Question

A 2kg block is sliding across a frictionless surface at 20m/s. The block collides with another block (inertia 1kg) that is at rest. The other block has glue on it and the two blocks stick together and move off at 3m/s.

(a)What is the energy dissipated in the collision?
(b)ls the collision isolated? Why or why not?
(c)What would be the total convertible energy in this collision, assuming it was isolated? (Do you want to change your answer to (b)?)

Answer 1

Answer:

Explanation:

mass of first block, M = 2 kg

mass of second block, m = 1 kg

initial velocity of first block, U = 20 m/s

initial velocity of first block, u = 0 m/s

final velocity of both the blocks, v = 3 m/s

(a) kinetic energy before collision

Ki = 1/2 MU² + 1/2mu²

Ki = 0.5 x 2 x 20 x 20 + 0

Ki = 400 J

Total kinetic energy after collision

Kf = 1/2 (M+m)v²

Kf = 0.5(2 + 1) x 3 x 3

Kf = 13.5 J

Energy dissipated, K = kf - ki

K = 400 - 13.5

K = 386.5 J

(b) the collision is isolated if the initial momentum is equal to final momentum

initial momentum = 2 x 20 + 0 = 40 Ns

final momentum = 3 x 3 = 9

As the initial momentum is not equal t the final momentum so the collision is not isolated.

(c) The convertible energy is 400 - 386.5 = 13.5 J