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?

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

Explanation:

Let's use the momentum conservation equation.

Initially, we have:

Where:

m(b) is the ball massv(ib) is the initial velocity of the ball

Now, the final momentum will be:

Where:

m(p) is the pin massv(fb) is the final velocity of the ballv(fp) is the final velocity of the pin

Then, using the equation of the conservation we have:

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

I hope it helps you!

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