Question

Two blocks $A$ and $B$ each of mass $m$, are connected by a massless spring of natural length Land 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 $mv^2/4$
• C. the maximum compression of the spring is $v\sqrt{(m/k)}$
• D. the maximum compression of the spring is v

Answer 1

Answer: (B), (D)

After collision between $C$ and $A, C$ stops while $A$ moves with speed of $C$ i.e. $v$ [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, $A$ and $B$ will move with same speed $v / 2$ (From conservation of linear momentum).

Let $x$ be the maximum compression in this position.
$\therefore$ KE of $A-B$ system at maximum compression
$=\frac{1}{2}(2 n)\left(\frac{v}{2}\right)^{2} $
or $K_{\max } =m v^{2} / 4$
From conservation of mechanical energy in two positions shown in above figure
or $\frac{1}{2} m v^{2}=\frac{1}{4} m v^{2}+\frac{1}{2} k x^{2}$
$\frac{1}{2} k x^{2}=\frac{1}{4} m v^{2} $
$ \Rightarrow \therefore x=v \sqrt{\frac{m}{2 k}}$