A ball with a mass of #2# #kg # and velocity of #5# # ms^-1# collides with a second ball with a mass of #7# #kg# and velocity of #- 4# #ms^-1#. If #40%# of the kinetic energy is lost, what are the final velocities of the balls?

Let's call the #2# #kg# ball '1' and the #7# #kg# ball '2' for convenience.

Momentum before the collision:

#p=m_1v_1+m_2v_2=2*5+7*(-4)=10-28=-18# #kgms^-1#

Kinetic energy before the collision:

#E_k=1/2m_1v_1^2+1/2m_2v_2^2#
#=1/2*2*5^2+1/2*7*(-4)^2=25+56=81# #J#

The momentum after the collision will be the same as before, the kinetic energy after the collision will be 60% of the value before (due to the 40% loss): #48.6# #J#

Momentum after the collision:

#p=m_1v_1+m_2v_2=2v_1+7v_2=-18# : call this Equation 1

Kinetic energy after the collision:

#E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2*2*v_1^2+1/2*7*v_2^2=48.6# : call this Equation 2

We have two equations in two unknowns, which means we can solve them. Rearrange Equation 1 to find a value for #v_1#:

#v_1=-7/2v_2-9#

Substitute into Equation 2:

#1/2*2*(-7/2v_2-9)^2+1/2*7*v_2^2=48.6#

#(-7/2v_2-9)^2+7/2v_2^2=48.6#

#(49/4v_2^2+63v_2+81)+7/2v_2^2=48.6#

#63/4v_2^2+63v_2+81=48.6#

#63/4v_2^2+63v_2+32.4=0#

This is a quadratic equation, which can be solved using the quadratic formula or whichever method your prefer:

This yields the value #v_2=-0.61# or #-3.4# #ms^-1#

Substituting these values into Equation 1 yields:

#v_1=-6.9 or2.9# #ms^-1#

The possible solutions for #(v_1,v_2)# are #(-0.61, -6.9)# or #(-3.4, 2.9)#.

The first is physically impossible, since the ball on the right would be moving faster to the left than the one to the left of it, so the solution is #(-3.4, 2.9)#.