-34 Js

or p = h/l = (6.63 x 10 ^-34Js)/1m

k = p²/2m = (6.63 x 10 ^-34)² / 2 x 9.1 x 10 ^-31 = 2.4 x 10 ^-37 J

= (2.4 x 10 ^-37) / 1.6 x 10 ^-19 = 1.5 x 10 ^-18 eV

Energy Q. A soft drink from Australia is labeled 'low Joule Cola'. The label says 100 ml yields1.7 kJ. The can contains 375 ml. Sally drinks the cola and then offsets this input of food energy by climbing stairs. How high would she have to climb if Sally has a mass of 65kg?

A. The energy given by 375ml of cola is 1.7? 375/100=6.375 kJ. In climbing the stairs the energy will be used up to gain potential energy mgh where m is the mass of the body, g is the acceleration due to gravity and h is the height above the ground. Thus, mgh=6375 gives h=10m.

Energy Q. A 750 kg car moving at 23m/s brakes to a stop. The brakes contain about 15 kg of iron that absorbs the energy. What is the increase in the temperature of the brakes?

A. To find the retardation produced in the car due to the application of brakes, we use the third equation of motion, v²-u²=2as where v is the final velocity, u is the initial velocity, s is the distance it covers before it comes to rest and a is the retardation. We get a=529/2s. Now force applied on the car to stop is F=ma=750? 529/2s.

Thus the work done in stopping the car is W=F? d=750? (529/2s) ? s=198375J. This work done is used up in increasing the temperature of the brakes. It can be calculated by W=ms? T. The specific heat of iron is 448 J/kg? C. Thus the increase in temperature of the brakes is 29.5 ? C.

Energy Q. If you have a 200kg lump of ice hanging from a roof at a height of 5 metres above the ground, a) if the ice were to detach and fall, what would its kinetic energy be at the "instant" before it touched the ground? b) How many grams of snow(at 0 degrees)would that amount of heat be able to melt? ( the heat of fusion of snow is 334J/g)

A. As the lump falls, the potential energy decreases and the kinetic energy increases. Just before the instant when it touches the ground, the energy is totally kinetic and is equal to the potential energy at the top. Thus K.E. = mgh = 200? 9.8? 5 = 9800 J.

Let us assume that this entire energy is converted to heat energy. We can use the formula Q = mL where Q is the amount of heat evolved or absorbed when change of state takes place, m is the mass of the substance that undergoes change of state and L is the latent heat. Thus, m = Q/L =9800/334 g.

Energy Q. In the chapter on Energy what does the valve present in the piston rod of the pump run by a windmill do?

A. When the piston rod is moving up there is a low-pressure region created between the piston and the bottom of the pump. Due to this the lower valve opens where as the valve in the piston remains closed. Water is sucked in from the ground.
When the piston rod is moving down due to the increasing pressure, due to compression, the lower valve closes whereas the valve in the piston opens and the water is accumulated above the piston. When the piston rod is moving up again the piston valve is closed and the accumulated water is pushed out through the opening in the pump.
This cycle is then repeated.

FLUIDS

Fluids Q. a) A fluid is rotating at constant angular velocity w about the central vertical axis of a cylindrical container. Show that the variation of pressure in the radial direction is given by dP/dr = p'w²r, where p' is the density of the fluid.

b) Take P=P" at the axis of rotation(r=0) and show that the pressure P at any point r is P=P"+1/2(p'w²r²).

c) Show that the liquid surface is of parabolic form; that is, a vertical cross section of the surface is the curve y=w²r²/(2g).

d) Show that the variation of pressure with depth is P=p'gh. (question sent by Reeve)

A. a) Centrifugal force on the rotating liquid F = mw²r. Pressure P = F/area . Area = lateral area of the cylinder = 2? rl where r is the radius of the cylinder and l is the length of the cylinder. Also, mass of the liquid in the container = density x volume = p' ? r²l. Combining these relations, we get, P = p'r²w²/2. Differentiating, we get, dP/dr = p'rw²

b) If pressure at the axis is P" then from a) pressure at any point r is P = P" + p'r²w²/2.

c) Pressure due to rotation = p'r²w²/2 and hydrostatic pressure = p'gy where y is the depth of the liquid. Since the liquid surface is at equilibrium, these two pressures must be equal. Equating, we get, y = r²w²/2g. This is the equation of parabola.

d) Pressure at any point at a depth h is the weight of the liquid above that depth divided by the area. It is P = mg/A = mgh/Ah = mgh/V = p'gh.

FORCE

Force Q. A car is moving on a plane. What is the direction of the frictional force?

A. If a car is moving on a plane, then since the bottom of the wheel is pushing backward against the road; the force of friction is in the forward direction.

In case of a freely rolling wheel, the force of friction is backward whereas in a driven wheel, the force of friction on the wheel exerted by the road is in the forward direction. In a driven wheel, the torque due to the axle is opposite to the torque due to the friction and the vertical force N which acts at a point ahead of the center. If the wheel does not slip, the maximum value of friction is the maximum value of static friction.

Force Q. A 17 kg crate is to be pulled a distance of 20m requiring 1210 J of work being done. If the job is done by attaching a rope and pulling with a force of 75 N, at what angle is the rope held?

A. The work done by a force F in moving a body by a distance d is given by the dot product, F.d=Fd Cos? where ? is the angle between the force and distance. Substituting, we get 1210=75? 20Cos? . Thus, Cos? =121/150.

Force Q. A 180 kg. horizontal beam is supported at each end. A 200 kg piano rests a quarter of the way from the end. What is the vertical force on each of the supports?
A. The total vertical force on the support which is closer to the piano will be [200 ? (3/4)] + 180/2 = 240kgwt and that on the other support will be [200 ? (1/4)] + 180/2 = 140kgwt.

Force Q. A 33N force acting at 90 degrees and a 44N force acting at 60 degrees act concurrently on point P. What is the magnitude and direction of a third force that produces equilibrium at point P?

A. The forces have to be resolved into components along the x axis and y axis. The component along the x axis is 44Cos60 = 22N and the component along the y axis is 33 + 44Sin60 = 71.1N. The resultant of these two forces is (22² + 71.1²)^1/2 = 74.4 N. The direction of this resultant force will be = Tan^-1(71.1/22) = 72.8 degrees to the x axis. The force which will keep the point P in equilibrium will be equal to 74.4 N in magnitude and in a direction opposite to the above direction.

Force Q. A 6'1" man swings a 48" golf club weighing 275 grams and launches a 20 gram ball at a 45 degree angle. How many miles per hour must the man swing the golf club for the ball to travel 400 yards? How many more yards will each additional 5 mph of club speed generate?

A. 1mph = 0.447m/s and 1 yard = 0.914m.

The ball is set into projectile motion. The range of projectile is given by the formula - R = u²Sin2q /g where u is the initial velocity of the projectile, q is the angle with which the body is projected and g is the acceleration due to gravity and R is the range or the horizontal distance that the projectile covers.

Now, R = 400 ? 0.914 = 365.6m and ? = 45 degrees.

Substituting, we get, u = 59.8m/s. This will be the initial velocity of the ball.

When the golf club comes and strikes the ball, it will transfer some of its kinetic energy to the stationary ball. If we assume that an elastic collision takes place, then since the ball is 13.75 times lighter than the golf club, it will transfer 13.75% of its kinetic energy to the ball (this can be shown mathematically).
Thus, (13.75/100)(1/2 m1 v² ) = 1/2 m2 u² where m1 is the mass of the golf club and m2 is the mass of the golf ball and v is velocity with which the club strikes the ball and u is the initial velocity of the ball. Calculating we get, v = 43.1m/s. So this is the velocity with which the golf club will have to be swung.

