Answer to Question #142029 in Physics for Ana

Question

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. a croquet player strikes the green ball and it gains an initial velocity of 2.26 m/s. It then strikes the blue ball which is initially at rest . After the collision the green ball has a velocity of 1.14 m/s is the same direction. If the ball roll on a frictionless surface and the the collision is head on what is the final velocity of the blue ball.

Answer 1

We can find the final velocity of the blue ball after the collision from the Law of Conservation of Momentum:

$$m_gv_{g,i}+m_bv_{b,i}=m_gv_{g,f}+m_bv_{b,f},$$

here, $m_g=0.525\ kg$ is the mass of the green ball, $m_b=0.482\ kg$ is the mass of the blue ball, $v_{g,i}=2.26\ \dfrac{m}{s}$ is the initial velocity of the green ball before the collision, $v_{b,i}=0\ \dfrac{m}{s}$ is the initial velocity of the blue ball before the collision, $v_{g,f}=1.14\ \dfrac{m}{s}$ is the final velocity of the green ball after the collision, $v_{b,f}$ is the final velocity of the blue ball after the collision.

Then, from this equation we can calculate $v_{b,f}$:

$$v_{b,f}=\dfrac{m_g(v_{g,i}-v_{g,f})}{m_b},$$ $$v_{b,f}=\dfrac{0.525\ kg\cdot(2.26\ \dfrac{m}{s}-1.14\ \dfrac{m}{s})}{0.482\ kg}=1.22\ \dfrac{m}{s}.$$

Answer:

$v_{b,f}=1.22\ \dfrac{m}{s}.$