A ball with a mass of #6# #kg # and velocity of #4# #ms^-1# collides with a second ball with a mass of #3# #kg# and velocity of #-5# #ms^-1#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

1572 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-03T10:25:12

The quadratic equation yields two solutions:

The first is #v_1=3.55# #ms^-1#, #v_2=-4.1# #ms^-1#

The second is #v_1=-1.55# #ms^-1#, #v_2=6.1# #ms^-1#

Let's call the #6# #kg# ball 'Ball 1", and its velocity after the collision '#v_1#'. Similarly, the #3# #kg# ball is 'Ball 2' and its mass '#v_2#'.

Momentum before the collision:

#p=mv+mv=6*4+3*(-5)=9# #kms^-1#

Momentum is conserved, so the momentum after the collision will be the same, #9# #kms^-1#.

Kinetic energy before the collision:

#E_k=1/2mv^2+1/2mv^2=1/2*6*4^2+1/2*3*(-5)^2=84# #J#

Kinetic energy is not conserved in this collision (it is partially inelastic). The energy afterward is #0.75*84=63# #J#.

Momentum after the collision:

#6v_1+3v_2=9#

Simplifying (divide by 3):

#2v_1+v_2=3# (call this Equation 1)

Kinetic energy after the collision:

#6/2v_1^2+3/2v_2^2=63#

Simplifying:

#3v_1^2+3/2v_2^2=63#

#v_1^2+1/2v_2^2=21# (call this Equation 2)

We have two equations in two unknowns.

Rearrange Equation 1 to find an expression for #v_2#:

#v_2=3-2v_1#

Substitute this into Equation 2:

#v_1^2+1/2(3-2v_1)^2=21#

#v_1^2+1/2(9-12v_1+4v_1^2)=21#

Multiply through by 2:

#2v_1^2+9-12v_1+4v_1^2=42#

#6v_1^2-12v_1+9=42#

#6v_1^2-12v_1-33=0#

This is a quadratic equation in #v_1#. Use the quadratic formula or other method to solve it.

Doing so yields #v_1=3.55# or #-1.55# #ms^-1#.

We can substitute this into our expression for #v_2#:

#v_2=3-2v_1= -4.1# #ms^-1# or #6.1# #ms^-1#