Question

Two blocks A and B, each of mass m, are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then,

• A. the kinetic energy of the A-B system, at maximum compression of the spring, is zero.
• B. the kinetic energy of the A-B system, at maximum compression of the spring, is $\frac{m{v}^2}{4}$.
• C. the maximum compression of the spring is $v\sqrt{\frac{m}{K}}$
• D. the maximum compression of the spring is $v\sqrt{\frac{m}{2K}}$

Answer 1

Answer: (B), (D)

The correct options are
B the kinetic energy of the A-B system, at maximum compression of the spring, is $\frac{m{v}^2}{4}$.
D the maximum compression of the spring is $v\sqrt{\frac{m}{2K}}$

A B and C are identical after collision of C with A linear momentum conservation. $mv=mvo+m{v}^{\prime }$

v' = v v' = speed of block A and C get rest.

at time of maximum compression both A and B moves with constant velocity

$mv=(m+m)v^{\prime }$

$v^{\prime }=\frac{v}{2}$

K.E of both block A and B is = $\frac{1}{2}m{v}^2+\frac{1}{2}mv{1}^2$

$=\frac{m{v}^2}{4}$

According energy conservation

$\frac{1}{2}{mv}^2=K.E+P.E\frac{1}{2}{mv}^2=\frac{m{v}^2}{4}+\frac{1}{2}{x}^2\frac{m{v}^2}{2}={Kx}^{2}x=\sqrt{\frac{m}{2k}v}$

x = maximum compression.