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

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whitneytr12

Answer:

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

Explanation:

Let's use the momentum conservation equation.

[tex]p_{i}=p_{f}[/tex]

Initially, we have:

[tex]p_{i}=m_{b}*v_{ib}[/tex]

Where:

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

Now, the final momentum will be:

[tex]p_{f}=m_{b}*v_{fb}+m_{p}*v_{fp}[/tex]

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:

[tex]m_{b}*v_{ib}=m_{b}*v_{fb}+m_{p}*v_{fp}[/tex]

[tex]6*3.5=6*2.77+0.7*v_{fp}[/tex]

[tex]6*3.5=6*2.77+0.7*v_{fp}[/tex]

[tex]v_{fp}=6.26 m/s[/tex]

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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