Question

On a frictionless horizontal air table, puck A (with mass 0.246 kg ) is moving toward puck B (with mass 0.372 kg ), which is initially at rest. After the collision, puck A has velocity 0.121 m/s to the left, and puck B has velocity 0.653 m/s to the right.

Part A

What was the speed vAi of puck A before the collision?

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vAi =

1.111.11

  m/s  

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Note that the two pucks are traveling in opposite directions after the collision.

Part B

Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.

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ΔK =

. 23.23

  J  

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Answer 1

Using momentum conservation

Pi = Pf

mA*vAi + mB*vBi = mA*vAf + mB*vBf

given that mA = 0.246 kg & mB = 0.372 kg

vAi = ? m/sec

vBi = 0 m/sec

vAf = -0.121 m/sec

vBf = 0.653 m/sec

So Using these values

0.246*vAi + 0.372*0 = 0.246*(-0.121) + 0.372*0.653

vAi = [0.246*(-0.121) + 0.372*0.653]/0.246

vAi = 0.866 m/sec towards the right

Part B.

Change in total KE will be

dKE = KEi - KEf

dKE = (0.5*mA*vAi^2 + 0.5*mB*vBi^2) - (0.5*mA*vAf^2 + 0.5*mB*vBf^2)

Using known values

dKE = (0.5*0.246*0.866^2 + 0.5*0.372*0^2) - (0.5*0.246*(-0.121)^2 + 0.5*0.372*0.653^2)

dKE = 0.011 J

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