Question

A ball with a mass of `2 kg` and velocity of `1 m/s` collides with a second ball with a mass of `7 kg` and velocity of `- 4 m/s`. If `40%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Linear momentum, the product of mass and velocity, is always conserved, so, when two objects collide,
`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, respectively, `u_1` and `u_2` are the initial velocities, and `v_1` and `v_2` are the final velocities.

Then, from the data available in the question,
`2*1-7*4=2v_1+7v_2`
`-26=2v_1+7v_2`

Now, it is given that, during the collision, `40%` of the kinetic energy is lost (presumably in the form of heat, sound, friction, etc.). So, the final amount kinetic energy is equal to `60%` of the original kinetic energy.

Then, construct another equation:
`0.6(1/2m_1u_1^2+1/2m_2u_2^2)=1/2m_1v_1^2+1/2m_2v_2^2`
`0.6(1/2*2*1^2+1/2*7*(-4)^2)=1/2*2v_1^2+1/2*7v_2^2`
`34.2=v_1^2+7/2v_2^2`

Now, we have a system of equations:
`{(2v_1+7v_2=-26, "Equation 1"),(v_1^2+7/2v_2^2=34.2, "Equation 2"):}`

From Equation 1, it can be seen that `v_1=-13-7/2v_2`. Substitute this into Equation 2:
`(-13-7/2v_2)^2+7/2v_2^2=34.2`
`63/4v_2^2+91v_2+134.8=0`

Using the quadratic equation,
`v_2=(-91±sqrt(91^2-4*63/4*134.8))/(2*63/4)`

This value is undefined, as we're taking the square root of `91^2-4*63/4*134.8=-211.4`, a negative number.

Hence, the situation described in the question is impossible.