A ball with a mass of 3 kg and velocity of 1 ms1 collides with a second ball with a mass of 5 kg and velocity of 8 ms1. If 25% of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Apr 30, 2016

The equations yield two sets of answers:

v1=0.8 ms1 and v2=6.92 ms1

OR

v1=8.46 ms1 and v2=2.32 ms1

Explanation:

Momentum is conserved in all collisions. In this instance, kinetic energy is not conserved: 25% is lost and 75% remains.

The initial momentum is given by:

p=m1v1+m2v2=31+5(8)
=340=37 kgms1

The final momentum will be the same as the initial momentum.

(if we define 'to the right' as the positive direction, the net momentum of the system is to the left)

The initial kinetic energy is given by:

Ek=12m1v21+12m2v22=12312+125(8)2
=32+160=161.5 J

The final kinetic energy will be 75% of the initial kinetic energy:

75100161.5=121.13 J

The expression for the final momentum is:

p=m1v1+m2v2=3v1+5v2=37 : call this Equation 1

The expression for the final kinetic energy is:

Ek=12m1v21+12m2v22=123v21+125v22
32v21+52v22=121.13 : call this Equation 2

We now have two equations in two unknowns, so we can solve the system.

Divide Equation 1 by 3 and rearrange to give:

v1=37353v2

Substitute this value for v1 into Equation 2:

32(37353v2)2+52v22=121.13

32(152.1+41.1v2+2.78v22)+52v22=121.13

228.15+61.65v2+4.17v22+2.5v22=121.13

Rearranging:

6.67v22+61.65v2+107.2=0

We can solve this quadratic equation using the quadratic formula (or other methods):

v2=61.65±61.65246.67107.226.67
v2=6.92 or 2.32 ms1

That is, the two possible solutions are that Ball 2, the 5 kg ball, travels to the left at either 6.93 ms1 or at 2.32 ms1. Remember that it was initially traveling to the left at 8 ms1.

Substituting these values back into Equation 1, we find that 3v1+5v2=37 which means v1=0.8 ms1 (when v2=6.92 ms1) or v1=8.46 ms1 (when v2=2.32 ms1).