MOTION WITH CONSTANT ACCELERATION
Consider
a particle with initial velocity u ms-1
and moved with a constant acceleration a
ms-2 covers a displacement s
from its initial position and attains a velocity v at time t.
Then
a = dv . dt , a is a constant
dt.
∫ a δt =
∫ δv . δt
δt
at + k = v
but
at t = 0, v=u
a(0) + k = v
k
= u
v = u + at
also v = δs
δt
u + at = δs
δt
∫
(u + at) δt = ∫ δs .δt
δt
ut + at2 + c = s
2
when
t = 0, s = 0, c = 0
S = ut + ½at2
From at v u
Substituting
in to ii)
;
S = ut + ½ (v - u)t
S
= ut + ½vt - ½ut
S
= ½ut + ½vt
2s
= ut + vt
2s
= (v + u)t
2as
= (v + u)at
2as
= (v + u)(v - u)
2as
= v2 u2
V2 = u2
+ 2as
NB:
The
equations apply only to situations involving constant acceleration. Problems
about non-uniform acceleration must be solved by other graphical methods or
calculus.
Always
choose the direction of your motion.
Example 1
A
particle moves in one line with a constant acceleration. Its initial velocity
is 6 ms-1 and its velocity after 8s is 10ms-1 and the
displacement of the particle after 16s.
Soln.
V = u + at, u = 6ms-1, v=10
T
= 8
10 = 6 + 8a
4
= 8
a
8
8
a = 0.5 ms-2
since
acceleration is constant, acceleration after 16s is 0.5ms-2
s = ut + ½at2
=6.16 + ½ x 0.5 x 162
=96 + 64 = 160m
Questions
A
stone is projected vertically upwards from the top of a cliff 20m high. After a
time of 3s, it passes the edge of a cliff on its way down. Calculate:
a)
speed of
projection
b)
speed
when it hits the ground
c)
times
when it is 30m above the top of the cliff
time when it is 5m above the ground.
2.
A particle x is projected vertically upwards from the ground with a velocity of
80ms-1. Calculate the maximum height reached by x. A particle y is
held at a height of 300m above the ground at the moment when x has dropped 80m
from its maximum height, y is projected downwards with of vms-1. The
particles reach the ground at the same time. Calculate the value of v.
FURTHER PARTICLE DYNAMICS.
Newtons
laws of Motion.
- Every body will remain at rest or
continue to move in a straight line at a constant speed unless an external
force acts on it.
- The rate of change of momentum of
a moving body is proportional to the applied force and takes place in the
direction of the force.
F
∞ m(v-u)
T
Basic
equation: F = ma
- To every action, there is an equal
and opposite reaction. If a body A exerts force on body B, then B exerts
an equal and opposite force on A. These forces between bodies are often
called reactions.
Problem
solving:
1.Body is at
rest on a rough inclined plane.
Mgsinα
= F
i (along plane)
Mgcosα
= R
ii(∟ to plane)
2.Body
sliding down rough plane at constant speed.
Forces
along the plane
Mg
sinα F = ma (resultant force)
At
constant speed a = 0
Mg
sinα F = m.0
Mg
sinα = F (frictional force)
I
R = mg cosα ii
3.Body
sliding down rough plane with acceleration.
∟
to the plane: R = mgcos α
along
the plane: mgsinα F=ma
Examples.
A
particle of mass 2kg is being lifted by a vertical force of 32N. Find its acceleration.
Resultant
force = (32 2g) = ma
= 32 2x9.8 = 2a
= 12.4 = 2a
= a = 6.2ms-2
2g
2. A
package of mass 8kg is being lowered by means of a vertical cable with a
downward acceleration of 2ms-2. Find the tension in the cable.
T
8g
T = ma
8g T = 8a 2ms-2
T = 8g 8a
8g
T = 8x9.8 8x2
T = 62.4N
3.A
boy of mass 30kg is ascending in a lift. Find the force exerted by the flow of
the lift on the boy when the lift is
a)
accelerating
at 1.2 ms-2
b)
moving
with constant speed
c)
decelerating
at 0.8ms-2.
R
a=
1.2
40g
R
- 40g = ma
R
= ma + 40g
= 40a + 40g
= 40x1.2 +40x9.8
R = 440N
b)
with constant velocity, a=0
R
40g = 0
R
= 40g
= 40x9.8
R = 392N
c)
deceleration at 0.8ms-2
R
40g = -ma
R
= -ma + 40g
= 40g 40a
= 392 32
R = 360N
Example
A
train of total mass 300 tonnes is traveling along a straight horizontal track
of constant speed of 54kmh-1. The resistances of the motion are 50N
per tonne. The rear coach of mass 50 tonnes becomes detached but the tractive
force of the engine remains the same. Calculate:
a)
the acceleration of the rest of the train
b)
the distance the rear track after becoming detached before.
(50x250) T(15000N)
300
Considering
the whole train
Total
mass = 300 tonnes
Total
resistance = (300x50)N
= 15,000N
Tractive
force = Resistance = 15000N
When
the rear coach gets detached, remaining mass = 250 tonnes.
