A ball with a mass of #6# #kg # and velocity of #7# #ms^-1# collides with a second ball with a mass of #4# #kg# and velocity of #- 8# #ms^-1#. If #15%# of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Oct 23, 2017

Momentum is conserved, but because this is an inelastic collision kinetic energy is not. The final velocities of the masses are:

#6# #kg# mass: #6.5# or #-4.5# #ms^-1#

#4# #kg# mass: #-7.25# or #9.25# #ms^-1#

Explanation:

Before the collision:

#p=m_1v_1+m_2v_2=6xx7+4xx(-8)#
#=42-32=10# #kgms^-1#

#E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx6xx7^2+1/2xx4xx(-8)^2#
#=275# #J#

After the collision #15%# of the energy is lost, so #85%# remains:

#0.85xx275=233.75# #J#

Because momentum is always conserved, the momentum of the system after the collision will still be #10# #kgms^-1#.

We can now set up a system of simultaneous equations.

After the collision:

#p=10=6v_1+4v_2#

#E_k=233.75=1/2xx6xxv_1^2xx4xxv_2^2=3v_1^2+2v_2^2#

We can use the first equation to express #v_2# in terms of #v_1#:

#v_2=(10-6v_1)/4=2.5-1.5v_1#

We can then substitute that expression into the second equation, so that it will be in only one variable:

#233.75=3v_1^2+2(2.5-1.5v_1)^2#

The algebra is a bit messy to do and mark up here, so I might leave that as an exercise for you to do. It's just math, not physics.

Because it's a quadratic equation it yields two solutions, #v_1=6.5# and #v_1=-4.5# #ms^-1#. These correspond to the #6# #kg# ball either bouncing back and therefore having a negative velocity or continuing in the direction it was initially moving at a slightly slower velocity.

Let's see what they lead to in terms of #v_2#:

#v_2=2.5-1.5v_1#

If #v_1=6# then #v_2=-7.25# #ms^-1#

If #v_1=-4.5# then #v_2=9.25# #ms^-1#

Both sets of solutions actually make sense.