Two blocks of masses m 1 = 2.05 kg and m 2 = 4.10 kg are each released from rest

Question

Two blocks of masses m1= 2.05 kg and m2= 4.10 kg are each released from rest at a height of h = 6.00 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.

y1f =

y2f =

v1i =

v2i =

Explanation / Answer

a) m1gh = m1v1i2/2

=> v1i = (2gh)1/2 = (2 * 9.81 * 6)1/2 = 10.85 m/s

Similarly, v2i = -10.85 m/s

b) Linear momentum is conserved.

=> m1v1i + m2v2i = m1v1f + m2v2f

=> 10.85(2.05 - 4.10) = 2.05v1f + 4.10v2f

=> v1f + 2v2f = -10.85 --- (1)

Also, since collision is elastic,

Velocity of approach = Velocity of separation

=> v1i - v2i = v2f - v1f

=> 10.85 - (-10.85) = v2f - v1f  

=> v2f - v1f = 21.7 --- (2)

Adding equations (1) and (2), we get,

3v2f = 10.85

=> v2f = 3.62 m/s

v1f = v2f - 21.7 = 3.62 - 21.7 = -18.1 m/s

c) mgy1f = mv1f2/2

=> y1f = v1f2/2g = 18.12/(2 * 9.81) = 16.7 m

Similarly, y2f = v2f2/2g = 3.622/(2 * 9.81) = 0.67 m

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