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

Answer:

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.