Question

A ball with a mass of `4` `kg` and velocity of `5` `ms^-1` collides with a second ball with a mass of `6` `kg` and velocity of `-1` `ms^-1`. If `20%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Momentum is conserved. Kinetic energy is not, but in this case we know that final `E_k` is 0.8 times initial `E_k`.

The final velocity of the `4` `kg` ball is `-2.38` `ms^-1` and the final velocity of the `6` `kg` ball is `2.73` `ms^-1`.

Explanation:

Let's call the `4` `kg` ball Ball 1 and the `6` `kg` ball Ball 2, just for the convenience of subscripts...

Before the collision:

Momentum:

`p=m_1v_1+m_2v_2=4xx5+6xx(-1)=14` `kgms^-1`

Kinetic energy:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx4xx5^2+1/2xx6xx(-1)^2`
`=50+3=53` `J`

After the collision:

Momentum:

`p=m_1v_1+m_2v_2=4v_1+6v_2`

Kinetic energy:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx4xxv_1^2+1/2xx6xxv_2^2`

We know that momentum is conserved, so the momentum after the collision will be `14` `kgms^-1`, as it was before the collision.

20% of the kinetic energy is lost to other forms of energy like heat and sound, so 80% will remain: `0.8xx53=33.9` `J`

That gives us two equations in two unknowns:

Equation 1: `4v_1+6v_2=14`

Equation 2: `2v_1^2+3v_2^2=33.9`

Rearranging Equation 1 to make `v_1` the subject:

`v_1=(14-6v_2)/4`

Substituting this into Equation 2:

`2((14-6v_2)/4)^2+3v_2^2=33.9`

I'll leave solving this equation for you - it's just algebra. It does involve solving a quadratic equation.

The solution is that `v_2=2.73` or `1.75` `ms^-1`

Substituting this back into Equation 1 and solving we find that `v_1=-2.38` or `3.5` `ms^-1` respectively.

To figure out which set of answers we want, let's imagine that left-to-right is the "+" direction and right-to-left is the "-" direction.

Initially, the `4` `kg` ball was moving left-to-right and the `6` `kg` ball was moving right-to-left. After the collision, we can imagine that they have both bounced in the opposite direction: the `4` `kg` ball is now traveling right-to-left and has a negative velocity, the `6` `kg` ball has a positive velocity.

The other set of answers doesn't really make physical sense, since it requires the left-hand ball to be traveling to the right faster than the right-hand ball.

Answer 2

`4kg " ball"=-1ms^-1 and 6kg " ball"=3ms^-1`
`4kg " ball"=1.52ms^-1 and 6kg " ball"=1.32ms^-1`

Explanation:

We notice that velocity of one ball is given as `5ms^-1` and of the other is given as `-1ms^-1`. This implies that both the balls are moving in a direction opposite to each other. both the balls have been given. Let `x` direction be defined by the positive velocity.

Before the collision

Momentum, in `x` direction only

`p_i=4xx5+6xx(-1)=14kgms^-1`

Kinetic energy `KE_i=1/2xx4xx5^2+1/2xx6xx(-1)^2`

`=50+3=53J`

After the collision
Let `v_(1x) and v_(2x)` be `x` components of respective velocities of `4kg and 6kg` balls and `v_(1y) and v_(2y)` be `y` components of their respective velocities.

Momentum:

`p_(fx)=4v_(1x)+6v_(2x)`
and `p_(fy)=4v_(1y)+6v_(2y)`
Since momentum is conserved, we have

`14=4v_(1x)+6v_(2x)` .....(1)
`0=4v_(1y)+6v_(2y)` .....(2)

Kinetic energy:

`KE_f=1/2xx 4(v_(1x)^2+v_(1y)^2)+1/2 xx6(v_(2x)^2+v_(2y)^2)`

Given is that `20%` kinetic energy is lost.
`:. KE_f=0.8xx53=42.4J`
This gives us

`42.4=2(v_(1x)^2+v_(1y)^2)+3(v_(2x)^2+v_(2y)^2)` .....(3)

Thus we have three equations in four unknowns. To find a solution lets assume that even after collision the balls move along the `x` direction only. Implies that equation (2) vanishes. And we have two equations to solve for two variables.

`4v_(1x)+6v_(2x)=14` ......(4)

`2v_(1x)^2+3v_(2x)^2=42.4` ....(5)

Dropping the suffix `x` we get

`4v_1+6v_2=14` ......(6)

`2v_1^2+3v_2^2=42.4` ....(7)

Substituting value of `v_1` from (6) in (7)
`v_1=(14-6v_2)/4`

`=>2((14-6v_2)/4)^2+3v_2^2=42.4`
`=>(196-168v_2+36v_2^2)+3v_2^2=42.4`
`=>39v_2^2-168v_2+153.6=0`
`=>13v_2^2-56v_2+51.2=0`
Solving the quadratic
`v_2=(-(-56)+-sqrt((-56)^2+4xx13xx51.2))/(2xx13)`
`v_2=(56+-sqrt(3136-4xx13xx51.2))/(26)`
`v_2=(56+-21.8)/(26)`
`v_2=3, 1.32`
From (6) corresponding values of
`v_1=-1, 1.52`