Now, since the golf club is swung in a circular movement, the taller the golfer and longer the club, the more will be the radius of the circle. From, v = ? r. So for a particular velocity v, the more the value of r, the less will be the value of ? or the angular velocity or the golfer will have to swing his arms more slowly.

For calculating the increase in the range, for each 5mph increase in the club speed, we will first have to calculate the increase in the initial velocity of the ball using the energy relation and then find the new range and the increase in range.

Force Q. A horse pulls a cart by a force F1 in the forward direction. From Newton's third law of motion, the cart pulls the horse by an equal force F2 in the backward direction. The sum of these forces is thus zero. Why should then the cart accelerate forward?

A. In the above argument we have not considered all the forces acting on the horse. While trying to pull a cart, the horse pushes the ground backward with a certain force F at an angle. The ground offers an equal reaction F in the opposite direction, on the feet of the horse. The forward component of this reaction when it exceeds F2 will accelerate the horse in the forward direction. Thus the horse will be able to pull the cart in the forward direction.  

Force Q. Does a force of 20KN acting on a tripod get distributed equally? This structure consists of three sticks with the same dimensions that are holding up a weight of 20KN in three faced triangular position.

A. Yes, the force will get distributed equally.

Force Q. Friction does not depend upon the surface area of contact. Then why should a tire be properly inflated?

A. A tire has a complex structure. In case it is not filled properly, there will be greater flexing of the tire walls. This will generate heat which in turn will adversely affect the ability of the tire to steer a vehicle properly. The steering of a vehicle is achieved by the stretching of the rubber in the tire to road contact patch. The heat transfer from the tire to a cold road will be very small as compared to the amount of heat generated due to the flexing of the tire. As a tire has a tread, the adhesion of tire to a road can't be explained in simple terms. This area, is in fact, a hi-tech area on which the tire manufacturers are doing much research.

Force Q. The broom balances at its CG. If you cut the broom in half at the CG and weigh each part of the broom, which end would weigh more?

A. Both parts of the broom will have equal weights. If it were not so, there will be a net torque on the broom when you try to balance it at its CG.

Force Q. Two people pull a box resting on a frictionless surface. One person pulls with a force of 10 N to the north. The other person pulls with a force of 15 N to the west. Find the magnitude and direction of the resultant using a graphical and algebraic approach.

A. Algebraic approach

The resultant force will be found by squaring the two given forces, adding them and then finding the square root.

Resultant force= square root of (100 + 225) = 18 N

Here 10 ² = 100 and

15 ² = 225

The graphical method would be to plot the forces on a graph and then to find the resultant as shown in the figure below.

 Force Q. What is the term moment of force measured in?

A. The turning effect of a force about the axis of rotation is called moment of force or torque due to the force. It is measured as the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation.

Its unit is Nm in SI units. It may be pointed out that no doubt, Nm is equivalent to joule (the unit of work), but joule is not used as the unit of the moment of force.

GRAVITY

Gravity Q. What is Universal law of gravitation. How do Kepler's laws apply to 2-d analysis? What formula describes the motion of a pendulum and SHM?

A. The Universal law of gravitation states that every body in this universe attracts every other body with a force which is directly proportional to the product of their masses and is inversely proportional to the square of the distance. Kepler's laws of planetary motion are by their very definition, applicable to 2-d motion of the planets around the sun. In 2-d analysis, they can be applied either by resolving the motion along the x and y directions or by using (r,?) coordinates.

The equation of SHM is F = -ky where F is the restoring force and y is the displacement from the mean position and k is a constant. The negative sign shows that the restoring force is always directed opposite to the displacement The equation for displacement in SHM is y = a Sin(wt + ? ) where y is the displacement of the particle undergoing SHM, a is the amplitude of vibration, w is the angular frequency and ? is the initial phase. These equations can describe the motion of a simple pendulum as it executes SHM. The time period of simple pendulum is given by T = 2? ? (l/g) where l is the length of the pendulum and g is the acceleration due to gravity.

Gravity Q. What is the force on a 1kg ball that is falling freely due to the pull of gravity ?

A. F = mg where m is the mass of the body and g is the acceleration due to gravity (9.8 m/s²). Substituting the value we get, F = 9.8 N.

LIGHT

Q An observer looks at a source through a grating having 1000 slits per cm. If the wavelength is 550nm, how many images of the source can be seen? (Hint: For what order would theta be + or - 90 degrees)

A.

d = grating spacing

deviation angle = 30 degs

l = wave length of light

n = spectral order

N = no of lines / cm = 1000

N= 1/d or d = 1/N = 1/10³ cm = 10^-3 cm = 10 ^-5 m

nl = d Sin q

l = 550 nm = 550 x 10 ^-9 m.

Maximum number of images will be seen for greatest angle of deviation i.e. q = + or - 90 degs

hence Sin q = + or - 1

therefore the order numbers for the visible spectrum for the grating are limited to

n < = d / l = 10 ^-5 / 550 x 10 ^-9 = 18.2

hence 18 complete orders will be observed, or 36 images;18 on each side of central maximum

Light Q. How does a rainbow form?

A. Rainbow is an arch of brilliant colors that appears in the sky when the sun shines during or shortly after a shower of rain. It forms in that part of the sky opposite the sun. If the rain has been heavy, the bow may spread all the way across the sky, and its two ends seem to rest on the earth.

The reflection, refraction, and diffraction of the sun's rays as they fall on drops of rain cause this interesting natural phenomenon. These processes produce all the colors of the color spectrum--violet, blue, green, yellow, orange, and red. However, the colors of a rainbow blend into each other so that an observer rarely sees more than four or five clearly. The width of each color band varies, and depends chiefly on the size of the raindrops in which a rainbow forms. Larger drops cause narrow bands.

Light Q. Is there a critical angle for light going from glass to water? What about water to glass?

A. Critical angle is the angle of incidence for which the corresponding angle of refraction is 90 degrees. Since the refracted ray is moving further away from the normal, the light ray should be travelling from denser medium to rarer medium. So there is a critical angle for light going from glass to water, but not from water to glass.

Light Q. What kind of film allows photographs to be taken under varied lighting conditions?

A. In photography, a permanent record of an image is formed on specially treated film or paper. In normal black and white photography a camera is used to expose a film or plate to a focused image of the scene for a specified time. This time will depend on the lighting conditions. When the light is not so bright, it will require longer exposure. The film or plate is coated with an emulsion containing silver salts and the exposure to light causes the silver salts to break down into silver atoms; where the light is bright dark areas of silver are formed on the film after development (by a mild reducing agent) and fixing. The negative so formed is printed, either by a contact process or by projection. In either case, light passing through the negative film falls on a sheet of paper also coated with emulsion. Where the negative is dark, less light passes through and the resulting positive is light in this area, corresponding with a light area in the original scene. Photographic emulsions are sensitive to ultraviolet and X-rays as well.

For color-photography, various methods are used. One common method is subtractive reversal system that utilizes a film with three layers of light sensitive emulsion, one responding to each of the three primary colors. On development a black image is formed where the scene is blue. The white areas are dyed yellow, the complementary color of blue, and the blackened areas are bleached clean. A yellow filter between this emulsion layer and the next keeps blue light from the second emulsion, which is green sensitive. This is dyed magenta where no green light has fallen. The final emulsion is red-sensitive and is given a cyan (blue-green) image on the negative after dying. When white light shines through the three dye layers the cyan dye subtracts red where it does not occur in the scene, the magenta subtracts green and the yellow subtracts blue. The light projected by the negative therefore reconstructs the original scene either as a transparency or for use with printing paper.

Light Q. Why is the sky blue?

