A softball of mass 0.220 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 4.8 m/s.

(a) Calculate the velocity of the target ball after the collision.

(b) Calculate the mass of the target ball.

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.

(a) Let's start by considering the conservation of momentum. The momentum before the collision is equal to the momentum after the collision. The momentum of an object is given by the product of its mass and velocity.

The initial momentum of the softball (mass = 0.220 kg, velocity = 6.5 m/s) is given by:
P_initial = mass * velocity = 0.220 kg * 6.5 m/s

Since the target ball is initially at rest, its initial momentum is zero:
P_initial_target = 0

After the collision, the softball moves backward and the target ball moves forward. Let V_target be the velocity of the target ball after the collision. The final momentum of the softball is given by:
P_final = mass * (-4.8 m/s) (since it moves backward)

The final momentum of the target ball is given by:
P_final_target = mass_target * V_target

According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
P_initial + P_initial_target = P_final + P_final_target

Substituting the given values, we have:
0.220 kg * 6.5 m/s + 0 = 0.220 kg * (-4.8 m/s) + mass_target * V_target

Simplifying this equation, we can solve for V_target:
V_target = (0.220 kg * 6.5 m/s - 0.220 kg * (-4.8 m/s)) / mass_target

(b) Now let's consider the conservation of kinetic energy. In an elastic collision, kinetic energy is conserved. The kinetic energy of an object is given by the equation:
KE = 1/2 * mass * (velocity)^2

The initial kinetic energy of the softball is given by:
KE_initial = 1/2 * (0.220 kg) * (6.5 m/s)^2

The final kinetic energy of the softball is given by:
KE_final = 1/2 * (0.220 kg) * (4.8 m/s)^2

Since kinetic energy is conserved, we have:
KE_initial + 0 = KE_final + 1/2 * mass_target * (V_target)^2

Substituting the given values, we can solve for mass_target:
0.5 * (0.220 kg) * (6.5 m/s)^2 = 0.5 * (0.220 kg) * (4.8 m/s)^2 + 0.5 * mass_target * (V_target)^2

Simplifying this equation, we can solve for mass_target.

By calculating both the velocity of the target ball after the collision and the mass of the target ball using these conservation principles, we can find the answers to parts (a) and (b) of the question.

To find the velocity of the target ball after the collision, we can use the conservation of linear momentum and the fact that the collision is elastic.

(a) Conservation of linear momentum states that the total linear momentum before the collision is equal to the total linear momentum after the collision.

The initial momentum before the collision is given by:
P_initial = m1 * v1_initial + m2 * v2_initial
= (0.220 kg) * (6.5 m/s) + (m2) * (0 m/s)
= 1.43 kg⋅m/s

Here, m1 is the mass of the incoming ball and v1_initial is its initial velocity.
m2 represents the mass of the target ball, and v2_initial is its initial velocity, which is 0 since it is initially at rest.

After the collision, the total momentum is given by:
P_final = m1 * v1_final + m2 * v2_final
= (0.220 kg) * (-4.8 m/s) + (m2) * (v2_final)

Since the collision is elastic, the total kinetic energy is conserved. Initially, only the incoming ball has kinetic energy, and after the collision, both balls will have kinetic energy.

The kinetic energy before the collision is given by:
KE_initial = (1/2) * m1 * (v1_initial)^2
= (1/2) * (0.220 kg) * (6.5 m/s)^2
= 0.74675 J

The kinetic energy after the collision is given by:
KE_final = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2
= (1/2) * (0.220 kg) * (-4.8 m/s)^2 + (1/2) * (m2) * (v2_final)^2

Since the collision is elastic, the kinetic energy before the collision is equal to the kinetic energy after the collision. Therefore, we have:

0.74675 J = (1/2) * (0.220 kg) * (-4.8 m/s)^2 + (1/2) * (m2) * (v2_final)^2

Simplifying the equation, we have:

0.74675 J = 5.7472 J + (1/2) * (m2) * (v2_final)^2

0 = 4.00045 J + (1/2) * (m2) * (v2_final)^2

Now we can solve for the velocity of the target ball after the collision, v2_final.

(b) To find the mass of the target ball, we will use the equation derived in part (a). Since the incoming ball's mass is given as 0.220 kg, we just need to solve for the target ball's mass, m2.

Let's plug in the values and calculate the solutions for both parts.