Question

# A block of mass m = 1.50 kg slides down a 30.0∘ incline which is 3.60...

A block of mass m = 1.50 kg slides down a 30.0∘ incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.20 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 = 1.50 kg after the collision.

Express your answer to three significant figures and include the appropriate units.

vm =

Part B

Determine the speed of the block with mass M = 6.20 kg after the collision.

Express your answer to three significant figures and include the appropriate units.

vM =

Part C

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

Express your answer to three significant figures and include the appropriate units.

dA =

Using Conservation of energy

0.5 mV^2=mgh

V=sqrt[2gh]=sqrt[2*9.8*3.6]
V=8.4 m/s

For an elastic collision, as per coefficient of resitution, relative velocity before and after are same ,so

V=V1+V2

By Conservation of Momentum

mV=MV2-mV1

mV=MV2-m(V-V2)

2mV =(M+m)V2
V2=2mV/(m+M)
V2 =2*1.5*8.4/(1.5+6.2)

V2=3.27 m/s

a) velocity of m1

V1=8.4-3.27 =5.13 m/s

b) velocity of m2

V2=3.27 m/s

c) considering 3rd equation of motion

h'=V1^2/2g =5.132/2*9.8

h'=1.34 m
Along the incline
d= h'/sin 30
d= 2.68 m
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Comment in case any doubt.. good luck

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