Two blocks of masses m1 = 1.80 kg and m2 = 4.43 kg are each released from rest a

Question

Two blocks of masses m1 = 1.80 kg and m2 = 4.43 kg are each released from rest at a height of h = 4.98 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision.


Part One:
Determine the velocity of the m1 = 1.80 kg block just before the collision.


Part Two:
Determine the magnitude of the velocity of the m2 = 4.43 kg block just after the collision.

Part Three:
Determine the magnitude of the velocity of the m1 = 1.80 kg block just after the collision.

Determine the maximum heights to which m1 rises after collision.

Explanation / Answer

velocity=2*g*h=2*9.8*4.98= 9.87 m/s is the initial velocities of both blocks

part 1)so velocity of m1=9.87 m/s just before collision

part 2)by conservation of momentum m1u1+m2u2=m1v1+m2v2

u1=u2

so (1.80+4.43)*9.87=1.80v1+4.43v2

61.4=1.80v1+4.43v2..........(1)

and in head on head collision v1-v2=u1-u2=0

so v1=v2

by substiting in equation 1

61.4=1.80v2+4.43v2

v2=9.85 m/s=v1

so v2=9.85 m/s

part 3)

v1=9.85 m/s

height it can reach=v^2/2g=4.95 m

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