Question

A ball with a mass of `4 kg` and velocity of `3 m/s` collides with a second ball with a mass of `5 kg` and velocity of `- 2 m/s`. If `20%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Ball 1: `\vec(v_(1_f)) = -2.11` m/s
Ball 2: `\vec(v_(2_f)) = 2.21` m/s

Explanation:

Collisions can be represented using momentum relations.

Momentum is defined as: `\vecp=m\vecv`

Using the values provided, we see the initial momenta are:
`\vec(p_(1_i)) = (4)(3) = 12` kg`*`m/s
`\vec(p_(2_i)) = (5)(-2) = -10` kg`*`m/s

The total net momentum of the system initially is:
`\vec(p_(Total_i)) = \vec(p_(1_i))+\vec(p_(2_i)) = 12 -10=2` kg`*`m/s

Since momentum is conserved, we know the magnitude of the momentum after the collision is also `2` kg`*`m/s.

Energy (kinetic) is defined as: `E = 1/2 m v^2`

Using the values provided, we see the initial energies are:
`E_(1_i) = 1/2(4)(3)^2 = 18` J
`E_(2_i) = 1/2 (5)(-2)^2 = 10` J

The total energy of the system initially is:
`E_(Total_i) = E_1 + E_2 = 18+10=28` J

Since we are told 20% of the energy is lost (80% remains), the energy is NOT conserved after the collision. The energy after the collision is:
`E_(Total_f) = 0.8E_(Total_i) = (0.8)(28) = 22.4` J

Now we can form some equations describing the post-collision system, which we can use to find the velocities of the balls.

They are magnitude of total momentum:
`4v_(1_f) + 5v_(2_f) = 2` kg`*`m/s

and total energy:
`2v_(1_f)^2 + 5/2 v_(2_f)^2 = 22.4` J

We can now solve this system of equations
(1) `4v_(1_f) + 5v_(2_f) = 2`
(2) `2v_(1_f)^2 + 2.5 v_(2_f)^2 = 22.4`

We can divide equation (1) by 4 and solve for `v_(1_f)`:
(3) `v_(1_f) = 0.5 - 1.25v_(2_f)`

Substituting (3) into (2) gives:
`2(0.5 - 1.25v_(2_f))^2 + 2.5 v_(2_f)^2 = 22.4`
`2(0.25-1.25v_(2_f)+1.5625v_(2_f)^2) + 2.5 v_(2_f)^2 = 22.4`
`0.5-2.5v_(2_f)+3.125v_(2_f)^2 + 2.5 v_(2_f)^2 = 22.4`
`5.625v_(2_f)^2-2.5v_(2_f)-21.9 = 0`

This quadratic can now be solved using the quadratic equation:
`v_(2_f)=(-b+-sqrt(b^2-4ac))/(2a)`
where `a=5.625`, `b=-2.5`, and `c=-21.9`

This gives: `v_(2_f) = 2.21` m/s and `v_(2_f) = -1.76` m/s

Since ball 2 originally travels in the negative direction, after the collision it will go in the positive direction. So we accept the root that is positive. Hence:
`v_(2_f) = 2.21` m/s

Now we can use this value in equation (1) to find the value of `v_(1_f)`:
`4v_(1_f) + 5(2.21) = 2`
`v_(1_f) = -2.11` m/s

It is good that this value is negative, as ball 1 was originally traveling in the positive direction. Post-collision, it should go the opposite direction.

Hence we have found the final velocities of the balls:

Ball 1: `\vec(v_(1_f)) = -2.11` m/s
Ball 2: `\vec(v_(2_f)) = 2.21` m/s