Question

A 1.2 kg ball moving with a velocity of 8.0 m/s collides head-on with a stationary ball and bounces back at a velocity of 4.0 m/s. If the collision is perfectly elastic, calculate (a) the mass of the other ball, (b) the velocity of the other ball after the collision, (c) the momentum of each ball before and after the collision, and (d) the kinetic energy of each ball before and after the collision.

Answer 1

Part A. Given that collision is perfectly elastic, so in elastic collision both momentum and kinetic energy remains conserved. Using momentum conservation:

Pi = Pf

m1*u1 + m2*u2 = m1*v1 + m2*v2

m1 = mass of ball 1 = 1.2 kg

m2 = mass of ball 2 = ?

u1 = Initial speed of m1 = 8.0 m/s

u2 = Initial speed of m2 = 0 m/s

v1 = final velocity of m1 = -4.0 m/s

v2 = final velocity of m2 = ?

So,

1.2*8.0 + m2*0 = 1.2*(-4.0) + m2*v2

m2*v2 = 1.2*12 = 14.4 kg.m/sec

Now in perfectly elastic collision, coefficient of restitution is 1, So

(v2 - v1)/(u1 - u2) = 1

v2 - v1 = u1 - u2

So,

v2 - (-4.0) = 8.0 - 0

v2 = 4.0 m/sec

Since m2*v2 = 14.4 kg.m/sec

So,

m2 = 14.4/4.0 = 3.6 kg

mass of 2nd ball = 3.6 kg

Part B.

Velocity of 2nd ball after collision, v2 = 4.0 m/s

Part C.

P1i = Momentum of 1st ball before collision = m1*u1 = 1.2*8.0 = 9.6 kg.m/sec

P2i = Momentum of 2nd ball before collision = m2*u2 = 3.6*0 = 0 kg.m/sec

P1f = Momentum of 1st ball after collision = m1*v1 = 1.2*(-4.0) = -4.8 kg.m/sec

P2f = Momentum of 2nd ball after collision = m2*v2 = 3.6*4.0 = 14.4 kg.m/sec

Part D.

KE1i = Kinetic energy of 1st ball before collision = (1/2)*m1*u1^2 = (1/2)*1.2*8.0^2 = 38.4 J

KE2i = Kinetic energy of 2nd ball before collision = (1/2)*m2*u2^2 = (1/2)*3.6*0^2 = 0 J

KE1f = Kinetic energy of 1st ball after collision = (1/2)*m1*v1^2 = (1/2)*1.2*(-4.0)^2 = 9.6 J

KE2f = Kinetic energy of 2nd ball after collision = (1/2)*m2*v2^2 = (1/2)*3.6*4.0^2 = 28.8 J