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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 energy lost by the colliding body A is :
  • a)
    1/9
  • b)
    8/9
  • c)
    4/9
  • d)
    5/9
Correct answer is option 'B'. Can you explain this answer?
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Body A of mass 4m moving with speed u collides with another body B of ...
Fractional loss of KE of ccolliding body
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Body A of mass 4m moving with speed u collides with another body B of ...
Concept: In an elastic collision, the total kinetic energy of the system is conserved. The momentum of the system is also conserved.

Given: 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.

To find: The fraction of energy lost by the colliding body A.

Solution:

Let the final velocities of bodies A and B be v1 and v2, respectively.

By conservation of momentum:
4mu = 2mv1 + 2mv2
2mu = mv1 + mv2

Solving for v1 and v2, we get:
v1 = (2u)/3
v2 = (u)/3

Kinetic energy of body A before collision = (1/2) * 4m * u^2 = 2mu^2
Kinetic energy of body A after collision = (1/2) * 4m * v1^2 = (4/9) * mu^2

Fraction of energy lost by body A = (Initial KE - Final KE) / Initial KE
= (2mu^2 - (4/9)mu^2) / 2mu^2
= (8/9)

Therefore, the fraction of energy lost by the colliding body A is 8/9.

Answer: Option B.
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Body A of mass 4m moving with speed u collides with another body B of ...
Body A of mass 4m moves with speed u and body B of mass 2m at rest.

so, initial linear momentum of system = 4mu + 2m × 0 = 4mu

Let final velocity of body A is v1 and body B is v2

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

as collision is elastic ,

so, v1 =

= (4m - 2m) × u/(4m + 2m) + 2(2m)×0/(4m+2m)

= 2mu/6m

= u/3

similarly, v2 =

= (2m - 4m) × 0/(4m + 2m) + 2(4m)u/(4m+ 2m)

= 8mu/6m

= 4u/3

so, initial energy of body A = 1/2 (4m) u² =2mu²

final kinetic energy of body A = 1/2 × (4m) × u²/9

= 2mu²/9

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

= 2mu²/9 - 2mu²

= 2mu² [ 1/9 - 1]

= -16mu²/9 [ here negative sign indicates kinetic energy lost after collision of body A]

so, fractional lost in kinetic energy = change in kinetic energy/initial kinetic energy

= (16mu²/9)/(2mu²)

= 8/9
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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 innature. After the collision the fraction of energy lost by the colliding body A is :a)1/9b)8/9c)4/9d)5/9Correct answer is option 'B'. Can you explain this answer?
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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 innature. After the collision the fraction of energy lost by the colliding body A is :a)1/9b)8/9c)4/9d)5/9Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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 innature. After the collision the fraction of energy lost by the colliding body A is :a)1/9b)8/9c)4/9d)5/9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 innature. After the collision the fraction of energy lost by the colliding body A is :a)1/9b)8/9c)4/9d)5/9Correct answer is option 'B'. Can you explain this answer?.
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