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 40.0 m/s and which is deflected 28.5 from its original direction. (See (Figure 1) .) Puck B acquires a velocity at a 45.0 angle to the original direction of A. The pucks have the same mass.

Part A

Compute the speed of puck A after the collision.

Part B

Compute the speed of puck B after the collision.

Part C

What fraction of the original kinetic energy of puck A dissipates during the collision?

Explanation / Answer

ma = mass of puck A = m

mb = mass of puck B = m

Vai = initial velocity of puck A = 40 m/s

Vbi = initial velocity of puck B = 0 m/s

Vaf = Final velocity of puck A

Vbf = Final velocity of puck B

Using conservation of momentum along y-direction

0 + 0 = m (Vaf) Sin28.5 - m Vbf Sin45

Vaf = 1.48 Vbf                                Eq-1

Along the X-direction

m (40) + 0 = m (Vaf) Cos28.5 + m Vbf Cos45

40 = (1.48 Vbf  ) Cos28.5 + Vbf Cos45

Vbf = 19.92 m/s

Vaf = 1.48 Vbf = 1.48 x 19.92 = 29.5 m/s

c)

fraction = ((0.5) m Vaf2 + (0.5) m Vaf2 ) /((0.5) m Vai2 ) = (29.52 + 19.922) /(40)2 = 0.792

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