Question

Two balls with masses of 2.50 kg and 5.60 kg travel toward each other at speeds of 14.0 m/s and 4.20 m/s, respectively. If the balls have a head-on inelastic collision and the 2.50-kilogram ball recoils with a speed of 8.40 m/s, how much kinetic energy is lost in the collision?

A cue ball traveling at 0.80 m/s hits the stationary 8-ball, which moves off with a speed of 0.27 m/s at an angle of 31° relative to the cue ball's initial direction. Assuming that the balls have equal masses and the collision is inelastic, what will be the speed of the cue ball?

Answer 1

Use conservation of momentum:

The 2kg ball has a momentum of (2.5 kg)(14 m/s) = 35 kg*m/s.
The 5.6kg ball has a momentum of (5.6 kg)(-4.20 m/s) = -23.52 kg*m/s.

So, the total initial momentum is (35 kg*m/s) + (-23.52kg*m/s) = 11.48 kg*m/s

The final momentum of the 2.5kg ball is (2.5 kg)(-8.40 m/s) = -21 kg*m/s

So, the momentum of the 5.66kg ball is (11.48kg*m/s) - (-21 kg*m/s) = 32.48 kg*m/s

Thus its velocity is (32.48 kg*m/s)/(11.48 kg) = 2.83 m/s

The initial energy is:
(1/2)(2.5kg)(14 m/s)^2 + (1/2)(5.6 kg)(4.20m/s)^2 = 294.4 J

The final energy is:
(1/2)(2.5 kg)(8.40 m/s)^2 + (1/2)(5.6 kg)(2.83m/s)^2 = 110.6 J

So, the energy lost is: (294.4J) - (110.6 J) = 183.775 J ...Answer.

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cue ball's initial velocity, u₁= 0.80 m/s
8-ball's initial velocity u ₂= 0

Since 8-ball's final velocity is 0.27m/s at 31° to cue's ball direction, the component along the cue ball's direction is, v₂= 0.27cos31 m/s

cue ball's final velocity = v₁

Since the collision is inelastic only the momentum is conserved

mu₁+ mu₂= mv₁+ mv₂

u₁+ u₂= v₁+ v₂
v₁= u₁+ u₂- v₂
v₁= 0.80+ 0 - 0.27cos31
v₁= 0.57 m/s......Answer.

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