A. Sky is the region of space visible from the earth. The sky consists of the atmosphere, which extends hundreds of kilometers above the earth. The atmosphere is composed chiefly of nitrogen and oxygen. In addition, the atmosphere contains tiny water droplets and ice crystals in the form of clouds and precipitation. Smoke, dust particles, and chemical pollutants may also fill the sky over cities.

The colors of the sky result from the scattering of sunlight by the gas molecules and dust particles in the atmosphere. Sunlight consists of light waves of varying wavelengths, each of which is seen as a different color. The shortest light waves appear blue and the longest red. The blue light waves are readily scattered by tiny particles of matter in the atmosphere, but the red light waves travel undisturbed unless they are struck by larger particles.

When the sky is clear, the waves of blue light are scattered much more than those of any other color. As a result, the sky appears blue. When the sky is full of dense clouds or smoke, the light waves of all colors are scattered, causing the sky to turn gray. At sunrise or sunset, sunlight must travel farther through the atmosphere than when the sun is overhead. Light waves of most colors are scattered. Undisturbed red light waves give the sun and sky near the horizon a red or orange appearance.

MASS

Mass Q. How are mass and inertia related?

A. Inertia A body continues to be in a state of rest or uniform motion unless some external force is applied to it. Galileo called this property of objects inertia. The word inertia, which means unchanging, has been derived from the Latin word inert. Inertia of a body may be defined as, the property of a body to resist any change in its state of rest or of uniform motion in a straight.

Measure of inertia From our common experience, we know that it is easier to move a small and light object than a large and heavy one. Thus, a large/ heavy body shows a greater resistance to change in its state of rest or of uniform motion. Thus, a heavier body has more inertia than a lighter body. Therefore we can say that the larger the mass, the larger is the inertia, and the smaller the mass, the smaller is the inertia. In other words we can say that the mass of a body is a measure of its inertia.

MATHEMATICS

Mathematics Q. What is the difference between hyperbola and parabola?

A. Parabola: When an object is thrown into the air at an angle with the horizontal and it falls back to earth, the curved path it follows is a parabola. It is an open plane curve formed by the intersection of a cone with a plane parallel to its side.

Hyperbola: It is the curve produced when a cone is cut by a plane that makes a larger angle with the base than the side of the cone does.

Mathematics Q. What is triangulation and how do you calculate it?

A. Triangulation is a geometrical method to measure the height of a distant object. Let AB be an object whose height h is to be measured. Let O be the observation point. By joining AOB, we form a triangle. Using an apparatus called sextant at O, we measure angle AOB = ? . This is called angle of elevation of the object.

In triangle AOB, tan? = AB/OB = h/x. Thus, h = x tan? . Here, x is the distance of the object from the observation point. If x is known then h can be calculated.

In certain cases, x is not known. The object is said to be inaccessible. In such a case, we hold the sextant at any point C and measure angle ACB = ? . The sextant is then moved to any other point D where CD = x is measured and angle ADB = ? ' is measured. Geometrically, one can show

h = x / (cot? - cot? ' ). Thus, h can be calculated.

Misc.

Misc. Q Hello, I was wondering about this problem, and expecting you to
tell me if my ideas about it may be right (even if only theoretically, in the hypothesis of being real an absolute unit), if you please:

The problem:

If we have to go from one point to other that is ten meters away, we must pass first by the 5 meters to the goal mark; then by the 2.5,then 1.25 and so on:
If we always must cross the middle point of what is left, we ever advance less than the total remaining distance...

Possibility 1:There is a minimal, absolute spatial measure, as Planck s length.

Possibility 2:There is not, and my question is about this hypothesis; in the abstract world of mathematics, we can solve the contradiction between an infinitely divisible quantity (let s say 1) and its finiteness by convention: we can have 0.999...and we say that is less than 1 although is an infinitely extensive number, and we can only achieve 1 when we establish by convention the indivisibility in some point of the scale (lets state from 0.9 the indivisibility in 0.01 and we do 0.9 + 0.01 x 10=1 and we got to the goal)


But as I see it, as long as the basic function of numbers is to describe the relation between singularity and plurality, to divide a number is in a sense the same as to multiply it, for example =1x10 or 1/10,as long as we now have a ten units quantity, and the difference of "size" between the constituting units of each case only has sense if we compare them to
each other as a frame of reference(i.e. to unify criteria concerning the minimal unit),and thus the numbers have no autonomous value apart from which is given to them by the frame of reference provided conventionally by establishing the indivisibility of the units in some point...

The role played by conventions in maths, as I see it, is played in the sensible world by the limitations of our senses (natural or artificial) which provide the minimal(and maximal) resolution(value of the unit) that constitutes the frame of reference for vaulting space:

As in the example of the apple on the table, an observer can claim its state of rest, but an orbital observer, as the Earth spins on its axis can affirm that the apple is moving as well, but the Earth revolutes the sun(observer 3),the sun the galactic core(observer 4),and so on...

If an ant and a man observe a room ,is this a room or a football stadium?

If I got this right, all of the apple observers can affirm a different dynamic state to the fruit, and both man and ant can claim their definition of the size of the observed space; and this valuations of movement and space only depend of the frame of reference, which is constituted in turn by our resolution, that provide the minimal units, therefore to change it is to change the value of space itself, as in the case of numbers.

But as long as any number different from 0 divided by other diverse from 0 will always produce a result different from 0 as well, we can augment our resolution as many times as we want, and the value of the minimal units will always be finite and therefore we will observe the object achieve the goal after passing the correspondent number of them that form the space to cover, no matter the resolution that we may use, AS LONG AS WE DON T CHANGE OUR RESOLUTION WHILE THE OBJECT IS MOVING TO COUNTERMEASURE THE ADVANCE OF THE OBJECT, because to do so is to change our frame of reference or the value of space itself:

So if the object advances 1/2 of the track and at the same time we double the resolution, he object would remain stationary; but if we improve the resolution by ten while doubling the advance...the object gets away from the goal¡¡¡

But as long as we remain stable(or in a disadvantaged relation for improvement of resolution)from the perceptive point of view, the object reaches the goal all right.

And we wouldn't need absolute units to contain magnitudes as we don t need them to claim the state of rest of the PC while the Earth is moving...

Has this any sense? Help...

A. Our views:

Let us look at the following problem to understand what you are saying : Two trains are headed towards each other on the same track, each having a speed of 30 km/hr. A bird that can fly at 60 km/hr flies off one train when they are 60km apart and leads directly for the other train. On reaching the other train it flies back to the first train and so on. The question is - how many trips can the bird make from one train to the other before they meet? And what is the total distance the bird travels?

If we solve this problem we can show that the bird will have to make infinite number of trips whereas it will travel a distance of 60 kms. This process will take one hour, which we can calculate from the fact that the relative velocity of the trains is 60 km/hr and the original distance between them is 60 kms.

But something which has to be done infinite number of times should require infinite number of times! When we talk of a process it is something that takes place in time. The trains after all do meet up. A mathematical inference is in conflict with fact.

Words, numbers and units are tools invented by human beings to be able to observe measure and express. They are isolated and discrete. But the experiences in space and time are not discrete. Space and time is a continuum. The processes that we observe-the slow and measured movement of the stars across the sky, the formation of day and night, the thoughts in the mind, they are continuous and smoothly flowing processes. To be able to explain and measure continuous processes using discrete numbers mankind has created tools like calculus. But we must keep it in mind that mathematical tools and theories can only be applied to mathematical facts!

Coming to the other point- state of rest or motion. Nothing is in absolute rest or in absolute motion. Motion is a combined property of the object under study and the observer. Unless there is an observer, to say that a body is in rest or in motion is meaningless. So we have to have a frame of reference of the observer, and the observer has to choose a convenient set of units. Then we can make observations by utilizing the mathematical tools.

