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Question

A moving particle of mass 'm' makes a head-on collision with a particle of mass '2m' initially at rest. If the collision is perfectly elastic, the percentage loss of energy of the colliding particle is


Solution

The correct option is C
Kinetic energy of projectile before collision Ki=12m1u21
Kinetic energy of projectile after collision Kf=12m1v21
Kinetic energy transferred from projectile to target DK = decrease in kinetic energy in projectile
ΔK=12m1u2112m1v21=12m1(u21v21) Fractional decrease in kinetic energy
ΔKK=12m1(u21v21)12m1u21=1(v1u1)2
We can substitute the value of v1 from the equation
v1=(m1m2m1+m2)u1+2m2u2m1+m2
If the target is at rest i.e. u2=0 then v1=(m1m2m1+m2)u1
From Equation (i) ΔKK=1(m1m2m1+m2)2     or  ΔKK=4m1m2(m1+m2)2
Percentage loss of energy=4m1m2(m1+m2)2×100
=4m×2m(2m+m)2×100=8009=88.9%
Hence the correct choice is (c).
 

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