Question

A 7.50 kg bowling ball moving6.42 m/s strikes a 1.60 kg bowlingpin at rest. After, the pin moves14.8 m/s at a -47.0° angle. Whatis the y-component of the ball'sfinal velocity?y-component (m/s)Enter

Answer 1

let the bowling ball initially is moving along the x-axis,

The whole process can be represented as,

Before the collision the net kinetic energy of the balls is,

`K_(iy)=(1)/(2)* m_1*(v_1\cos (90^o))^2+(1)/(2)* m_2*(0)^2`

where the values of the variables are given as,

`\begin{gathered} m_1=7.5\text{ kg} \\ v_1=6.42ms^(-1) \\ m_2=1.6\text{ kg} \end{gathered}`

As the value of cos(90) is zero.

Substituting the known values,

The value of the net kinetic energy with the velocity along the y-axis before the collision is,

`K_(iy)=0`

The velocity of the second ball is at rest in the initial state, thus its value is taken as zero.

After the collision of the balls, the net kinetic energy is,

`\begin{gathered} K_(fy)=(1)/(2)* m_1*(v^(\prime)_1\sin (\alpha))^2+(1)/(2)* m_1*(v^(\prime)_2\sin (-47^o))^2 \\ K_(fy)=(1)/(2)* m_1*(v_(1fy))^2+(1)/(2)* m_1*(v^(\prime)_2\sin (-47^o))^2 \end{gathered}`

Negative sign here indicating the direction of the final balls motion will be opposite to the each other.

The values of the variables are given as,

`v^(\prime)_2=14.8ms^(-1)`

Substituting the known values,

The value final kinetic enegry of the balls with velocity along the y-axis is,

`\begin{gathered} K_(fy)=(1)/(2)*7.5*(v_(1fy))^2+(1)/(2)*1.6*(14.8*\sin (360^o-47^o)^2 \\ K_(fy)=3.75*(v_(1fy))^2-93.72 \end{gathered}`

Here 360 degree is used because the pin ball direction is downward of the positive x-axis.

Then by the law of conservation of energy along the y-axis,

`\begin{gathered} K_(iy)=K_(fy) \\ 0=3.75*(v_(1fy))^2-93.72 \\ (v_(1fy))^2=(93.72)/(3.75) \\ (v_(1fy))^2=25 \\ v_(1fy)=5ms^(-1) \end{gathered}`

Thus, the y component of the final velocity of the bowling ball of mass m1 is 5 meter per second.