These observations have meaning only in that particular frame of reference using that particular set of units.

Misc. Q. Below is a description of a weaver stick lift. The current world record is 10lbs. I am curious what formula could a use to calculate how much force (torque?) is the person lifting trying to overcome. I understand that the farther out the weight goes the harder it gets, but I don't know how to calculate it. Does the angle matter? In the example below the person is trying to hold the stick parallel to the ground if it was angled slightly up or down would it make a difference?

Weaver Stick Lift

Stick dimensions: 42 inches long. At one end, place a notch 1/2 inch from end. The weight will be placed in notch. Thirty six (36) inches from center of notch, mark a line on stick. This will be the foremost position of the hand. Place some sort of bracket (angle brackets will work )at this point, leaving 5 1/2 inches for the gripping surface. The gripping surface may be taped, for thickness, with non-stick tape.
Place the stick on a surface, even with the lifter's hand when hanging straight down. The stick must be lifted approximately parallel to the floor. The stick must be lifted straight up from the lifting surface, with no rocking of the stick prior to lifting. The lifting hand and arm must remain free of the body, and the heel of the hand must remain on the top of the stick. If the hand twists around the stick, the lift is not allowed. The entire weight must be free of the surface and under control. The lift ends on command.
The lift may also be made by reversing the grip and grasping the stick with the little finger towards the weight, instead of the thumb towards the weight. The body may be bent during this method of lift.

A.

The formula in this case will be as under

(Downward force by the heel of the hand) x (Width of the hand) = (Weight) x (36 inches)

If the angle of the stick is different it will make a difference as the muscles of the hand which will come into play will change. Also there will be a reduction in the distance at which the weight and the downward force will be acting with reference to the reaction. In a competitive sport we would have to keep the parameters of competition exactly the same.

If we make the weight go further out to the left it would lead to an increase in the distance from 36 inches to a higher figure say 40 inches. This would mean that the right hand side of the equation will become a larger value. As a result the left hand side would have to also increase if the equation is to remain equal. This means the effort required would have to increase.

MOTION

Motion Q. A ball is dropped from the top of a tower h meters high and another ball is projected upwards simultaneously from the bottom. They meet when the upper ball has covered 1/n of the distance. Show that when they meet , the velocities are in the ratio 2:n-2 and the velocity of the lower ball is [ngh/2]1/2.

A. Upper ball: h/n=(1/2)gt² or t=[(2h)/(ng)]1/2. And (v1)=gt=[(2gh)/n]1/2. For the lower ball (h-h/n)=(u2)t +(1/2)gt². Substitute for t and simplify for (u2). (u2)=[ngh/2]1/2. By substituting in the (v2)=(u2)+gt solve for (v2). Take the ratio (v1) : (v2).

Motion Q 16. Please give some good numericals for the chapters: Describing motion; Force and acceleration; Gravitation; Waves; Simple Pendulum; Work, Power and Energy.

A. Describing Motion:

1. A body is moving along a circular path of radius R. What will be the distance and displacement of the body when it completes half a revolution? (Ans. Distance=¶R, displacement=2R)

2. Find the initial velocity of a car which is stopped in 10s by applying brakes. Retardation due to brakes is 2.5m/s².(Ans.25m/s)

3. An aeroplane taking off from a field has a run of 500m. What is the acceleration and takeoff velocity if it leaves the ground 10s after the start?(Ans. a=10m/s², v=100m/s)

4. A stone is dropped into a well. The splash is heard after 2s. How deep is the well? (Ans. 19.6m)

5. A cycle is running at a speed of 10m/s. If the radius of each wheel of the cycle is 45cm then calculate the angular velocity of the wheels. (Ans. 22.2rad/s)

6. A car moves through 20km with a speed of 40km/hr and the next 20 km with a speed of 60 km/h. Calculate the average speed.(Ans. 48km/h)

Force and Acceleration:

1.A bullet of mass 20g travelling with a velocity of 20m/s penetrates a sand bag and comes to rest in 0.05s. Find the distance through which it penetrates in the sand bag and the value of the retarding force of the sand bag. (Ans. -400m/s², 8N)

2.A certain force exerted for 1.2s raises the speed of an object from 1.8m/s to 4.2m/s. Later this same force is applied for 2s. How much does the speed change in 2s?

3.Calculate the rate of change of momentum of a body of mass 1kg when a force applied to it changes its velocity from 10m/s to 25m/s in 2s ?(Ans.7.5N)

4.A body of mass 20kg moves uniformly over a frictionless horizontal surface through a distance of 10m. Find out the work done.

5.A force of 0.6N acting on a body increases its velocity from 5m/s to 6m/s in 2s. Calculate the mass of the body.

6.A car weighing 3000kg travelling at a speed of 108km/h collides with a building and is stopped in 0.8s. What is the impulse exerted on the car?

7.A ball weighing 500g is thrown vertically up with a speed of 10m/s. What will be its momentum initially and at the highest point?

8.A particle of mass 0.5kg is kept at rest. A force of 2N acts on it for 5s. Find the distance moved by the particle in this 5s and the next 5s.

9. A feather of mass 20g is dropped from a height. It is found to fall down with a constant velocity. What is the net force acting on it?.

Motion Q. A 10kg brick falls from a height of 2m. What is its momentum as it reaches the ground?

A. We can use the third equation of motion to find the velocity of the brick as it reaches the ground, v²-u² = 2gh. Substituting, we get, v=6.28 m/s. Thus, the momentum is p=mv=10×6.28=62.8 kg m/s.

Motion Q. A 110 kg football player running at 3 m/s collides head on with a 55 kg referee by accident.  This collision gives an impulse to the referee at the expense of the player.  Irrespective of how large the impulse is, how will the magnitude of the change in the referee’s velocity during the collision compare with that of the player? Explain carefully.

A. Let m₁ , u₁ , v₁ , be the mass, initial velocity and final velocity of the player and m₂ , u₂ , v₂ , be the mass, initial velocity and final velocity of the referee. Then

m₁ = 110 kg, u₁ = 3 m/s , v₁ ,= 0 m/s

m₂ ,= 55 kg, u₂ = 0 m/s, v₂ = not known

m₁ u₁ , ₊ m₂ u₂ , = m₁ v₁ , ₊ m₂ v₂

110 x 3 + 0 = 0 + 55 x v₂

v₂ = 6 m/s

change in referee's velocity = 6-0 = 6 m/s

change in player's velocity= 0 -3 = -3 m/s

change in referees velocity as compared to player's velocity = 6-(-3) = 9 m/s

Motion Q. A ball rolling west at 3.0 m/sec has a mass of 1.0 kg. It collides with a second ball whose mass is 2.0 kg and is stationary. After the collision, the first ball moves, south. Determine the momentum and speed of each ball after the collision.

A. We assume that the collision is elastic. We apply the conditions of conservation of momentum and kinetic energy. After the collision the first ball moves south. Thus, the second ball will move in the north west direction at an angle ? to the negative x axis so that the y component of the momentum of the second ball will equalize the final momentum of the second ball and the x component will give the total final momentum equal to the initial momentum. If m₁ and m₂ are the masses of the two balls, u₁ and u₂ are the initial velocities of the two balls (u₂=0) and v₁ and v₂are the final velocities of the two balls. Writing the equations for conservation of momentum along the x and y axes, we get,

m₁u₁ =m₂v₂Cos? and m₁v₁ = m₂v₂Sin?

Applying the condition for the conservation of kinetic energy, we get

1/2m₁u₁² = 1/2m₁v₁² + 1/2m₂v₂².

