A block of mass m = 2.00 kg slides down a 30.0 incline which is 3.60 m high. At

Question

A block of mass m = 2.00 kg slides down a 30.0 incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.20 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.

Part A.)Determine the speed of the block with mass m = 2.00 kg after the collision.

Part B.)Determine the speed of the block with mass M = 7.20 kg after the collision.

Part C.)Determine how far back up the incline the smaller mass will go.

Explanation / Answer

speed of block at bottom

mgh = 0.5mv^2

v = sqrt(9.81*3.6/ 0.5)

v = 8.40 m/s

conserve momentum here,

m1u1 + m2u2 = m1v1 + m2v2

2*8.4 + 0 = 2*v1 + 7.2v2 --------1

consere KE as collison is elastic

0.5m1u1^2 = 0.5m1v1^2 + 0.5 m2v2^2 ---------2

on solving 1 and 2

A).v1 = (m1-m2)u1/(m1+m2)

v1 = (2-7)*8.4 / 9 = -4.66 m/s { opposite to its initial motio }

B.)v2 = 2m1u1/(m1+m2)

v2 = 2*2*8.4 / 9 = 3.73 m/s

c.)

conserve energy

mgh= 0.5mv^2

h = 0.5*4.66^2/9.81 = 1.106 m

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