Chemistry, asked by jayanthbabu3799, 1 year ago

A ball of mass m1 experiences a perfectly elastic collision with a stationary ball of mass m2. What fraction of the kinetic energy does the striking ball lose if, (i) it recoils at right angle to the original direction of motion (ii) the collision is head on one

Answers

Answered by Answers4u
56

(i) Mass m1 undergoes elastic collision

Mass m2 is at rest and let the recoiling angle be denoted by A.

Momentum:

m1*u = m2*v2*CosA

m1*v1 = m2*v2*SinA

Squaring and adding: m1^2 (u^2 + v1 ^2 ) = m2^2*v2^2

Equate intial and final K.E

1/2*m1*u^2 = 1/2*m1*v1^2 + 1/2*m2*v2^2

m2 (u^2 - v1^2 ) = m1(u^2 + v1^2 ) (v1/u)^2

= (m2 - m1 / m2 + m1)

Fraction of kinetic energy lost:

(0.5*m1*u^2 - 0.5*m1*v1^ 2 ) / 0.5*m1*u 2 1 - (v1/u) 2

Therefore, 1 - (v1/u)^2 = 1 -(m2 - m1 / m2 + m1)

Fraction of KE lost = 2m1 / (m1 + m2)

(ii) For head-on collision,

Momentum is m1*u = m1*v1 + m2*v2

Solving in the similar way,

Fraction of KE lost = 4m1*m2/(m1+m2)^2​


Answered by Sidyandex
7

All things considered, since the impact is versatile and we realize that the motor vitality when the crash stay rationed so it doesn't make a difference the balls crash happen at right point or 90 degree and furthermore the head on impact additionally does not make a difference,

So there will be no loss of dynamic vitality in both of the causes as the inquiry as of now said the impact is superbly flexible.

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