Question

A ball with a mass of `7 kg` and velocity of `4 m/s` collides with a second ball with a mass of `2 kg` and velocity of `- 6 m/s`. If `15%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

`v_(1f)=0.801"m"//"s"`
`v_(2f)=7.600"m"//s"`

Explanation:

There may be a simpler way to solve this, but here's what I came up with:

Momentum is conserved in all collisions. In an inelastic collision, momentum is conserved as always, but energy is not; part of the kinetic energy is transformed into some other form of energy. Therefore, we have an inelastic collision.

`vecp=mvecv`

The equation for momentum.

We can use momentum conversation and kinetic energy to find the final velocities of the balls.

Momentum conservation:

`DeltavecP=0`

`=>vecp_f=vecp_i`

For multiple objects, we use superposition as with forces:

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

So we have:

`=>color(crimson)(m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f))`

We are given the following information:

• `|->"m"_1=7"kg"`
• `|->"v"_(1i)=4"m"//"s"`
• `|->"m"_2=2"kg"`
• `|->"v"_(2i)=-6"m"//"s"`
• `"kinetic energy lost:" 15%`

We can begin by calculating the momentum before the collision:

`P_i=m_1v_(1i)+m_2v_(2i)`

`=(7kg)(4m/s)+(2kg)(-6m/s)`

`=color(blue)(16" kgm"//"s")`

Note that momentum is a vector quantity and the positive value indicates direction. Let's set to the right is the positive direction.

By momentum conservation, the total momentum after the collision should also be `16"Ns"` (directed toward the right)

`=>color(blue)(P_f=m_1v_(1f)+m_2v_(f2)=16" kgm"//"s")`

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((7"kg")(4" m"//"s")^2+(2"kg")(-6" m"//"s")^2)`

`K_i=1/2(136("kgm"^2)/"s"^2)`

`K_i=68("kgm"^2)/"s"^2`

`=color(darkblue)(68"J")`

We are given that `15%` of the kinetic energy is lost in the collision, so we can calculate what the final kinetic energy of the system should be.

`15%=0.15`

`=>0.15*68"J"`

`=10.2"J"`

`:.10.2` joules of kinetic energy are lost during the collision.

`=>68"J"-10.2"J"=57.8"J"`

`:.57.8` joules of kinetic energy remain after the collision.

This tells us that:

`color(darkblue)(K_f=1/2m_1(v_(1f))^2+1/2m_2(v_(2f))^2=57.8"J")`

We now have two equations expressing final momentum and kinetic energy:

`color(lightblue)(m_1v_(1f)+m_2v_(f2)=16" kgm"//"s")`

`color(lightblue)(1/2(m_1(v_(1f))^2+m_2(v_(2f))^2)=57.8"J")`

As the masses of the objects do not change, we can fill them in and simplify:

`color(skyblue)(7v_(1f)+2v_(f2)=16" kgm"//"s")`

`color(skyblue)(7(v_(1f))^2+2(v_(2f))^2=115.6)`

We now have two equations and two unknowns. We can now manipulate one of these equations to solve for either `v_(1f)` or `v_(2f)` and substitute this into the second equation. Neither looks particularly friendly, so I will solve the first equation for `v_(1f)`.

`7v_(1f)+2v_(2f)=16`

`=>7v_(1f)=16-2v_(2f)`

`=>color(crimson)(v_(1f)=1/7(16-2v_(2f)))`

Substituting into the second equation, we now have everything in terms of `v_(2f)`:

`7(1/7(16-2v_(2f)))^2+2(v_(2f))^2=115.6`

Simplifying:

`1/7(256+4(v_(2f))^2-64v_(2f))+2(v_(2f))^2=115.6`

Im going to multiply both sides by 7 to get rid of the fractions:

`256+4(v_(2f))^2-64v_(2f)+14(v_(2f))^2=809.2`

`=>18(v_(2f))^2-64v_(2f)+256=809.2`

`=>color(crimson)(18(v_(2f))^2-64v_(2f)-553.2=0`

We have a quadratic equation which we can solve using the quadratic formula:

For an equation of the form `ax+by+c=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Therefore:

`v_(2f)=(64+-sqrt((-64)^2-4(18)(-553.2)))/(2*18)`

`=>(64+-sqrt(4096+39830.4))/(36)`

`=>color(skyblue)((64+-sqrt(43926.4))/(36))`

We therefore have two answers for `v_(f2)` and would also have two answers for `v_(f1)`. However both answers may not be possible.

Since this is an elastic collision and we expect the balls to move off in the opposite direction that they approached each other from, we expect a negative velocity for `"m"_1` and a positive velocity for `"m"_2`.

Only one of the calculated velocities will work, and so `color(crimson)(v_(2f)=7.600"m"//s")`

We can use the above answer to solve for the final velocity of the first ball, `v_(1f)`:

`7v_(1f)+2v_(f2)=16`

`=>7v_(1f)+2(7.600)=16`

`=>7v_(1f)=0.801`

`=>color(indigo)(v_(1f)~~0.801"m"//"s")`

As stated above, we expect the balls to move off in the opposite direction that they approached from, so we would expect a negative velocity for `"m"_1`. Instead, we get a small positive value, which implies that in the collision, the first ball was not hit hard enough to change direction, but was hit hard enough to slow down significantly. Considering the relatively small amount of kinetic energy lost and the significant change in the magnitude (and direction) of the second ball's velocity, this seems reasonable.

Therefore:

• `color(lightgrey)(v_(1f)=0.801"m"//"s")`
• `color(skyblue)(v_(2f)=7.600"m"//s")`

Note that these answers can be checked with momentum conservation. Simply calculate the final momentum using the velocities determined and confirm it is `16" kgm"//"s"`.