Question

A 6.2 kg bowling ball is thrown straight down a lane at 8.4 m/s and makes a head on collision with a 1.6 kg bowling pin initially at rest after the collision the bowling ball is moving in the same direction but has a velocity of only 6.7 m/s what is the speed of the bowling pin after the collision

Answer 1

We can find the speed of the bowling pin after the collision from the Law of Conservation of Momentum:

$$m_bv_{b,i}+m_pv_{p,i}=m_bv_{b,f}+m_pv_{p,f},$$

here, $m_b=6.2\ kg$ is the mass of the bowling ball, $m_p=1.6\ kg$ is the mass of the bowling pin, $v_{b,i}=8.4\ \dfrac{m}{s}$ is the initial speed of the bowling ball before the collision, $v_{p,i}=0\ \dfrac{m}{s}$ is the initial speed of the bowling pin before the collision, $v_{b,f}=6.7\ \dfrac{m}{s}$ is the final speed of the bowling ball after the collision, $v_{p,f}$ is the final speed of the bowling pin after the collision.

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

$$v_{p,f}=\dfrac{m_bv_{b,i}-m_bv_{b,f}}{m_p},$$ $$v_{p,f}=\dfrac{6.2\ kg\cdot 8.4\ \dfrac{m}{s}-6.2\ kg\cdot 6.7\ \dfrac{m}{s}}{1.6\ kg}=6.6\ \dfrac{m}{s}.$$

Answer:

$v_{p,f}=6.6\ \dfrac{m}{s}.$