Question

A ball with a mass of `2 kg` and velocity of `2 m/s` collides with a second ball with a mass of `1 kg` and velocity of `- 4 m/s`. If `20%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

`v_(1f)=-4/sqrt5m/s`, `v_(2f)=8/sqrt5m/s`

Decimal approximations:

`v_(1f)~~-1.79m/s`, `v_(2f)~~3.58m/s`

Explanation:

In all collisions, momentum is conserved. In an inelastic collision, momentum is conserved, but part of the kinetic energy is transformed into some other form of energy. Therefore, we have an inelastic collision.

Momentum conservation:

`vecp_f=vecp_i`

where `vecp=mvecv`

For multiple objects:

`vecP=vecp_(t o t)=sumvecp=vecp_1+vecp_2+...`

We can calculate the momentum before the collision:

`vecP_i=m_1vecv_(1i)+m_2vecv_(2i)`

Where `m_1=2kg`, `m_2=1kg`, `vecv_1=2m/s`, and `vecv_2=-4m/s`

`vecP_i=(2kg)(2m/s)+(1kg)(-4m/s)`

`vecP_i=4(kgm)/s-4(kgm)/s=vec0`

Therefore, by momentum conservation, the total momentum after the collision should also be zero.

`vecP_f=m_1v_(1f)+m_2v_(f2)=vec0`

We can also calculate the initial kinetic energy before the collision:

`K=1/2mv^2`

`K_i=1/2m_1(v_(1i))^2+1/2m_2(v_(2i))^2`

`K_i=1/2((2kg)(2m/s)^2+(1kg)(-4m/s)^2)`

`K_i=1/2(24(kgm^2)/s^2)`

`K_i=12(kgm^2)/s^2=12Nm=12J`

We are given that `20%` of the kinetic energy is lost in the collision, so we can calculate what the final kinetic energy of the system should be.

`20%=0.20=1/5`

`1/5*12=12/5`

`12/5` joules of kinetic energy are lost during the collision.

`12-12/5=48/5`

`48/5` joules of kinetic energy remain after the collision.

This tells us that:

`K_f=1/2m_1(v_(1f))^2+1/2m_2(v_(2f))^2=48/5`

We now have two equations:

`m_1v_(1f)+m_2v_(f2)=0`

`1/2(m_1(v_(1f))^2+m_2(v_(2f))^2)=48/5`

As the masses of the objects do not change:

`2v_(1f)+v_(f2)=0`

`2(v_(1f))^2+(v_(2f))^2=96/5`

We now have two equations and two unknowns. From the first equation, we can rearrange to find that `v_(f2)=-2v_(1f)`. Substituting this into the second equation:

`2(v_(1f))^2+(-2v_(1f))^2=96/5`

We can solve for the final velocity of the first object, `v_(1f)`:

`2(v_(1f))^2+4(v_(1f))^2=96/5`

`6(v_(1f))^2=96/5`

`(v_(1f))^2=16/5`

`v_(1f)=+-sqrt(16/5)=+-4/(sqrt5)~~+-1.79m/s`

Now that we know the possible values of `v_(1f)`, we can put this value back into the first equation to solve for `v_(2f)`:

`2(4/sqrt5)+v_(f2)=0`

`v_(f2)=-8/sqrt5~~-3.58m/s`

OR

`2(-4/sqrt5)+v_(f2)=0`

`v_(f2)=8/sqrt5~~3.58m/s`

Therefore, the possible exact answers are:

`v_(1f)=4/sqrt5m/s`, `v_(2f)=-8/sqrt5m/s`

`v_(1f)=-4/sqrt5m/s`, `v_(2f)=8/sqrt5m/s`

Decimal approximations:

`v_(1f)~~1.79m/s`, `v_(2f)~~-3.58m/s`

`v_(1f)~~-1.79m/s`, `v_(2f)~~3.58m/s`

I assume that the balls bounce off of each other and move in directions opposite those which they traveled in initially, so:

`v_(1f)=-4/sqrt5m/s`, `v_(2f)=8/sqrt5m/s`