The surfaces are frictionless. The tracks are 60 degree from horizontal. A 425 k

Question

The surfaces are frictionless. The tracks are 60 degree from horizontal. A 425 kg mass is released from rest on a track at a height 3.9 m above a horizontal surface at the foot of the slope. It collides elastically with a 509 kg mass initially at rest on the horizontal surface, as shown in the figure. The mass 509 kg slides up a similar track. The acceleration of gravity is 9.8 m/s2. What is the speed of the block 509 kg immediately after the collision? Answer in units of m/s

Explanation / Answer

for 425kg block mgh = ,.5mv^2 => v = square root(2*9.8*3.9) = 8.74m/sec Now after the collision let the velocity of 509 kg block be A and that of 425kg block be B Velocity of apporach = velocity of separation => 8.74 = A-B => B= A-8.74 by momentum conservation 509*A + 425*B = 425*8.74 => 509A + 425(A -8.74) = 425*8.74 => A = 7429/(509+425) = 7.95m/sec

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