Question

A softball of mass 0.220 kg that is moving with a speed of 8.0 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 6.4 m/s.

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

(b) Calculate the mass of the target ball

Answer 1

Given data:

1. Mass of the ball = m1 = 0.220 Kg

2. Speed = v1 = +8.0 m/s    (+ since n positive direction)

3. After colision: v2 = -6.4 m/s

To calculate:

a)the velocity of the target ball after the collision

b)mass of the target ball

a) For this part, we can use law of conservaton of momentum.

i.e. final momentu = intial momentum.

Now, let initial velocities be v1 and v2 and final velocities be u1 and u2

m1v1 + m2v2 = m1u1 + m2u2

(0.220 kg) * (8.0m/s) + m2*0 = (0.220 kg) * (-6.4 m/s) + m2u2

1.76 = -1.408 + m2u2

m2u2 = 3.168           ... (1)

b) We will get back to equation above later.

Now, since this is an elastc collision, enegy also is conserved. Thus,

                   ....(2)

   ......(3)

now use equation (1) in above equation.

This is the velocity of the target ball.

b) Mass of the target ball:

We have equation (3)

(m2u2²) = 5.0688

This is the mass of the target ball.