Question

a moving ball of mass m undergoes a head on collision with another stationary ball of mass 2m show that the colliding ball loses 8/9 of its total energy after collision

Answer 1

we have to show that the colliding ball loses 8/9 of its total energy after collision

proof : first of all, find final velocity of balls.

let v1 and v2 are final velocities of ball of mass m and 2m respectively.

here, u1 = u and u2 = 0

using formula,

v2 = 2m1u1/(m1 + m2) - (m1 - m2)/(m1 + m2)u2

= 2mu/(m + 2m) - (m - 2m)/(m + 2m) × 0

= 2u/3

and , v1 = (m1 - m2)u1/(m1 + m2) + 2m2u2/(m1 + m2)

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

= -u/3

initial kinetic energy of colliding ball, K.E_i = 1/2 mu²

final kinetic energy of colliding ball, K.E_f = 1/2 m(-u/3)² = 1/2 mu²/9

so, loss in kinetic energy of colliding ball = K.E_i - K.E_i

= 8/9 1/2 mu²

= 8/9 K.E_i

hence, it shows that the colliding ball loses 8/9 of its total energy after collision.

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Answer 2

$\huge\bold\purple{Answer:-}$

first of all, find final velocity of balls.

let v1 and v2 are final velocities of ball of mass m and 2m respectively.

here, u1 = u and u2 = 0

using formula,

v2 = 2m1u1/(m1 + m2) - (m1 - m2)/(m1 + m2)u2

= 2mu/(m + 2m) - (m - 2m)/(m + 2m) × 0

= 2u/3

and , v1 = (m1 - m2)u1/(m1 + m2) + 2m2u2/(m1 + m2)

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

= -u/3

initial kinetic energy of colliding ball, K.E_i = 1/2 mu²

final kinetic energy of colliding ball, K.E_f = 1/2 m(-u/3)² = 1/2 mu²/9

so, loss in kinetic energy of colliding ball = K.E_i - K.E_i

= 8/9 1/2 mu²

= 8/9 K.E_i

hence, it shows that the colliding ball loses 8/9 of its total energy after collision.