Physics, asked by jeni1852, 1 year ago

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..​


jeni1852: please anyone help me

Answers

Answered by arvishaali2004
19

Here is ur answer....

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

Hope it helps!!!!!!

Pls mark my answer as Brainliest!!

Answered by TheUnsungWarrior
1

Answer:

After the collision the fraction of energy lost by the colliding body A is 8/9.

Explanation:

Given;-

   Mass of body A = 4m

    Mass of body B = 2m

    Velocity of body A = u

    Velocity of body B = 0

We know that, in a head on & elastic collision, the fraction of kinetic energy lost by a body to another or the fraction of kinetic energy transferred from one body to another is given by ;-

       Energy transferred = 4n/ (1+n)²

Given that the mass of A is twice that of B. Putting the given values in the above formula, we get;-

        Energy transferred = 4 × 2/ ( 1 + 2

       Energy transferred = 8/9

Hope it helps! ;-))

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