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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 BYJU'S Exam Prep
Updated on: September 25th, 2023
(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 ‘As final velocity be v1 and body ‘Bs 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.