Now, m₁=1kg, u₁ = -3m/s, and m₂ = 2kg. Substituting these values and solving the above equations, we get, v₁=v₂=3 ^1/2 . Thus the momenta of the two balls are the product of their mass and the speed .

Motion Q. A ball rolling west at 3.0 m/sec has a mass of 1.0 kg. It collides with a second ball whose mass is 2.0 kg and is stationary. After the collision, the first ball moves, south. Determine the momentum and speed of each ball after the collision.

A. We assume that the collision is elastic. We apply the conditions of conservation of momentum and kinetic energy. After the collision the first ball moves south. Thus, the second ball will move in the north west direction at an angle ? to the negative x axis so that the y component of the momentum of the second ball will equalize the final momentum of the second ball and the x component will give the total final momentum equal to the initial momentum. If m₁ and m₂ are the masses of the two balls, u₁ and u₂ are the initial velocities of the two balls (u₂=0) and v₁ and v₂are the final velocities of the two balls. Writing the equations for conservation of momentum along the x and y axes, we get,

m₁u₁ =m₂v₂Cos? and m₁v₁ = m₂v₂Sin?

Applying the condition for the conservation of kinetic energy, we get

1/2m₁u₁² = 1/2m₁v₁² + 1/2m₂v₂².

Now, m₁=1kg, u₁ = -3m/s, and m₂ = 2kg. Substituting these values and solving the above equations, we get, v₁=v₂=3 ^1/2 . Thus the momenta of the two balls are the product of their mass and the speed .

Motion Q. A boat heading due north crosses a wide river with a speed of 10 m/s relative to the water. The river has uniform speed of 5 m/s due east. Find the velocity of the boat w.r.t. a stationary ground observer.

A. We will have to use vectors to solve this problem. The velocity of the boat w.r.t. water is denoted by v_bw and velocity of water w.r.t. ground is denoted by v_wg and velocity of boat w.r.t. ground is denoted by v_bg. Then we have, v_bw + v_wg = v_bg. Since v_bw is along the north direction, it can be denoted along the positive y-axis and v_wg is along the east direction, it can be represented along the positive x-axis, the two vectors forming the perpendicular and base of the right angled triangle. The resultant of the two vectors will be given by the hypotenuse. Thus, v_bg = (10 ²+ 5²)^1/2= 125^1/2= 11.18 m/s.

The direction of the boat with respect to the ground will be tan^-110/2 = at an angle of 63.4° to the x-axis.

Motion Q. A car A is towing car B with a fixed towing connection at 90 MPH. Car B starts his engine and accelerates to 90 MPH. How many MPH will cars A and B be travelling at that time? Why?

A. Once car B has accelerated to 90 MPH, the cars A and B will travel at a speed of 90MPH with no tension in the tow bar. As the speed of the car B increases, the tension in the tow bar will decrease, reaching a value 0 when the car B has attained the speed of 90MPH. The speed of the system of cars A and B together will be given by the speed of the center of mass which is equal to (m1v1 + m2v2)/(m1 + m2). Assuming that the cars have the same mass, their speed will be simply 90MPH, where v1=v2=90MPH.

Motion Q. A car drives straight off the edge of a cliff that is 54m high. The police at the scene of the accident note that the point of impact is 130m from the base of the cliff. How fast was the car travelling when it went over the cliff?

A. For motion in two dimensions, we can resolve the motion into motion along x-axis and y-axis and look at the two components independently.

Looking at the motion in the vertical direction, when the car just takes off from the cliff, it has got only initial horizontal velocity and zero initial vertical velocity. However, there is no acceleration in the horizontal direction and the acceleration in the vertical direction is acceleration due to gravity, g(9.8m/s²). The distance covered in the horizontal direction is 130m and that in the vertical direction is 54m.

Vertical motion: We apply the equation of motion, h = ut + 1/2 gt². Here u = 0 and h = 54. Substituting we get, t = 3.3s.

Horizontal motion: Since there is no acceleration in the horizontal direction, s = vt where s is the distance covered and v is velocity in the vertical direction. Substituting for time from above, we get v = s/t = 130/3.3 = 39.4m/s. So the car was travelling with a speed of 39.4m/s when it went over the cliff.

Motion Q. A jet fighter is travelling horizontally at 111metres per second at an altitude of 3.00x10squared metres, when the pilot accidentally releases an outboard fuel tank.
(a) How much time elapses before the tank hits the ground?
(b) What is the velocity of the tank just before it hits the ground?

A. We will follow the logic of the previous question.

Vertical motion: initial velocity u = 0 and vertical distance h = 300m. Applying the equation of motion, h = ut + 1/2 gt², we get t = 7.8s.

For finding the final velocity in the vertical direction, we apply the equation of motion, v² - u² = 2gh. Substituting, we get, v_y = 76.7m/s.

Now, the velocity in the horizontal direction will remain 111m/s as there is no acceleration in the horizontal direction. Thus, v_x = 111m/s.

The velocity with which the tank hits the ground, v = (v_x² + v_y²)^1/2 = 134.9 m/s and the direction will be as follows: ? = tan^-1v_y/v_x where ? is the angle made with the x-axis.

Motion Q. An object weighing 1,000 pounds is attached to a rope and fed through a suspended pulley. The other end of the rope is wrapped around a 12" drum which is fixed on a rotatable shaft. The weight is allowed to fall (say 500 feet), spinning the shaft. How fast will the object fall? How fast will it spin the shaft (with only minor bearing resistance)? How much horse power will it produce at the shaft and does this change at different points of travel?

A. Let the speed of the object be v when it descends through a height h. So is the speed of the rope and hence of a particle at the rim of the drum. The angular velocity of the wheel is v/r and its kinetic energy at this instant is 1/2 I(v/r)² where r is the radius of the drum and I is moment of inertia of the drum. If the drum is a solid cylinder, then I = 1/2mr² where m is the mass of the drum and r is the radius of the drum. Using the principle of conservation of energy, the gravitational potential energy lost by the object must be equal to the kinetic energy gained by the object and the drum. Thus,

Mgh = 1/2Mv² + 1/2 I (v/r)² where M is the mass of the object. Or v = [2Mgh/(M+I/r²)]^1/2.  

The angular velocity of the spinning shaft will be ? = v/r.

The power produced at the shaft will be given by P = ? ? where ? is the torque which is rotating the drum and ? is the angular velocity of the rotating drum. ? will be given by rT where T is the tension in the rope. Yes, the power produced will change at different points of travel because as the object has fallen by different heights, the velocity of the object and hence velocity of the drum will be different. Thus, the angular velocity will be different. The power can be expressed in Horse Power by using the relation, 1HP = 746 watts.

Motion Q. For 1/p of the distance between two stations a train is uniformly accelerated and for 1/q it is uniformly retarded. It started from rest from the first station and comes to rest at the second. Prove that the ratio of the greatest velocity to the average velocity is 1+1/p+1/q.

A. Let total distance be x. Then x/p=1/2(a1)(t1)² and v=(a1)(t1). Rearranging these we get (t1)=2x/pv . Now the distance for which the train moves with uniform speed is x(1-1/p-1/q). Thus t2=(x/v)(1-1/p-1/q). For the x/q part of the distance, -v²= -2(a2)(x/q) and 0=v-(a2)(t3). Rearranging these we get (t3)=2x/vq. Now average velocity is total distance/total time or x/[(t1)+(t2)+(t3)]. Substitute for the values of time from above and the highest velocity attained is v and then take the ratio. The result is [1+1/p+1/q]

Motion Q. Give examples to demonstrate Newton's Laws of Motion.

A. First Law: Take a tumbler. Place a piece of cardboard on it. On top of the cardboard place a coin. Now flick the cardboard away with your finger. The coin will not follow the cardboard but drop into the tumbler. This is because there is no force acting on the coin and it continues to stay in a state of rest.

