A block of mass m = 1.90 kg slides down a 30.0?incline which is 3.60 m high. At

Question

A block of mass m = 1.90 kg slides down a 30.0?incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.60 kg which is at rest on a horizontal surface (Figure 1) . (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored.

1.Determine the speed of the block with mass m = 1.90 kg after the collision.

2.Determine the speed of the block with mass M = 7.60 kg after the collision.

3.Determine how far back up the incline the smaller mass will go.

Explanation / Answer

Velocity ofass m= 1.90kg is given by energy conservation, potential energy of mass m=1.90 kg converted in to kinetic energy as it slide down.

v1 = (2gh)1/2 = (2*9.8*3.6)1/2= 8.4m/s

A) Give collision is elastic, velocity of mass m=1.9 kg after collision is given by

vm =[ (m-M)v1]/(m+M) = [(1.9-7.6)*8.4]/(1.9+7.6)= -5.04 m/s

Negative sign indicate that mass m will start moving in opposite direction after collision.

B) velocity of mass M after collision is given by

vM = 2mv1/(m+M) = 2*1.9*8.4/(1.9+7.6)= 3.36 m/s

C) Height climb by small mass d is given by energy conservation, kinetic energy of mass m converted into potential energy.

D = vm2/2g= (5.04)2/(2*9.8) = 1.296 m = nearly 1.3 m

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