A 2 kg block slides along a frictionless table at a speed of 1 m/s and collides
ID: 1976670 • Letter: A
Question
A 2 kg block slides along a frictionless table at a speed of 1 m/s and collides elastically with a 0.2 kg block which is initially at rest. The surface of the table is 1 m above the floor.(a) What is the speed of the 2 kg block after the collision?
m/s
(b) What is the speed of the 0.2 kg block (initially at rest) after the collision?
m/s
(c) Where how far from the base of the table does the 2 kg block land on the floor?
m
(d) Where how far from the base of the table does the 0.2 kg block land on the floor?
m
Explanation / Answer
Mass of the blocks m = 2 kg M = 0.2 kg Initial speeds u = 1 m/ s U = 0 Height of the table h = 1m Let the speed of the blocks after collision be v and V . For eleastic collsion coeffcieint of restitution e = 1 ( V - v ) / ( u - U ) = 1 V - v = u - U = 1 v = V - 1 -------( 1) From law of conservation of momentum, mu +MU = mv + MV 2 + 0 = 2v + 0.2 V 2 = 2(V -1) + 0.2 V = 2V - 2 + 0.2 V 4 = 2V + 0.2 V V = 1.8181 m/ s From eq( 1) , v = 0.8181 m/ s (a)The speed of the 2 kg block after the collision v = 0.8181 m/ s (b) The speed of the 0.2 kg block (initially at rest) after the collision V = 1.8181 m/ s(c) Required distance of 2 kg block land on the floor R = v [ 2h / g ] = 0.8181 [2/9.8] = 0.3695m (d) Required distance of 0.2 kg block land on the floor R ' = V [2h / g] = 1.8181 [ 2/9.8] = 0.8213 m
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