Question

A 3.6 kg block moving with a velocity of 4.3 m/s makes an elastic collision with a stationary block of mass 2.1 kg.

(a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision. 1.1315 m/s (for the 3.6 kg block) 5.43 m/s (for the 2.1 kg block)

(b) Check your answer by calculating the initial and final kinetic energies of each block. 33.282 J (initially for the 3.6 kg block) J (initially for the 2.1 kg block) J (finally for the 3.6 kg block) J (finally for the 2.1 kg block) Are the two total kinetic energies the same?

Answer 1

a) Velocity of the block of mass 3.6 kg after collision = 1.13 m/s

Velocity of the block of mass 2.1 kg after collision = 5.43 m/s

b) Initial energy of the 3.6 kg block = 33.282 J

Final energy of the 3.6 kg block = 2.3 J

Initial energy of the 2.1 kg block = 0J

Final energy of the 2.1 kg block = 30.96 J

The two total kinetic energies are the same = 33.30 J

Step-by-step explanation:

Check the attached files for the complete solution and explanations.