Second Law: A cricketer moves his hands back when taking a catch. This to increase the time period over which the velocity of the ball reduces from a high value to zero. Or in other words the acceleration is reduced. Thus the force transferred to the cricketer's hand is reduced.

Third Law: When a rocket is launched, the downward action of the exhausted gases results in the upward reaction on the rocket, causing it to rise.

Motion Q. How do you solve two dimensional trajectory problems?

A. Trajectory problems may be put into two categories. Horizontal projection and angular projection. During its motion the object covers horizontal distance due to horizontal velocity and vertical distance due to vertical velocity. So each problem can be simplified into two one dimensional problems by taking the components of position, velocity and acceleration along the horizontal and vertical directions. Let us consider the case of horizontal projection. The object is projected with initial horizontal velocity u. Since the velocity of the object in the horizontal direction is constant so the acceleration a_x along horizontal direction is zero. If the initial position of the object was ( x₀, y₀) the position of the object at any time t along the horizontal direction i.e. along the x-axis is given by x = x₀ + u_xt + 1/2 a_xt² , u and a_x = 0. If we put the origin of the co-ordinate system at the initial position then the co-ordinates of the initial position become (0,0). Our equation becomes x = 0 + ut + 1/2 (0)t ² = ut. This means t = x / u.

Let us take the downward direction as positive. Thus the acceleration in the vertical direction is g (9.8 m / s ²). The position of the object at any time t is given by y = y₀ + u_yt + 1/2 a_yt ². Here y ₀ = 0 , u_y = 0 and a_y = g . The equation then becomes y = 0 + (0) t + 1/2 g t ² = 1/2 g t ² = 1/2 g ( x / u) ² = ( 1/2 g / u ² ) x ² = k x ² where ( ( 1/2 g / u ² ) = k, a constant. This is the equation of a parabola, which is symmetric about the y-axis. Hence the path of the projectile projected horizontally from a certain height from the ground is a parabolic path. The above conditions can be substituted in the equation of motion v = u + at , separately for motion along the x and the y axis and the horizontal and the vertical can be calculated at any instant. The resultant velocity v is given by v = (v _x ² + v _y ² )^1/2 .

Motion Q. How fast must a roller coaster travel around a vertical circular track with radius 10m if it is not to fall from the track?

A. The minimum speed at the lowest point of the circular track, so that it completes the vertical circular track is (5gr)^1/2 where r is the radius and g is the acceleration due to gravity (9.8m/s²). Substituting the values we get minimum speed=22.14 m/s.

Motion Q. How long would it take for a ball to fall from a 70 feet tower?

A. This is a problem of free fall. We use the equation of motion, h = ut + 1/2 gt² to solve it. Here, h is the height by which the body falls, u is the initial velocity, t is the time taken to fall by a distance h and g is the acceleration due to gravity. Now, u = 0 and h = 70 feet. By substituting we can get the answer.

Motion Q. If you fire a bullet from a level gun and drop a bullet at the same time, both the bullets will hit the ground at the same time. Explain.

A. Both the bullets have the same acceleration in the vertical direction which is the acceleration due to gravity. Also both the bullets have zero initial velocity in the vertical direction. The bullet fired from the gun has an initial velocity in the horizontal direction but it will not affect the motion in the vertical direction. Since, both the bullets have to cover the same vertical distance, they will take the same time to do to it and will hit the ground at the same time.

A. If a is the acceleration of the falling mass and T is the tension in the string, then ma = mg - T. Since the mass falls by a distance 2m starting from rest, we can find the acceleration by using the equation of motion s = ut + 1/2at². Substituting t = 2.5s we get a = 0.64 m/s². Substituting the first equation, we get, T = 3.664N.

(b) Using the equation of motion v = u + at, the velocity with which it hits the ground is 1.6m/s.

(c) Using the relation v = rw, initial angular velocity of the pulley when the mass hits the ground is 10.7 rad/s. The pulley finally come to a stop, thus the final w = 0. Using the relation w = w0 + ? t, we get angular acceleration ? as 0.357 rad/s² where t = 30s.

(d) Now, Torque = I ? where I is the moment of inertia. And torque = rF where r is the rotation of the pulley and F is the tangential force acting on it which is equal to the tension in the string. Substituting, we get, I = 1.54 kgm²

(e) Torque to frictional resistance = rF = 0.16×3.664 = 0.55 Nm.

Motion Q. Starting from rest at t=0, a wheel undergoes a constant angular acceleration. When t=3.7 s, the angular velocity of the wheel is 7.4 rad/s. The acceleration continues until t=37 s when it abruptly ceases. Through what angle does the wheel rotate in the interval t= 0 to 74 s?

A. According to equation for rotational motion, w = w0 + ? t where w is the angular velocity after time t , w0 is the initial angular velocity and ? is the angular acceleration. Substituting, w=7.4, w0=0 and t=3.7, we get ? =2 rad/s². Using the same relation, we find the angular velocity w1at the end of 37s. Thus, w1=2×37=74 rad/s². We can find the angle covered in the first 37s using the relation, ? = w0t + 1/2? t². Thus, ? 1=1369 rad. After 37s since there is no acceleration, ? 2= w1t = 74(74-37). Total angle executed is ? = ? 1 + ? 2.

Motion Q. What are Newton's Laws of Motion?

A. First Law: A body will continue to be in a state of rest or uniform motion unless an external force is applied on it. This is also known as the law of inertia.

Second Law: Rate of change of momentum of a body is equal to the external force applied to it.

Third Law: To every action there is an equal and opposite reaction and they act on different bodies.

Motion Q. What is the difference between average speed and average velocity?

A. Please refer to the topic difference between speed and velocity on this site

Motion Q. What is the minimum speed the pilot of an aircraft should have so as to successfully loop a vertical loop without falling at the top of the loop? Also what is the thrust acting on the aircraft?

A. The forces acting on the aircraft are - the weight mg acting downwards and the force F by the air upward. Let the aircraft be at any position of the vertical loop such that the radius vector of the position of the aircraft makes an angle ? with the vertical. The weight mg can be resolved into two components- mgCos? and mgSin? . The net force acting on the aircraft will be F - mgCos? which will provide the necessary centripetal force mv²/r where r is the radius of the loop. Thus, F = mgCos? + mv²/r. Now, F will be minimum when the value of Cos? is minimum which will be -1 and it will be minimum at the highest point of the loop.

F(min) = mv²/r - mg. The aircraft will be able to complete the loop only if F(min) > 0. If v' is the velocity at the highest point, then mv'²/r ? mg or v' ? ? (gr) . This is the minimum value of velocity at the highest point.

If we apply the principle of conservation of energy, then total mechanical energy at the lowest point = total mechanical energy at the highest point.

1/2mv² = 1/2mv'² + mg(2r). If we substitute the value for v', we get, v ? ? (5gr). Thus, for looping the loop the minimum speed the aircraft should have at the lowest point should be ? (5gr). Also the maximum thrust on the aircraft will be at the lowest point. F(max) = mv²/r + mg.

Motion Q. When a ball is moving on a smooth plane, it will have acceleration only when there is an external force. Can it be said that the above statement is wrong because when ball is on a smooth inclined plane, it still has acceleration?

A. To understand this question, let us first understand Newton's first law. It says that if the vector sum of all the forces acting on a body is zero then and only then the body remains unaccelerated (i.e. remains at rest or moves with uniform velocity).

