Two blocks of masses m 1 = 1.96 kg and m 2 = 4.66 kg are each released from rest

Question

Two blocks of masses m1 = 1.96 kg and m2 = 4.66 kg are each released from rest at a height of h = 5.20 m on a frictionless track, as shown in the figure, and undergo an elastic head-on collision.

A) Determine the velocity of the m2 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)

B) Determine the velocity of the m1 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)

C) Determine the maximum height to which m2 rises after the collision.

D) Determine the maximum height to which m1 rises after the collision.

Explanation / Answer

Using conservation of energy for Block m1 :

m1 g h = (0.5) m1V1i2

V1i = sqrt(2gh) = sqrt(2 x 9.8 x 5.20) = 10.1 m/s

similarly , V2i = sqrt(2 x 9.8 x 5.20) = 10.1 m/s

V1i is to the right and V2i is to the left hence

V1i = 10.1 m/s     and   V2i = -10.1 m/s

Using conservation of momentum :

m1 V1i + m2 V2i = m1 V1f + m2 V2f

(1.96 + 4.66) 10.1 = 1.96 V1f + 4.66 V2f                       

1.96 V1f + 4.66 V2f = 66.86                   eq-1

also V2f - V1f = V1i - V2i = 10.1 - (-10.1) = 20.2 m/s

V2f = V1f + 20.2                Eq-2

Solving Eq-1 and eq-2

V1f = -4.12 m/s

V2f = 16.08 m/s

To calculate the heights, we again use the formula

V1f = sqrt(2gh1f )

-4.12 = sqrt( 2 x 9.8 h1f )

h1f = 0.87 m

similarly

V2f = sqrt(2gh2f )

-4.12 = sqrt( 2 x 9.8 h2f )

h2f = 13.2 m

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