Question

Darna rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the collision. The pin went flying forward at 13.2 m/s. Determine the post-collision speed of the ball.

Answer 1

During collision momentum is conserved

Momentum before collision =Momentum after collision

$m_{ball}*u_{ball}+m_{pin}*u_{pin}=m_{ball}*v_{ball}+m_{pin}*v_{pin}$

$7.05kg*8.24m/s+1.52kg*0m/s=7.05kg*v_{ball}+1.52kg*13.2m/s$

$7.05*8.24+1.52*0=7.05*v_{ball}+1.52*13.2$

$58.092+0=7.05*v_{ball}+20.064$

$(58.092-20.064)/7.05=v_{ball}$

ANSWER: ${\color{Red} v_{ball}=5.394m/s}$

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