| 1) (a) Two tennis balls each with a mass of 0.30 kg collide with each
other. After the collision, the first ball moves to the |
| right with
a velocity of 4.0 m/s and the second ball moves to the left with a velocity
of 5.0 m/s. Given that the |
| velocity
of the first ball before the collision is 5.0 m/s to the left, determine
the velocity of the second ball before |
| collision. |
| (b) Is the collision elastic? Justify your
answer using calculations. |
| |
| (a) m1 =
0.30 kg m2
= 0.30 kg |
| v1
= -5.0 m/s
v2 = ? |
| v1'
= 4.0 m/s
v2' = -5.0 m/s |
| |
| Δp
= 0 |
| pi
= pf |
| m1v1
+ m2v2 = m1v1' + m2v2' |
0.30
kg • -5.0 m/s + 0.30 kg • v2 = 0.30
kg • 4.0 m/s + 0.30 kg • -5.0 m/s |
| v2
= 4.0 m/s to the right |
| |
| (b) ΔKE
= 0 |
| KEi
= KEf |
| 1/2m1v2
+ 1/2m2v22 = 1/2m1v1'2
+ 1/2m2v2'2 |
| 1/2•0.30
kg•(-5.0 m/s)2 + 1/2•0.30 kg•(4.0 m/s)2
= 1/2•0.30 kg•(4.0 m/s)2 + 1/2•0.30 kg•(-5.0
m/s)2 |
| 6.2
J = 6.2 J |
| |
| You
may be wondering why moving to the right is a positive velocity and to
the left a negative velocity. |
| There
is no particular reason as long as you are consistent in your signs within
the same problem. The |
| same
results would be obtained if you reversed the sign convention in the problem.
In this case |
| v2
= -4.0 m/s which would still mean the first mass before the collision
would be moving to the right with |
| a
velocity of 4.0 m/s. Of course for the kinetic energy calculation it would
not make any difference |
| because
the velocities are squared. |
| |
| 2) (a) Two carts with masses of 2.0 kg and 0.90 kg are held together
by a compressed spring. When released the |
| 2.0 kg cart
moves to the left with a velocity of 6.0 m/s. Determine the velocity of
the 0.90 kg cart. |
| (b) Before the event (any change to the system
can be thought of as an event), the two carts are stationary meaning |
| that before the
event the kinetic energy is zero. After the release, each cart is moving
meaning that after the event |
| the kinetic energy
is non-zero. Is the conservation of energy being violated here? Justify
your reasoning. |
| |
| (a) m1 =
2.0 kg m2
= 0.90 kg |
| v1
= 0 m/s
v2 = 0 |
| v1'
= -6.0 m/s v2'
= ? |
| |
| Δp
= 0 |
| pi
= pf |
| m1v1
+ m2v2 = m1v1' + m2v2' |
0 = 2.0 kg • -6.0 m/s + 0.90 kg •
v2' |
| v2
= 13 m/s to the right |
| |
| (b) The conservation
of energy is not being violated. You must be careful of how you
define the system. In |
| this
problem, the system is made up of the two masses and the spring. The compressed
spring |
|
possesses elastic potential energy that is converted into kinetic
energy of the masses after the event. |
| |
| 3) A 4.0 kg mass is moving with a velocity of 12 m/s and collides with
a stationary mass of 2.0 kg. Calculate the |
| velocity of each mass after the collision. |
| |
| m1
= 4.0 kg m2
= 2.0 kg |
| v1
= 12 m/s
v2 = 0 |
| v1'
= ?
v2' = ? |
| |
| Δp
= 0 |
| pi
= pf |
| m1v1
+ m2v2 = m1v1' + m2v2' |
4.0 kg • 12 m/s + 0 = 4.0 kg • v1'
+ 2.0 kg • v2' |
| 24
m/s = 2.0 • v1' + v2' (1) |
| |
| Not
a pretty sight! One equation with two unknowns. However, life is good
because we are told that this |
| is
an elastic collision which means that the conservation of kinetic energy
is valid. |
| |
|
ΔKE
= 0 |
| KEi
= KEf |
| 1/2m1v2
+ 1/2m2v22 = 1/2m1v1'2
+ 1/2m2v2'2 |
1/2
• 4.0
kg • (12 m/s)2 + 0 = 1/2 • 4.0 kg
• v1'2 + 1/2 • 2.0 kg •
v2'2 |
| 288
m2/s2 = 2.0 • v1'2
+ v2'2 (2) |
| |
| The
problem becomes doable because equations (1) and (2) represent simultaneous
equations. If we |
| solve
(1) for v2'
and substitute into (2), we will be left
with one equation with one unknown. |
| |
| It
is worthy to note that you will need to solve a quadratic equation. If
you have a TI graphing |
| calculator,
you may find the TI-83
Solver useful. If you don't
have a calculator, you may find the |
| Quadratic
Equation Solver useful. |
| |
| The
actual detailed chug and plug can be found at Plug
And Chug For Elastic Collisions. |
| |
| By
the way, v1' = 4.0 m/s and v2' = 16 m/s. |