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. 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 with A. Then the :

• A. Kinetic energy of the A - B system, at maximum compression of the spring, is zero
• B. Kinetic energy of the A - B system, at maximum compression of the spring, is $m{v}^2/4$
• C. Maximum compression of the spring is $v\sqrt{mk}$
• D. Maximum compression of the spring is $v\sqrt{m/k}$

Answer 1

Answer: (B), (D)

Kinetic energy of the A - B system, at maximum compression of the spring, is $m{v}^2/4$
Maximum compression of the spring is $v\sqrt{m/k}$

At maximum compression block A and B have same velocity(v')

By momentum conservation,

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

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

By energy conservation,

$\frac{1}{2}m{v}^2=\frac{1}{2}m(\frac{v}{2}{)}^2+\frac{1}{2}m(\frac{v}{2}{)}^2+\frac{1}{2}k{x}^2$......(1) where, $=\frac{1}{2}m(\frac{v}{2}{)}^2+\frac{1}{2}m(\frac{v}{2}{)}^2$=K.E of A and B block

$=\frac{m{v}^2}{4}$=K.E of AB system

Considering (1),

$\frac{1}{2}m{v}^2=\frac{m{v}^4}{4}+\frac{1}{2}k{x}^2$

$k{x}^2=\frac{m{v}^2}{2}$

$x=v\sqrt{\frac{m}{2k}}$