Question

Two blocks $A$ and $B$ each of mass $m$ are connected by 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
(1) the maximum compression of the spring is $v \sqrt{m / k}$
(2) the maximum compression of the spring is $v \sqrt{m / 2 k}$
(3) the kinetic energy of the A-B system at maximum compression of the spring is zero
(4) the kinetic energy of the $A-B$ system at maximum compression of the spring is $\frac{m v^{2}}{4}$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (C)

Initially $A$ and $B$ are at rest on smooth surface. Then $C$ moves with velocity $v$, collides elastically with $A$ and stops.
As a result of head-on collision between $C$ and $A$ block. $A$ acquires a velocity $v$ and moves towards $B$.
It compresses spring $L$ which pushes $B$ towards right $A$ goes on compressing the spring till the velocity acquired by $B$ becomes equal to the velocity of $A$.
Let this velocity be $v'$. This state occurs when the spring is in a state of maximum compression.
Let $x$ be the maximum compression in spring at this stage.
By conservation of linear mementum, we get
or $m v=m v'+m v'$ or $v'=\frac{v}{2}\,\,\,...(i)$
By conservation of mechanical energy, we get
$\therefore \frac{1}{2} m v^{2}=\frac{1}{2} m v^{'2}+\frac{1}{2} m v^{'2}+\frac{1}{2} k x^{2}$
or $\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{2} k x^{2}$
$\therefore \frac{1}{2} m v^{2}=\frac{1}{4} m v^{2}+\frac{1}{2} k x^{2}$
or $\frac{k x^{2}}{2}=\frac{m v^{2}}{4}$
or $x=v \sqrt{\frac{m}{2 k}}$
At maximum compression of the spring, the kinetic energy of $A-B$ system is
$=\frac{1}{2} m v^{'2}+\frac{1}{2} m v'^{ 2}=m v^{'2}=m\left(\frac{v}{2}\right)^{2}=\frac{m v^{2}}{4}$