A. M1*V1 + M2*V2 = M1V + M2V.
5*10^4 * 2.50 + M2*0 = 5*10^4V+4*10^4V,
7.5*10^4 + 0 = 9*10^4V,
V = 0.833 m/s.
B. Before collision: KE1 = 0.5*M1*V1^2.
KE1 = 0.5*5*10^4 * (2.50)^2 = 3.125*10^4 Joules.
After collision: KE2 = 0.5*M1*V^2 + 0.5*M2*V^2,
KE2 = 0.5*5*10^4*V^2 + 0.5*4*10^4*V^2,
KE2 = 4.5*10^4V^2 = 4.5*10^4*(0.833)^2 = 3.123*10^4 Joules.
KE1 = KE2, Kinetic energy conserved!
Therefore we have an elastic collision.
Use the equations below:
P=mv
Ek=1/2mv^2
3) A railway carriage of 5.00 x 10^4 moves along a track at 2.50ms-1. It collides with a second, stationary, carriage with a mass of 4.00 X 10^4 kg and the carriages join together.
A) Calculate the initial velocity of the coupled carriages after the impact.
B) Calculate the change in kinetic energy during the collision and hence determine if the collision is elastic or inelastic and why.
10 marks
1 answer