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An iron ball of mass m=50 gm falls from a height of h1=5 m and rises upto h2=3.2 m after colliding with the horizontal surface. If the time of contact with the surface is Δt=0.02 s, find the average contact force(Newtons) exerted on the ball by the horizontal surface. Take g=10 m/s2.

A
25 N
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B
60 N
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C
45 N
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D
95 N
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Solution

The correct option is C 45 N
The change in linear momentum in vertical direction (y-direction) during collision is:


Δp=mv2mv1
=m(v2v1)
=m[v2^j(v1^j)]=m(v1+v2)^j

The average force during the collision is
F=ΔpΔt=m(v1+v2)^jΔt..............i

Using kinematic equation:
v2=u2+2as

Putting a=g and s=+h; where s= Displacement of ball after colliding with surface, we get:

v1=2gh1, v2=2gh2

or F=m(2gh1+2gh2)^jΔt

F=(501000)(2×10×5)+2×10×3.20.02^j=45^j N

During collision, the horizontal surface pushes the ball up with an average force of 45 N.

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