Question

A 6.0 kg bowling ball moving at 3.5m/s to the right makes a collision, head-on, with a stationary 0.70 kg bowling pin. If the ball is moving 2.77 m/s to the right after the collision, what will be the velocity ( magnitude and direction) of the pin?

Answer 1

Answer:

The velocity of the pin will be 6.26 m/s in the right direction.

Explanation:

Let's use the momentum conservation equation.

$p_{i}=p_{f}$

Initially, we have:

$p_{i}=m_{b}*v_{ib}$

Where:

• m(b) is the ball mass
• v(ib) is the initial velocity of the ball

Now, the final momentum will be:

$p_{f}=m_{b}*v_{fb}+m_{p}*v_{fp}$

Where:

• m(p) is the pin mass
• v(fb) is the final velocity of the ball

• v(fp) is the final velocity of the pin

Then, using the equation of the conservation we have:

$m_{b}*v_{ib}=m_{b}*v_{fb}+m_{p}*v_{fp}$

$6*3.5=6*2.77+0.7*v_{fp}$

$6*3.5=6*2.77+0.7*v_{fp}$

$v_{fp}=6.26 m/s$

Therefore the velocity of the pin will be 6.26 m/s in the right direction.

I hope it helps you!