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 29.5 from its original direction. (See (Figure 1) .) Puck B acquires a velocity at a 43.5 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

In the y direction its initial momentum of A is zero.
but its component of final momentum in y direction is m vA sin 29.5
The component of final momentum of B in the y direction is – vB sin 43.5
Hence vA sin 29.5 -vB sin 43.5 = 0
have vA/vB = sin 43.5 / sin 29.5 = 1.4
vA =1.4vB. -------------1

In the x direction, the initial momentum of A was uA = 40m m/s
Its component of final momentum is m vA sin 29.5
The component of final momentum of B in the x direction is mvB cos 43.5
40 = vA cos 29.5 + vB cos 43.5
From 1
40 = 1.4vB * cos 29.5 + vB cos 43.5
40 = 1.21 vB + 0.725 vB
40 = 1.935 vB
vB = 20.67 m/s

vA = 1.4vB
vA = 1.4 *20.67
vA == 28.93. m/s

============================
K.E of A after collision 0.5mvA^2
K.E of A before collision 0.5mu^2

Their ration = (vA/u) ^2= (28.93/40)^2 = 0.5 = 50% is the energy remaining after collision

Energy lost is 1-0.5=0.5 = 50%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.