Two blocks of masses m 1 = 2.00 kg and m 2 = 4.80 kg are each released from rest

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.80 kg are each released from rest at a height of h = 4.10 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which m1 and m2 rise after the collision.

Explanation / Answer

(a) From conservation of energy

(1/2)m1u12 =m1gh1

u1 = [2gh1]1/2 = [2*9.8*4.1]1/2

u1 = v1i =8.96 m/s

From conservation of energy

(1/2)m2u22 =m2gh2

u2 = [2gh2]1/2 = [2*9.8*4.1]1/2

u2 = v2i = 8.96 m/s

(b) u1 =8.96 m/s , u2 = -8.96 m/s

From conservation of monetum and kinetic energy in elastic collision

v1 = [(m1-m2)u1/(m1+m2)] +[2m2u2/(m1+m2)]

v1 = [(2-4.8)*8.96/(2+4.8)] +[- 2*4.8*8.96/(2+4.8)]

v1 = -3.67-12.65

v1 =v1f = -16.32 m/s

From conservation of monetum and kinetic energy in elastic collision

v2 = [(m2-m1)u2/(m1+m2)] +[2m1u1/(m1+m2)]

v2 = [-(4.8-2)*8.96/(2+4.8)] +[2*2*8.96/(2+4.8)]

v2 = -3.67+5.27

v2 =v2f = 1.6 m/s

(c) from v^2 -u^2 = 2gh

0 - v1f2 =-2gy1f

y1f = v1f2 /2g = (-16.32)2 /(2*9.8)

y1f = 13.59 m

from v^2 -u^2 = 2gh

0 - v2f2 =-2gy2f

y2f = v2f2 /2g = (1.6)2 /(2*9.8)

y2f = 0.131 m

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