Forces
acting
T R = ma
15000
12500 = 250x103 a
2500
=
a
250x103
a = 0.01ms-1
Considering
the detached coach
50x103Kg
R
The only force acting on the coach is the resistance of (50x50)N
-R = ma
50
x 50 = 50 x 103a
a
= - 50 x 50
50 x 103
= - 0.05ms‑2
if
a = - 0.05ms-2, v=0 s = ?, u = 54kmh-1 = 15ms-1
v2 = u2 + 2as
02 = 152 + 2 x
0.05s
0.1s = 225
0.1
0.1
s
= 2250m.
Questions:
1. A car accelerates uniformly from rest to 60km/h in 30 seconds. Find the distance it travels in this time.
2. A cyclist travels 1.25km as he accelerates uniformly at a rate of k ms-2 from a speed of 15km/h to 30km/h. Find the value of k.
4. A particle is projected vertically upwards from a point O with a speed 25ms-1. Find the maximum height reached by the particle and the time that elapses before it returns to O.
5. A body on a bridge throws a pebble vertically upwards at a speed of 6ms-1. After 2 seconds it hits the water below. Find the speed at which the pebble hits the water and its initial height above the water.
6. The points O, A, B, and C lie in a straight line such
that AB = 28m and BC = 72m. A particle moving with
constant acceleration starts at rest from rest at O and passes through A, B, and
C, its velocities at B and C being 9ms-1 and 15ms-1 respectively. Find
the velocity of the particle at A and
the time it takes to travel from A to
C.
6. A particle moving in a straight line with constant acceleration travels 10m in 2 seconds, then a further 22m in the next 2 seconds. Find the further distance traveled in 2 more seconds and the speed of the particle at the end of this 6-second interval of time.
7. A train, being brought to rest with uniform retardation, travels 30 m in 2 seconds, then a further 30m in 4 seconds. Find a retardation of the train and the additional time it takes to come to rest.
8. A particle is projected vertically upwards from a point A. Given that it rises 15m in the third second of its motion, find its initial speed and the maximum height above A that it reaches.
9. A bus sets off from a bus station P. It accelerates uniformly for T1 seconds, covering a distance of 300 m. It travels at a speed of V km/h for T2 seconds, covering a further distance of 1250m. It then decelerates uniformly for T3 seconds, coming to rest at a bus stop Q. If the total time taken in 3 minutes and 2T1 = 3T3, find the distance from P to Q and the values of T1, T2, T3 and V.
10. A train stops at a station A, It then accelerates at 0.1ms-2 for 5 minutes, reaching a speed of V ms-1. It continues at this speed for 12 minutes, then the brakes are applied for 3 minutes, bringing the train to rest with uniform retardation at a station B. Find the value of V and the distance AB.
11. A train stops at two stations 24 km apart. It takes 3 minutes to accelerate uniformly to a speed of 40ms-1 then maintains this speed until it comes to rest with uniform retardation in a distance of 1200m. Find the time taken for the journey.
12. Two stations
A and B are 10km apart. A train travelling at constant speed of 144km/h
passes station A at 10.00 hours. At a
distance d km from B the brakes are applied, producing a
constant retardation of 0.4ms-2. If the train comes to rest at
station B, find the value of d and the time at which the train reaches B.
13. A particle accelerates from rest at a constant rate of 3ms-2 to a speed of V ms-1. It continues to move at that speed for a certain time, then decelerates at a constant rate of 1.5ms-2. If the total time taken is one minute and the total distance travelled is 1 km, find the value of V.
14. A car travelling along a straight level road at a constant speed of 54kmh-1 passes a second car as it tries to accelerate from rest at a uniform rate of 0.5ms-2. Find the time that elapses and the distance covered when the second car draws level with the first.
15. A ball is thrown vertically upwards from a point A with a speed of 20ms-1. At the same instant, a second ball is dropped from a point B which is 60m vertically above A. Find the time which elapses before the two balls meet and their height above A at this instant.
16. A stone is dropped from the top of a tower. After one second, another stone is thrown vertically upwards from the same point at a speed of 15ms-1. If the stones reach the ground simultaneously, find the height of the tower.
QUESTION 2:
1. An engine exerting a constant tractive
force is pulling seven 10-tonne trucks and giving them an acceleration of
0.1m/sec2 on the level, the frictional and other resistances to
motion being 60 N/tonne. If one truck is uncoupled and the engine exerts the
same tractive force, show that the acceleration is increased by about 27 per
cent.
2. A lift with a mass of 1,200 kg is raised
from rest by a cable with a tension of 1350g N. After a time the tension drops
to 1,000g N, and the lift comes to
rest at a height of 25 m above its starting point. Find the height at which the
tension changes, the greatest speed of the gift and the total time.
3. A lorry travelling on the level at 40km/h
can be stopped by its breaks in a distance of 16m. Find the speed from which it
can be brought to rest in the same distance when descending a hill whose angle
of the slope is sin-1 (1/15).
4.
A plumb line in a ship is seen to be
inclined at an angle of 1o 30 to the vertical. Find the
acceleration of the ship.
