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Question

# 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 NThe change in linear momentum in vertical direction (y-direction) during collision is: Δ→p=m→v2−m→v1 =m(→v2−→v1) =m[v2^j−(−v1^j)]=m(v1+v2)^j ⇒The average force during the collision is →F=Δ→pΔt=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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