Question

Ball 1 is moving toward you at 11 m/s , and you decide to throw ball 2 at it to make it reverse its velocity. The balls collide head-on, and the coefficient of restitution for the collision is 0.90. Ball 1 has an inertia of 0.500 kg and ball 2 has an inertia of 0.650 kg . Call the direction of motion in which we throw ball 2 the +x direction.
1.How fast must ball 2 be traveling in order to reverse the velocity of ball 1?
2.What is the initial relative speed of the two balls?
3.How do the balls initially move?
4.What is their reduced inertia?
5.What percentage of the original kinetic energy is convertible?
6.What is the x-component of the final velocity of the ball 1 immediately after the collision
7.What is the x-component of the final velocity of the ball 2 immediately after the collision?

Answer 1

1. Let $v_{1s}=-11, v_{2s}$ be the velocities of balls before collision. Let $v_{1f}=11, v_{2f}$ be the velocities of balls after collision. From the momentum conservation law obtain:

$m_1v_{1s}+m_2v_{2s}=m_1v_{1f}+m_2v_{2f}$

From the definition of coefficient of restitution: $0.9 = |\dfrac{v_{1f}-v_{2f}}{v_{1s}-v_{2s}}|.$

Substituting values to the first equation : $0.5\times(-11) +0.65v_{2s}=0.5\times11+0.65v_{2f}\\ 0.65v_{2s}-0.65v_{2f}=11$

Substituting values to the second equation:

$0.9\times (-11)-0.9v_{2f}=11-v_{2s}\\ v_{2s}-0.9v_{2f}=20.9$

By solving these two equations obtain:

$v_{2s} = 56. 7, v_{2f}=39.8$

Thus, ball 2 mast be traveling with the velocity 56.7 m/s.

2. The initial relative speed of the two balls is: $v_{2s} - v_{1s} = 56.7 - (-11) =67.7 m/s.$

3 . balls initially move towards each other with velocities 11 m/s and 56.7 m/s.

4. Reduced inertia is: $\mu =\dfrac {m_1m_2} {m_1+m_2} =\dfrac {0.5\times 0.65} {0.5+0.65}=0.28 kg$

6. The x-component of the final velocity of the ball 1 immediately after the collision is $v_{1f}=11 m/s.$

7. The x-component of the final velocity of the ball 2 immediately after the collision is $v_{2f}= 39.8 m/s.$