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 $v\sqrt{m/2k}$
(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 $v^{\prime }$. 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 $mv=m{v}^{\prime }+m{v}^{\prime }$ or $v^{\prime }=\frac{v}{2}...(i)$
By conservation of mechanical energy, we get
$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}m{v}^{{}^{\prime }2}+\frac{1}{2}m{v}^{{}^{\prime }2}+\frac{1}{2}k{x}^2$
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 $x=v\sqrt{\frac{m}{2k}}$
At maximum compression of the spring, the kinetic energy of $A-B$ system is
$=\frac{1}{2}m{v}^{{}^{\prime }2}+\frac{1}{2}m{v}^{\prime 2}=m{v}^{{}^{\prime }2}=m{\left(\frac{v}{2}\right)}^2=\frac{m{v}^2}{4}$