A ball with a mass of 3 kg is rolling at 4 ms^-1 and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

1 Answer
Jan 15, 2016

In an elastic collision, both momentum and kinetic energy are conserved. The velocity of the 3kg ball is 0.9 or 2.7 ms^-1 and the velocity of the 4kg ball is 2.7 or 1.1 ms^-1.

Explanation:

Momentum is conserved in all collisions. Kinetic energy is conserved in elastic collisions but not in inelastic or partially elastic collisions.

The initial momentum of the whole system is p = mv for the 3kg ball: the 4kg ball has zero momentum because its has zero velocity.

p=mv=3*4=12 kgms^-1

The total momentum after the collision will be the same after the collision. If we call the 3kg ball 1 and the 4kg ball 2, the final momentum will be given by:

p = m_1v_1 +m_2v_2 = 3v_1 +4v_2 = 12 (call this Equation 1)

The total kinetic energy before the collision will be E_k=1/2mv^2 for the 3 kg ball only - the 4 kg ball has zero kinetic energy because it has zero velocity.

E_k = 1/2mv^2 = 1/2*3*4^2 = 24 J

Since kinetic energy is conserved, the final kinetic energy will be the same, and will be given by:

E_k = 1/2m_1v_1^2 + 1/2m_2v_2^2 = 3/2v_1^2+4/2v_2^2 = 24

Multiply both sides by 2 to make it tidier:

3v_1^2 + 4v_2^2 = 48 (call this Equation 2)

We now have two equations and two unknowns, so we can solve them as simultaneous equations. Rearranging Equation 1 to express v_2 in terms of v_1:

v_2 = (12-3v_1)/4

Substituting this into Equation 2:

3v_1^2+4((12-3v_1)/4)^2 = 48

I'll leave the algebra as an exercise for the reader, but this solves to give v_1=0.9ms^-1 or 2.5 ms^-1. Substituting this back into Equation 1 gives v_2=2.7 ms^-1 or 1.1 ms^-1, respectively.

Substituting these values should confirm that both momentum and kinetic energy were conserved.