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' |
|
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 = |
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' |
|
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
|
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. |