Question

a trolley A of mass 2kg at 5 meter per second collide with a stationary trolley B of mass 3kg after the collision the two travel together at 2 meter per second find (a) what is the momentum of A before the collision (b) what is the momentum of A after the collision (c) what is the kinetic energy of A before collision (d) what is the kinetic energy of each trolley after collision (e) during the collision A kinetic energy gained by B is less than the kinetic energy gained by A, how much kinetic energy is un accounted for?​

Answer 1

a) The momentum of trolley A before the collision is (2 kg) * (5 m/s) = 10 kg*m/s.

b) The momentum of trolley A after the collision is (2 kg) * (2 m/s) = 4 kg*m/s.

c) The kinetic energy of trolley A before the collision is 1/2 * (2 kg) * (5 m/s)^2 = 25 J.

d) The kinetic energy of trolley A after the collision is 1/2 * (2 kg) * (2 m/s)^2 = 2 J. And the kinetic energy of trolley B after the collision is 1/2 * (3 kg) * (2 m/s)^2 = 3 J.

e) The amount of kinetic energy that is unaccounted for is the difference between the initial kinetic energy of trolley A and the final kinetic energy of trolley A and trolley B. This is 25J - (2J + 3J) = 20 J.

It's important to note that the total momentum of the system is conserved during the collision, meaning that the momentum of trolley A before the collision is equal to the total momentum of trolley A and B after the collision. Also the total kinetic energy of the system is not conserved during the collision as some of the energy is transferred to internal energy of the system and other forms of energy.