Question

A cue ball traveling at 0.60 m/s hits the stationary 8-ball, which moves off with a speed of 0.20 m/s at an angle of 33° 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?

An explosion breaks an object initially at rest into two pieces, one of which has 1.2 times the mass of the other. If 7600 J of kinetic energy were released in the explosion, how much kinetic energy did the heavier piece acquire?

Answer 1

1)

Given,The speed of cue ball, u = 0.60 m/s

The speed of 8 ball after collision, v₂ = 0.20 m/s

The angle, = 33 ^o

By using the conservation of momentum

The initial momentum = final momentum

In x -axis:

mu = mv₁cos + mv₂ cos

u = v₁cos + v₂ cos

0.60 = v₁cos + 0.20 * cos 33

v₁cos = 0.60 - 0.167

v₁cos = 0.433 -------> (1)

In y - axis:

mv₁ sin + mv₂ sin = 0

v₁ sin = - v₂ sin

v₁ sin = - 0.20 sin 33

v₁ sin = - 0.109 ------> (2)

The deflection in the path of the cue ball is, Tan = v₁ sin / v₁cos

Tan = - 0.109 / 0.433

Tan = -0.252

= -14.14 ^o clockwise with + x axis

The speed of the cue ball after the collision is:

From (1),

v₁cos = 0.433

v₁ = 0.433 / cos

= 0.433 / cos(-14.14)

= 0.447 m/s