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Question

An iron ball of mass m=50g falls from a height of h1=5m and rises upto h2=2m after colliding with the horizontal surface. If the time of the glass half is Δt=0.02s, find the average contact force exerted on the ball by the horizontal surface.

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Solution

The change in linear momentum during collision is
ΔP=mv2mv1
=m(v2v1)
=m[v2^j(vi^j)]=m[v1+v2]^j
The average force during the collision is
F=ΔpΔt=m(v1+v2)^jΔt
We can calculate v1 and v1 by using kinematics.
using v2=u2+2as
Putting a=g and s=h, we get
v1=,2gh1,v2=2gh2

Or F=m(2gh2+2gh2)^jΔt

=(501000)2×10×5+2×10×3.20.02^j=45^jN

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


1028992_982470_ans_c5d3580eaad24317b11393952ccb883c.PNG

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