Let us consider a body moving on a smooth horizontal surface with a uniform velocity. The forces on the body are the gravitational force exerted by the earth and the contact force or the normal reaction exerted by the plane surface on the body. These forces are equal and opposite and they cancel out. Thus the net force acting on the body is zero and the body remains unaccelerated. Now, let us consider the body moving on a smooth inclined surface. Gravitational force acting on the body can be resolved into two components -mgCos? perpendicular to the inclined surface and mgSin? parallel to the inclined surface where m is the mass of the body, g is the acceleration due to gravity and ? is the angle of inclination of the incline with the horizontal. The contact force which is exerted by the surface on the body will be equal and opposite to mgCos? . If we take a resultant of all the forces acting on the body then mgSin? is the resultant force on the body acting down the inclined plane. Thus the body will have acceleration.

OPTICS

Optics Q. How (in which position) exactly a prism is used for refraction/dispersion experiments?

A. The prism should be placed in such a manner that if we view the prism from the top, we see one of the triangular faces of the prism. Then one angle of the prism can be taken as the refracting angle. Thus the refracting edge will become vertical. The incident ray falls on one of the faces which contains one of the sides making the refracting angle. The refracted ray emerges from the face containing the other edge of the refracting angle. The image showing the dispersion of white light through a glass prism is shown on the index page of edumore.netfirms.com.

Optics Q. (1) Locate the image of an object placed 10.0cm from a concave spherical mirror of radius 15.0cm.

(2)A mouse 6.0cm tall stands 12cm in front of a concave mirror whose radius of curvature is 18cm. Determine the size and location of the mouse's image.

(3)Locate the image of an object placed 8.0cm in front of a concave mirror whose radius of curvature is 20.0cm.

(4)A mouse 7.5cm tall stands at a point 4.0cm in front of a concave mirror whose focal point is 6.8cm. Describe the mouse's image in the mirror.

(5)A butterfly rests on a pin 8.6cm in front of a convex mirror of focal length 22.5cm. Where is its image?

(6)The convex rear view mirror on a car door has a focal length of 185cm. The image of an oncoming car is at 3.87m. What is the actual position of the car?

(7)An ant 5.00mm long is placed 2.0cm from a converging lens whose focal length is 3.85cm. How large and at what distance from the lens will the ant appear to be?

(8)A student looks at a mouse 8.05cm tall standing 21.8cm from a converging lens of focal length 12.5cm. Describe the image of the mouse as seen through the lens.

(9)In focusing a slide photograph on a screen, the convex lens of a projector has to be moved to a distance of 23.2cm from the slide. The focal length of the lens is 22.4cm. At what distance is the screen from the lens? How large will a figure 2.8mm tall on the slide appear to be on the screen?

(10)What is the focal length of a lens that produces a real image three times as large as the object if the distance between image and object is 1.0m?

A. We have to know the lens and the mirror formulae and the sign convention to solve the above problems.

1/v + 1/u = 1/f : mirror formula

1/v - 1/u = 1/f : lens formula

m = -v/u for a spherical mirror and m = v/u for a lens and it is positive for a convex lens when the image formed is virtual and is negative for a real image; it is always positive for a concave lens and f = R/2 where v is the image distance, u is the object distance, f is the focal length, R is the radius of curvature and m is the magnification.

Sign convention is that all distances measured in the same direction as the incident light are taken as positive and in the opposite direction are taken as negative. In case of mirrors, distances are measured from the pole and in case of lenses, the distances are measured from the optical center. Thus, the focal length of a concave mirror will be negative and that of a convex mirror will be positive. Also, the focal length of a convex lens will be positive and that of a concave lens will be negative.

(1) f = -15/2cm and u = -10cm. Substituting in the mirror formula, we get, v = -30cm. Thus the image is real and inverted and at a distance of 30cm from the mirror and on the same side as the object.

(2) R = -36cm, f = -9cm and u = -12cm

Substituting in the mirror formula, we get v = -36cm. Thus, image is on the same side as the object at a distance of 36cm from the mirror.

Magnification m = -(-36/-12) = -3

Thus size of the image is 3×6 = 18cm.

(3) Same method as (1)

(4) Same method as (2)

(5) As per the sign convention for a convex mirror, f = 22.5cm, and u = -8.6cm. Substituting in the mirror formula, 1/v = 1/22.5 + 1/8.6, we can find v.

(6) f = 185cm and v = 3.87cm as the image formed in a convex mirror is always virtual thus v will be positive. Substitute in the mirror formula to find u.

1/u = 1/185 - 1/3.87

(7) u = -2cm, f = 3.85cm. Applying the lens formula, 1/v = 1/f + 1/u = 1/3.85 - 1/2

Thus, v = -4.17cm or the image is virtual. Magnification m = 4.17/2 = 2.1 and the size of the image is 5×2.1 = 10.5mm

(8) Same as (7)

(9) u = - 23.2 cm, f = 22.4 cm. Substituting in the lens formula, v = 6.49 m. The magnification will be

m = - 649 / 23.2 = - 28. Thus, the size of the image of an object of size 2.8 mm will be 7.84 cm.

(10) If the image is real, then it is a convex lens. We have v = 1 - u. Now, v / u = (1 - u) / u = 3. Solving we get u = 1/4 m and v = 3/4 m. For a real image u is negative and v is positive. Substituting in the lens formula, we get f = 18.75 cm.

Optics Q. A block of glass has a critical angle of 39? . What is the index of refraction of the glass?

A. According to Snell's law, Sin i/Sin r = refractive index of second medium w.r.t. to the first medium where incident ray is coming from the first medium and is getting refracted in the second medium. Also, i is the angle of incidence and r is the angle of refraction.

In the problem, light is passing from the medium glass to the medium air at the angle of incidence being equal to the critical angle. In such a case, the angle of refraction is equal to 90? .
Thus, Sin 39? /Sin 90? = refractive index of air w.r.t glass = 1/? where ? is the refractive index of glass w.r.t. air. We get, ? = 1/Sin 39? .

Optics Q. How do we determine the radii of curvature for a converging glass lens(n=1.52) in order for it to have a focal length of f= 20cm. Assuming R1=R2.

A. The focal length of a convex or a concave lens can be calculated by using the lens formula,

1/f = ( n - 1 )( 1/R1 - 1/R2 ) where f is the focal length of the lens, n is the refractive index of the material of the glass lens and R1, R2 are the radii of curvature of the curved surfaces of the lens. For a convex lens, R1 is positive and R2 is negative. For a concave lens, R1 is negative and R2 is positive.

Here, f = 20, n = 1.52, R1 = R and R2 = - R. Substituting into the lens makers formula, R = 20.8 cm.

Optics Q. When the transmission axes of two Polaroid films are perpendicular to each other, what is the percentage of incident light which will pass the two films?

A. When incident light falls on the first Polaroid film, it allows only those vibrations to pass through which are parallel to its own transmission axis. The emerging light has vibrations confined to one plane only which is perpendicular to the transmission axis of the second Polaroid film. Thus no light will pass through the second Polaroid film.

Optics Q. Why would it be impossible to obtain interference fringes in a double slit experiment if the separation of the slits is less than the wavelength of light used?

The above diagram shows the double slit experiment. The condition for maxima in this is dSin? =m? where d is the slit separation, ? is the angle S'SO and ? is the wavelength of the monochromatic source of light and m is an integer. If d is less than ? , then since m is an integer, Sin? becomes greater than 1 which is not possible. Thus it is impossible to obtain interference fringes in a double slit experiment if the separation of the slits is less than the wavelength of the light used. Moreover, if d is less than ? , then S and S' stop being two coherent sources but rather behave like a single source.

PARTICLES

Q What is the speed of a particle
a) whose kinetic energy is equal to twice its rest energy?
b) Whose total energy is equal to twice its rest energy?

