Question

Your friend is up next at the bowling alley. Suppose when they take their turn, the ball collided with the same pin, but this time the collision is a glancing collision. After the collision, the pin scatters at an angle of 12 degrees below the horizontal while the ball scatters at an angle of 6 degrees above the horizontal. What is the magnitude of the velocity of the bowling ball and bowling pin if the mass of the pin is 0.68 kg, the mass of the bowling ball is m= 1.6 kg, and the initial velocity is 6m/s.

Answer 1

For this problem, you have to set two equations that maintain constant the momentum in x and y axis.

`\begin{gathered} \sum_{n\mathop{=}0}^(\infty)Px=(6m/s)(1.6kg)=(V1\cdot cos(6))(1.6kg)+(V2\cdot cos(12))(0^.68kg) \\ \sum_{n\mathop{=}0}^(\infty)Py=0=(V1\cdot sin(6))(1.6kg)-(V2\cdot sin(12))(0.68kg) \end{gathered}`

`\begin{gathered} 9.6=(V1\cdot1.59)+(V2\cdot0.665) \\ 0=(V1\cdot0.167)-(V2\cdot0.141) \\ \\ V1=V2\cdot0.847 \end{gathered}`

Then you find the relation between the two values and solve the equations system.

`\begin{gathered} 9.6=V2\cdot0.847\cdot1.59+V2\cdot0.665 \\ 9.6=V2(1.346+0.665)=V2\cdot2.011 \\ V2=4.77m/s,\text{ the speed of the pin} \\ \\ V1=V2\cdot0.847=4.043m/s,\text{ the speed of the ball} \\ \end{gathered}`