Question

A ball of mass 0.305kg that is moving with a speed of 5.7m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.3m/s.

Calculate the velocity of the target ball after the collision.

Calculate the mass of the target ball.

Answer 1

if the collision is elastic we can use both momentum and energy conservation

momentum conservation tells us the momentum before equals the momentum after; the momentum before is

0.305kg x 5.7 m/s = 1.7385kgm/s

so momentum after = -0.305kg x 3.3m/s + mv

notice that the momentum of the softball is negative since it is moving in the direction opposite its initial velocity, and m and v represent the mass and velocity of the second ball, so we have

1.7385=-1.0065+mv

or mv=2.745kgm/s

energy conservation:

1/2 (0.305kg)(5.7m/s)^2=1/2(0.305kg)(3.3m/s)^2 + 1/2 mv^2

4.954=1.66+1/2mv^2

6.586= mv^2

mv^2=mv(v) = 2.745 v so we know that

2.745v = 6.586 or v=2.399m/s

if mv=2.745

=>m=2.745/2.399=1.144kg

the second ball is 1.144kg and has a speed of 2.399 m/s in the original direction of the ball