A

Rest energy E_o = m_o c²

E = K + E_o

E = total energy in absence of Potential Energy e = mc²

K = Kinetic energy

E_o = m_oc² = Rest energy

(i) K = 2 E_o = 2 m_o c²

E = 2E_o + E_o = 3 E_o

mc² = 3 m_o c²

m_o/(1- v²/c²)^{-0. 5} =3 m_o

3(1 - v² / c²) ^{- 0. 5} = 1

squaring both sides

9(1 - v² / c²) = 1

hence v² / c² = 8/9

or v = .943 c

(ii) E = 2 E_o

Þ 2 E_o = K + E_o

K = E_o

K = (x - 1) m_o c² where x = 1 / (1 - v² / c²) ^{- 0.5}

(x-1)m_o c² = m_o c²

{1 / (1 - v² / c²) ^{- 0.5}} - 1 = 1

1 / (1 - v² / c²) ^{- 0.5} = 2

1 - v² / c² = 1/4

v = 0.866c

Particles Q. A particle of mass m starts at rest and slides down a frictionless track. It leaves the track horizontally, at an of height of 1.25m from the ground . It strikes the ground 1m away(horizontally) from the point of projection. At what point above the ground did it start ?

A. Since the particle leaves the track horizontally, it can not be sliding down a plane incline. In case of a plane incline the projected particle will have components of initial vertical as well as horizontal velocity. In such a case additional information that is angle of the incline should be given to solve the problem. Thus the incline has to be a kind of concave incline with the curve gradually becoming horizontal. To solve this problem, we look at the horizontal and vertical motion separately.

Vertical motion: Initial velocity = 0. Applying the equation h=ut +1/2gt², h=1.25m, u=0, g=10, we get t=0.5s

Thus time for horizontal motion also is 0.5s. Horizontal distance x=1m. Applying x= ut, we get initial horizontal velocity u=2m/s. Note that there is no acceleration in the horizontal direction.

Now gravitational force is a conservative force. And the surface is frictionless. Thus the loss in potential energy in coming down the track will be equal to the gain in the kinetic energy. If it comes down by a vertical height of h1 as it comes down the concave incline then mgh1= 1/2mv². This gives h1=0.2m. Thus total height from the ground is 1.25+0.2 = 1.45m.

Particles Q. Each of the protons in a particle beam has a kinetic energy of 3.25 ? 10^-15 J. What are the magnitude and direction of the electric field that will stop these protons at a distance of 1.25m?

A. The work done in stopping the protons will be equal to the change in its kinetic energy. The kinetic energy will change from the given value to zero. Thus, W.D = ? K.E or F.d = ? K.E or qEd = 3.25 ? 10^-15 where q is the charge on a proton and E is the strength of electric field and d is the distance upto which the proton will move before coming to a stop. Now, q = 1.6? 10 ^-19 coulombs and d = 1.25m. Substituting the values we get E = 16250 N/coulombs. The direction of electric field has to be opposite to the direction of the particle beam.

Particles Q. An electron is moving at .06 m/s through a magnetic field of .5 T. Calculate the deflecting force on the electron.

A. The force on a particle having charge q moving with velocity v through a magnetic field induction B is given by the relation, F = q (v × B) = q v B sin? where ? is the angle between the velocity v and direction of magnetic field. In the question above, let the angle be 90? since no value is given, q = 1.6 ? 10^-19 coulombs, v is .06m/s and B is .5T. Substituting, we get F = 4.8 ? 10^-21 N.

SOUND

Sound Q. A noisy machine in a factory produces a decibel rating of 65dB. How many identical machines could you add to the factory without exceeding the 95-dB limit?

A. The intensity level L( in bel ) of a sound of intensity I is represented by

L = log I/I₀ where I₀ is the zero level of intensity or threshold of hearing. Also, 1 decibel = 1/10 bel.

Let the intensity of sound produced by one machine be I. Then, 6.5 = log I/I₀ . Let I' be the intensity of sound when x number of machines are used to produce decibel rating of 95dB. Then, 9.5 = log I' /I₀ .

Taking the antilog of the above equations, I/I₀ = 10^6.5 and I'/I₀ = 10^9.5. Taking the ratio of these two relations, we get, I/I' = 10³ or I' = 1000 I or the intensity of sound produced could be 1000 times that produced by one machine. This means that 999 more machines can be used.

Sound Q. Explain the terms the fundamental mode of vibration and resonant air columns.

A. Organ pipes are musical instruments which are used for producing musical sounds by blowing air into the pipe. Longitudinal stationary waves are formed on account of superimposition of incident and deflected longitudinal waves. The fundamental mode of vibration is a simplest mode of vibration and the frequency produced in this mode is the lowest frequency that can be produced and is called the fundamental frequency that can be produced. For example in a closed organ pipe ( closed at one end ) in its fundamental mode of vibration the open end acts as an antinode. This is because the air can move freely there. The closed end acts as a node because air cannot move too and fro there. If L is the length of the air column then since the distance between an antinode and node is a quarter wavelength, L = ? /4 or ? = 4L. Thus the fundamental frequency is v / 4L, where v is the velocity of the wave.

Resonant air column It is an apparatus which consists of a long cylindrical tube filled with water having its lower end joined by a rubber tubing to a moveable reservoir of water. The cylindrical water tube is fixed along a meter rod and the level of water in it can be lowered or raised with the help of a reservoir.

A tuning fork of known frequency is gently struck against a rubber pad and is held horizontally at the mouth of the water tube. At the same time the level of water in the tube is lowered till a loud sound is heard. This increase in intensity of sound is due to resonance between the tuning fork and the air column inside the water tube. This happens when the compressions and rarefactions sent down by the vibrating tuning fork reinforce each other by reflection from the water surface and at the mouth of the water tube. The first resonance position is the fundamental mode of vibration. Lower water levels will give us the next harmonics.

WAVES

Waves Q. The radar systems used by police to detect speeder are sensitive to the Doppler shift of a pulse of radio waves. Discuss how this device sensitive to the Doppler measures the speed of a car?

A. The Doppler effect is used in radar to provide information regarding the speed of moving targets by measuring the frequency shift between the emitted and the reflected radiation. A transmitter produces pulsed radio frequency radiation. It is fed to a movable aerial from which it is transmitted as a beam. When the beam strikes the moving vehicle a part of the energy of the radiation is reflected back to the aerial. Signals received by the aerial are passed to the receiver, where they are amplified and detected. There will be a shift in frequency of the reflected wave and emitted wave due to the Doppler effect. The apparent frequency of the reflected wave is given by

F = f ( 1 - v/c ) where v is the speed at which the source and the observer are moving apart and c is the speed of electromagnetic radiation, f is the real frequency or the frequency of the emitted signal

The output of the detector is usually displayed on a cathode ray tube. The apparent frequency is measured and thus the speed of the vehicle is calculated.

A heterodyne device may also be used in which beats are produced by superimposing the emitted radio wave over the reflected (from the vehicle) radio wave. In the heterodyne wave meter, a variable frequency local oscillator is adjusted to give predetermined beat frequency with the incoming reflected wave, enabling the frequency of the reflected wave which has had Doppler shift to be determined. Thus the speed of the vehicle can be determined.

WORK

Work Q. Explain the principle of a screw jack.

A. A screw jack is a device based on the simple machine "the screw". A screw is a special case of an inclined plane. You may think of it as an inclined plane, which has been rolled up around an axis. Just as it is easier to take a load up an inclined plane rather than to pull it straight up, similarly, by using a screw we can lift up an object more easily. When we use a screw jack, we convert the rotatary motion of the screw into the vertical motion of the jack. The principle of the inclined plane allows us to lift a much heavier load as compared to the effort applied for rotating the screw.