The surfaces are frictionless. A 397 kg mass is released from rest on a track at

ID: 1960134 • Letter: T

Question

The surfaces are frictionless. A 397 kg mass is released from rest on a track at a height

2.7 m above a horizontal surface at the foot of the slope. It collides elastically with a 483

kg mass initially at rest on the horizontal surface, as shown in the figure. The mass 483

kg slides up a similar track. The acceleration of gravity is 9.8 m/s.

a) What is the speed of the block 483 kg immediately after the collision?

b) To what maximum height above the horizontal surface will the mass 483 kg slide?

just before collision the velocity of 397kg is, v = (2*g*h) = 7.2746m/s

just after collision let the velocities of 397kg and 483kg be v1 and v2 respectively

now applying conservation of linear momentum along horizontal direction just before and after the collision we get

397*7.2746 = 397*v1 + 483*v2

since the collision is elastic from coefficient of restitution we get

1 = (v2-v1)/v

or v2 - v1 =v

solving these two equations we get v2 =6.5636m/s

maximum height reached by it is v22/2g = 2.198m

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