Question

A ball with a mass of `1 kg` moving at `9 m/s` hits a still ball with a mass of `9 kg`. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Answer 1

`v_(2f)=1"m"//"s, " K_("lost")=36"J"`

Explanation:

We can solve this problem using energy conservation and momentum conservation. 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, and we can use this to calculate the velocity of the second ball and the energy lost in the collision.

`color(darkblue)(vecp=mvecv)`

The equation for momentum, where `m` is the mass of the object and `v` is the object's velocity.

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))`

Energy conservation

`DeltaE=0`

`=>DeltaU+DeltaK+color(crimson)(DeltaE_(th))=DeltaE=0`

We have no potential energy in this problem, so we have:

`=>DeltaK+DeltaE_(th)=0`

`=>DeltaK+ E_(thf)-cancel(E_(thi))=0`

`=>DeltaK+E_(thf)=0`

`=>color(darkblue)(DeltaK=-E_(th))`

Therefore, the change in kinetic energy is equal to the energy lost in the collision as heat.

We are given the following information:

• `|->"m"_1=1"kg"`
• `|->"v"_(1i)=9"m"//"s"`
• `|->"v"_(1f)=0`
• `|->"m"_2=9"kg"`
• `|->"v"_(2i)=0`

Velocity of second ball:

Factoring in our zero values, we now have:

`m_1v_(1i)+cancel(m_2v_(2i))=cancel(m_1v_(1f))+m_2v_(2f)`

`=>m_1v_(1i)=m_2v_(2f)`

Which we can solve for `v_(2f)`:

`color(darkblue)(v_(2f)=(m_1v_(1i))/m_2)`

Substituting in our known values:

`v_(2f)=((1"kg")(9"m"//"s"))/(9"kg")`

`=color(darkblue)(1"m"//"s")`

This answer can be checked by comparing the momentum before and after the collision. They should both be equal by momentum conservation.

Kinetic energy lost:

`DeltaK=-E_(th)`

We can calculate the change in kinetic energy to find the energy lost as heat.

`DeltaK=K_f-K_i`

`=>=1/2m(v_f)^2-1/2m(v_i)^2`

Initially, only the first ball is moving, so it is the only source of kinetic energy in the system.

`=>K_i=1/2(1"kg")(9"m"//"s")^2`

`=>K_i=40.5"J"`

Finally, only the second ball is moving.

`=>K_f=1/2(9"kg")(1"m"//"s")^2`

`=>K_f=4.5J"`

Therefore, we have:

`-DeltaE_(th)=4.5"J"-40.J"`

`color(darkblue)(DeltaE_(th)=36"J")`

`:.36"J"` of kinetic energy was lost as heat in the collision.