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3 years ago

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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 ba

ll is moving 2.77 m/s to the right after the collision, what will be the velocity ( magnitude and direction) of the pin?

1 answer:

devlian [24]3 years ago

4 0

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!

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