5. A block of wood of mass 120 kg rests on a
smooth horizontal plane. On the upper face of the block, which is horizontal,
rests a mass of 10kg. The coefficient of friction between the mass and the
block is Ό. If the mass is pulled horizontally with a force of 4g N, find the accelerations of the mass
and the block. Find also how far the block will have moved when the mass has
moved 10cm along the block.
6. A train travelling uniformly on the level
at 72km/h begins an ascent of 1 in 75. The tractive force which the engine
exerts during the ascent is constant and equal to 24.5 kN, the resistance ()due
to friction, e.t.c
is constant and equal to 14.7kN, and the mass of the whole train is 255 tonnes. Show that the train will come to a standstill after climbing for 2.3km.
7. A train of 180 tonnes starts from rest with an engine pull of 29.4 kN and makes a run of 2 km from one station to rest at the next. At the instant of maximum speed, the steam is shut off and the brakes applied, producing an effective coefficient of friction of 1/30. Neglecting frictional resistance other htan those due to breaks, prove that the time occupied is about 192 sec and the maximum speed about 75 km/h.
8. Find in N per 1,000 kg the force exerted by the breakss of a train travelling at 60 km/h, which will bring it to rest in half a kilometre, and find the time during which the breaks act.
9. A bullet of mass 30 g is fired in to a fixed block of wood with a velocity of 294 metres per second and is brought to rest in 1/150 sec. Find the resistance exerted by the wood, supposing it to be uniform.
10. A shell of mass 20 kg is fired horizontally through three screens placed at equal intervals 100m apart. The velocity of the shell at the first screen is 400 m/sec and the time of flight between the first and the second screens is 0.27 sec. Assuming the retardation due to air resistance to be constant, find the retardation force and the time of flight between the second and third screens.
11. A body whose true weight was 13 N appeared to weigh 12 N when weighed by means of a spring balance in a moving lift. What was the acceleration of the lift at the instant of weighing?
12. A ship of 10,000 tonnes slows, with engines stopped from 6 knots to 5 knots in a distance of 300 m; assuming the resisatance to be uniform, calculate its value in Newtons. (A knot may be assumed to be a speed of 0.515 m/sec.)
13. A force equal to a weight of 1,000g N acts for 3 seconds on a mass of 5,000 kg. Find the velocity produced and the space passed over, stating the units in which the results are measured.
14. A and B are two railway stations 1,000 m apart on a level track. A train starting from rest at A is uniformly accelerated for one minute and reaches a speed of 36 km/h. This speed is mantained constant until the train is 80 m from B, when, steam being shut off, it is uniformly retarded by the brakes so as to stop at B. Find by means of a speed-time graph, or otherwise the acceleration and the retardation, and the time for the journey from A to B.
CONNECTED PARTICLES
Two particles connected by a light inextensible string passing over a smooth frictionless pulley are called connected particles. The tension in the string is the same through out its length, so each particle is acted upon by the same tension. Problems concerned with connected particles usually involve finding the acceleration of the system and the tension in the string.
To solve these problems:
1. Draw a clear diagram showing the forces on each particle and the common acceleration.
2. Write down the equation of motion for each particle.
3. Solve the two equations simulteneously to find the acceleration and the tension in the string.
Common Situations.
1. One particle on a smooth horizontal table as shown.
M2 > M1
a
R
T
a
T
M1g
M2g
For M1
R = M1g I
T = M1a ...ii
For M2
M2g T = M2a iii
M2g M1a = M2a
Ϋ M2g = ( M2 + M1)a
Ϋ a = M2g .
M2 + M1
2. One particle on a rough horizontal table as shown.
a
R
T
F
a
T
M1g
M2g
For M1
R = M1g ..i)
T F = M1a .ii
T = M1a + F = M1a + mR
= M1a + mM1g
For M2
M2g T = M2a iii
Ϋ M2g M1a - mM1g = M2a
Ϋ M2g mM1g = (M2 + M1)a
Ϋ a = (M2 - mM1)g
M2 + M1
Ϋ T = M2M1g ( 1 + m )
M2 + M1
3. One particle
on a smooth inclined plane as shown. M2 > M1
a
R T
M1 T
a
m1gsinq
m1gcosq m2
q
m2g
m1g
For m1
T m1gsin q = m1a I)
R = m1g cos q ..ii
For m2
M2g - T = m2a .iii
4. One particle on a rough inclined plane as shown.
R
T
T
F
M2g
M1g
For m1
T m1gsinq - F = ma
R = m1g cos q
For m2
M2g T = m2a
5. Two particles of mass m1 and m2 with m1 > m2 are connected by a light inextensible string passing over a fixed light frictionless pulley. Find the common acceleration of the mass and tension T in the string if the system is moving.
a
a T T
m2g
m1g
m1g T = m1a (i)
T m2g = m2a ii
(i) + (ii)
m1g m2g = m1a + m2a
(m1 m2)g = (m1
+ m2)a
m1 + m2
m1 + m2
a = (m1 m2) g
m1 + m2