Question

A ball with a mass of `3 kg` moving at `15 m/s` hits a still ball with a mass of `17 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

The speed of the second ball is `=2.65ms^-1`
The loss in kinetic energy is `=277.8J`

Explanation:

We have conservation of momentum

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`u_2=v_1=0`

So,

`m_1u_1=m_2v_2`

`v_2=m_1/m_2*u_1`

`=3/17*15=2.65ms^-1`

The kinetic energy, before the collision, is

`KE_1=1/2m_1*u_1^2=1/2*3*15^2=337.5J`

The kinetic energy, after the collision, is

`KE_2=1/2m_2*v_2^2=1/2*17*2.65^2=59.7J`

The loss in kinetis energy is

`DeltaKE=KE_1-KE_2`

`=337.5-59.7=277.8J`

Answer 2

`45/17\ "m"/"s"~~2.647\ "m"/"s"`

`4725/17~~277.941\ "J"`

Explanation:

Remember that momentum `mv` (`m` is mass and `v` is velocity) is always conserved.

If two objects collide, their initial momentum and their final momentum are always the same, or `m_1u_1+m_2u_2=m_1v_1+m_2v_2`, where `m_1` and `m_2` are the masses of the two objects (in this case, 3 and 17, respectively), `u_1` and `u_2` are their initial velocities (15 and 0, respectively), and `v_1` and `v_2` are their final velocities (0 and `v_2`, which we are trying to find).

We substitute the values given to us in the question into this formula to get `3*15+17*0=3*0+17*v_2`. Solving for `v_2`, we get `45/17\ "m"/"s"~~2.647\ "m"/"s"`.

To find the amount of kinetic energy lost, we find the initial kinetic energy minus the final kinetic energy. Kinetic energy can be found using the formula `(mv^2)/2`, where `m` is mass and `v` is velocity.

The initial kinetic energy of the first and second ball is `(3*15^2)/2=675/2\ "J"` and `(17*0^2)/2=0\ "J"`, respectively. The final kinetic energy of the first and second ball is `(3*0^2)/2=0\ "J"` and `(17*(45/17)^2)/2=2025/34\ "J"`, respectively. The difference is `675/2-2025/34=4725/17~~277.941\ "J"`.