A block of mass m = 1.50kg slides down a 30.0? incline which is 3.60 m high. At

Question

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

Determine the speed of the block with mass m = 1.50kg after the collision.

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

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

Explanation / Answer

By Conservation of energy

(1/2)mV2=mgh

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

For an elastic collision 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) =2*1.5*8.4/(1.5+6.2)

V2=3.27 m/s

a)

V1=8.4-3.27 =5.13 m/s

b)

V2=3.27 m/s

c)

h'=V12/2g =5.132/2*9.8

h'=1.34 m

so

S=h'/sin(o) =1.34/sin30

S=2.68 m up incline

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