Change in kinetic energy question?
Done i and ii but not iii
- az_lenderLv 71 month ago
Let's skip the 10^(-23), because it won't matter in the answers to (a) and (b).
(7.5 kg)(42 m/s) + (2.5 kg)(20 m/s)
= (7.5 kg)(32.4 m/s) + (2.5 kg)(v2) =>
(7.5 kg)(9.6 m/s) = (2.5 kg)(v2 - 20 m/s) =>
v2 - 20 m/s = 3(9.6 m/s) = 28.8 m/s =>
v2 = 48.8 m/s.
Before the collision, the velocity of A relative to B is 22 m/s. After the collision, the separation velocity is 48.8 m/s - 32.4 m/s = 16.4 m/s. This is a somewhat inelastic collision, where the coefficient of restitution (16.4/22) is less than 1.
(c) Calculate the sum of (1/2)mv^2 for the two particles before the collision. And then do the same for after the collision. The difference is the change in total kinetic energy. Just don't forget to put the 10^(-23) back into the units.