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..
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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!!!!!!
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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)²
Energy transferred = 8/9
Hope it helps! ;-))