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You are bowling for a spare. There is just one pin left, and you make a nice play. As the 5.1 kg ball approches at a velocity of 3.40 m/s north it hits the pin! The 1.5 kg pin is sent into motion and the bowling ball continues to roll at a velocity of 0.87 m/s north. What is the final velocity of the pin?

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Answer:11.56 m/sExplanation:The momentum of the ball is transferred into the pin so the transfer of momentum has the formula, m1*v1 = m2 * v2substituting the values,5.1 * 3.40 = 1.5 * V217.34 = 1.5 * V2V2 = 17.24 / 1.5V2 = 11.56 m/s

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