Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head on and elastic in nature. After the collision, the fraction of kinetic energy lost by the colliding body A is?

By Ritesh|Updated : November 13th, 2022

(a) 1/9

(b) 8/9

(c) 4/9

(d) 5/9

After the collision, the fraction of kinetic energy lost by the colliding body A is 8/9. Given that body 'A' of mass 4m moves with speed u and body ‘B’ of mass 2m is at rest.

Therefore, the initial linear momentum of the system = 4mu + 2m x 0 = 4mu.

Let body 'A"s final velocity be v1 and body 'B"s final velocity be v2

Then, the final linear momentum of the system = 4mv1 + 2mv2

We know that as collision is elastic, e = 1

So v1 = (4m - 2m)u/ (4m + 2m) + [2 (4m) u]/ (4m + 2m) = 8mu/6m = 4u/3

So, initial kinetic energy of body 'A' = ½ (4m) u2 = 2mu2

Final kinetic energy of body 'A' = ½ (4m) u2/9 = 2mu2/9

So, change in kinetic energy = final kinetic energy - initial kinetic energy

Substituting the values we get:

= 2mu2/9 - 2mu2

= 2mu2 [1/9 - 1]

= -16mu2/ 9 [Here negative sign indicates that kinetic energy is lost after collision]

So, fractional loss in kinetic energy:

= [16mu2/ 9]/ (2mu2)

= 8/9

Summary:

Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head on and elastic in nature. After the collision, the fraction of kinetic energy lost by the colliding body A is?

Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head-on and elastic in nature. After the collision, the fraction of kinetic energy lost by the colliding body A is 8/9.

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