Q.285. Elastic collision.

Question 285.
An object X which is moving with a velocity v on a smooth horizontal surface collides elastically with another
object Y which is moving with a velocity v/4 in the opposite direction. If the masses of X and Y are m and 2m respectively, what is the velocity of Y after the collision?

Answer 285.
Let u = velocity of the object X after the collision in the direction of v
and w = velocity of the object Y after the collision in the direction of v

For elastic collison, both the linear momentum and the kinetic energy are conserved.
=> mv - 2mv/4 = mu + 2mw
=> v/2 = u + 2w
=> v = 2u + 4w ... ( 1 )
and
(1/2)mv^2 + (1/2)(2m)(v/4)^2 = (1/2)mu^2 + (1/2)(2m)w^2
=> 9v^2/8 = u^2 + 2w^2
=> 9v^2 = 8u^2 + 16w^2 ... ( 2 )

From eqn. ( 1 ), 16w^2 = (v - 2u)^2 = v^2 - 4vu + 4u^2
Plugging in eqn. ( 2 ),
9v^2 = 8u^2 + v^2 - 4uv + 4u^2
=> 12u^2 - 4vu - 8v^2 = 0
=> 3u^2 - vu - 2v^2 = 0
=> 3u^2 - 3uv + 2uv - 2v^2 = 0
=> 3u (u - v) + 2v (u - v) = 0
=> (u - v) (3u + 2v) = 0
=> u = - (2/3) v ... [because u ≠ v]

From eqn. ( 1 ),
v = - (4/3) v + 4w
=> w = 7v/12
Answer: Velocity of Y after the collision is 7v/12 in the direction opposite to its initial direction of motion.

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