A hockey puck B rests on a smooth surface of ice and is struck by a second puck

Question

A hockey puck B rests on a smooth surface of ice and is struck by a second puck A , which was originally traveling at v= 40.0m/s and is deflected, at angle1 = 27.0 degrees from its original direction. Puck B acquires a velocity at angle2 = 47.0 degrees to the original direction of A. The pucks have the same mass. Compute the speed of puck A after the collision. Compute the speed of puck B after the collision. What fraction of the original kinetic energy of puck A dissipates during the collision?

Explanation / Answer

Total Momentum before collision = m*va + m*vb = m*40 + m*0 = 40m

Total Momentum in original direction after collision = m*va'*Cos27 + m*vb'*Cos47

Total momentum in perpendicular direction to original direction after collision = m*va'*Sin27 - m*vb'*Sin47

Momentum conservation: 40m = m*va'*Cos27 + m*vb'*Cos47...........(1)

and 0 = m*va'*Sin27 - m*vb'*Sin47...........(2)

From (2) we get, vb' = 0.62*va'

Putting it in (1) we get, va' = 30.43 m/s

So, vb' = 0.62*30.43 = 18.87 m/s

Initial KE = 1/2*m*40^2 = 800m

Final KE = 1/2*m*va'^2 + 1/2*m*vb'^2 = 641m

KE loss = 800m - 641m = 159m

Fraction of KE loss = 159m/800m = 0.1987

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