In a billiard game, a cue ball of mass 180 g moving to the left at 4.75 m/s collides head-on with a stationary target ball of 165 g. If the collision between the balls is elastic,


(a)
Show that the coefficient of restitution is equal to 1.

(b)
Show that the total kinetic energy
of the system is conserved before and after the collision.

(c)
If the collision between the
billiards lasts for 0.085 s, what is the magnitude and direction of the net force exerted on the target ball?

To solve these questions, we'll use principles of conservation of momentum and conservation of kinetic energy.

(a)
The coefficient of restitution (e) is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach between two colliding objects. In an elastic collision, the coefficient of restitution is equal to 1.

To verify this, we can calculate the relative velocity of approach and separation.

Relative velocity of approach (Va) = velocity of cue ball - velocity of target ball before collision
Va = -4.75 m/s - 0 = -4.75 m/s

Relative velocity of separation (Vs) = velocity of cue ball - velocity of target ball after collision
Vs = -4.75 m/s - 0 = -4.75 m/s

Now, we can calculate the coefficient of restitution using the formula:
e = Vs / Va = -4.75 m/s / -4.75 m/s = 1

Therefore, the coefficient of restitution is equal to 1 in this case.

(b)
To show that the total kinetic energy of the system is conserved before and after the collision, we need to calculate the initial and final kinetic energies and compare them.

Initial kinetic energy (KEi) = 1/2 * mass of cue ball * (velocity of cue ball)^2 + 1/2 * mass of target ball * (velocity of target ball)^2
KEi = 1/2 * 0.180 kg * (4.75 m/s)^2 + 1/2 * 0.165 kg * (0 m/s)^2
KEi = 0.508 kg*m^2/s^2

Final kinetic energy (KEf) = 1/2 * mass of cue ball * (final velocity of cue ball)^2 + 1/2 * mass of target ball * (final velocity of target ball)^2
Since the collision is elastic, the final velocities of the cue ball and the target ball are switched.

KEf = 1/2 * 0.180 kg * (0 m/s)^2 + 1/2 * 0.165 kg * (4.75 m/s)^2
KEf = 0.508 kg*m^2/s^2

Comparing the initial and final kinetic energies, we can see that KEi = KEf. Therefore, the total kinetic energy of the system is conserved before and after the collision.

(c)
To find the net force exerted on the target ball during the collision, we can use the impulse-momentum principle. The impulse (J) is equal to the change in momentum.

The change in momentum (Δp) = (momentum of target ball after collision) - (momentum of target ball before collision)

Since the cue ball is initially moving to the left and comes to rest after the collision, the target ball receives all the momentum from the cue ball.

The momentum of the cue ball before collision = (mass of cue ball) * (velocity of cue ball before collision)
Pcue_ball_before = (0.180 kg) * (-4.75 m/s) = -0.855 kg*m/s

The momentum of the target ball before collision = (mass of target ball) * (velocity of target ball before collision)
Ptarget_ball_before = (0.165 kg) * 0 = 0 kg*m/s

The momentum of the target ball after collision = (mass of target ball) * (velocity of target ball after collision)
Since the collision is elastic, the target ball will acquire the velocity of the cue ball after the collision.

Ptarget_ball_after = (0.165 kg) * (-4.75 m/s) = -0.78375 kg*m/s

Now, we can calculate the change in momentum:
Δp = Ptarget_ball_after - Ptarget_ball_before
Δp = (-0.78375 kg*m/s) - (0 kg*m/s)
Δp = -0.78375 kg*m/s

Since the collision lasts for 0.085 s, we can use the definition of force (F) to find the magnitude and direction of the net force:
F = Δp / Δt

F = -0.78375 kg*m/s / 0.085 s

Calculating the magnitude of the net force will give the answer to part (c).