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After being stuck by a bowling ball, a 1.8 kg bowling pin slides to the right at 3.6 m/s and collides head-on with another 1.8 kg bowling pin initially at rest. What is the final velocity of the second pin if the first pin moves to the right at 0.54 m/s after the collision

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Answer:

Step-by-step explanation:

We shall apply law of conservation of momentum at the time of collision

Momentum before collision of both the pins = 1.8 x 3.6 + 0 = 6.48 kg m/s

Momentum after the collision of both the pins = 1.8 x .54 + 1.8 x v where v is velocity of second pin

So , according to law

6.48 = 1.8 x .54 + 1.8 x v

6.48 = .972 + 1.8 x v

v = 3.06 m /s to the right .

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