Two identical blocks A and B, each of mass m resting on a smooth floor are connected by a light spring of natural length Land spring constant K, with the spring in its natural length. A third identical block of mass m moving with a speed v along the line joining Aand B collides with A. The maximum compression in the spring is

Options

A. `vsqrt((m)/(2K))`
B. `msqrt((v)/(2K))`
C. `sqrt((mK)/(v))`
D. `(mv)/(2K)`

Answer

A

Explanation

Because the masses of A and Care identical and, initially the body A is not in motion, hence when the body .C. collides elastically to body A, it itself gets stopped and transfers its momentum ‘mv.and K.E.(`1//2 m^2)` totally to body A. In this position, B is stationary, and spring is in its normal state. Hence, the instantaneous momentum of the system is my and KE is `1//2 mv^2`. Now, the spring is compressed and block B also comes in motion. Suppose the instantaneous velocity of A and B after the collision be `v^2` and compression of the spring is x.As no external force acts on the system, we have according to law of conservation of momentum and energy,
`mv = mv. + mv. " or " v = 2 v. (1) `
Also , `1/2 mv^2 = 1/2 mv^2 + 1/2 mv.^2 + 1/2 Kx^2 " or " v^2 = 2v.^2 + (Kx^2)/(m)`
From eqn. (1) `v. = v/2 therefore v^2 = (v^2)/(2) + (Kx^2)/m`
or `(kx^2)/m = (v^2)/(2) " or " x = v sqrt(m/(2K))`.