Billiard balls in headon collision with one at rest

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In an elastic collision between two billiard balls of equal mass, where one ball travels at 4.0 m/s and collides head-on with a stationary ball, the first ball comes to a complete stop after the collision. The principle of conservation of momentum applies, leading to the conclusion that the second ball will take on the initial velocity of the first ball. Therefore, the speed of the second ball after the collision is 4.0 m/s. This outcome is consistent with the conservation of kinetic energy in elastic collisions. The problem can be solved using the provided equations for momentum and the given initial conditions.
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a billiard ball traveling 4.0m/s has an elastic head on collision witha billiard ball of equal mass that is initially at rest. the first ball is at rest after the collision. What is the the speed of the second ball after the collision?





[b(] Kim
 
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4 m/s
 
Because the collision is elastic, kinetic energy is conserved. Since the first ball has completely stopped, its kinetic energy is 0, so the second ball takes the first's kinetic energy.
 
We must conserve momentum:


m_1 v_{1i}+m_{2}v_{2i}=m_1 v_{1f}+ m_2 v_{2f}
you are given
v_{1i}=4\frac m s
and
m_1=m_2

You should be able to complete the problem.
 
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