Question

Tasha rolls a 7.0 bowling ball down down the alley for the league championship. One pin is still standing, and Jeanne hits it head-on with a velocity of 9.0 m/s . The 2.0pin acquires a forward velocity of 14.0m/s. . What is the new velocity of the bowling ball

Answer 1

Answer:5$ms^{-1}$

Explanation:

Let $u_{1}$ be the velocity of the bowling ball before collision.

Let $u_{2}$ be the velocity of the pin before collision.

Let $v_{1}$ be the velocity of the bowling ball after collision.

Let $v_{2}$ be the velocity of the pin after collision.

Let $m_{1}$ be the mass of the bowling ball.

Let $m_{2}$ be the mass of the pin.

According to the conservation of momentum,

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Given,

$m_{1}=7Kg\\m_{2}=2Kg\\u_{1}=9ms^{-1}\\u_{2}=0ms^{-1}\\v_{2}=14ms^{-1}$

$7(9)+2(0)=7(v_{1})+2(14)$

$7v_{1}=35\\v_{1}=5ms^{-1}$

New velocity of bowling ball is $5ms^{-1}$