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

Question

A block of mass rn = 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. (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored. Determine the speed of the block with mass m = 2.00 kg after the collision. Express your answer to three significant Figures and include the appropriate units. Determine the speed of the block with mass At M = 7.20 kg after the collision Express your answer to three significant figures and include the appropriate units. Determine how far back up the incline the smaller mass will go. Express your answer to three significant figures and include the appropriate units.

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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