Question

Two identical blocks A and B, each of mass $m$ resting on smooth floor are connected by a light spring of natural length $L$ and the spring constant $K$, with the spring at its natural length. A third identical block C (mass $m$) moving with a speed $v$ along the line joining A and B collides with A. The maximum compression in the spring is:

• A. $v\sqrt{\frac{m}{2k}}$
• B. $m\sqrt{\frac{v}{2k}}$
• C. $\sqrt{\frac{mv}{2k}}$
• D. $\frac{mv}{2k}$

Answer 1

Answer: (A)

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

On compression, considering collision to be instantaneous and using linear momentum conservation since no external force is acting.

$\Rightarrow$ Velocity of A after impact $V_A=v$

Now spring will compress until $V_A>V_B$ and maximum compression will be when $V_A=V_B$.

So using moment conservation on system $"A+B+spring"$,

$\Rightarrow$ Total linear momentum of the system will remain constant.

$mv=m(V_A\text{when max comp}+V_B\text{when max compress})$

$\therefore$ $mv=m(V_{\text{max compress}}+V_{\text{max compress}})$

$\Rightarrow V_{maxcompress}=\frac{v}{2}$

Total K.E of block A & B at max compression $K.E_{total}=2\times \frac{1}{2}m(V_{\text{max compress}}{)}^2=\frac{m{v}^2}{4}$

Initial K.E (after collision) $K.E_i=\frac{1}{2}m{v}^2$

Difference in kinetic energy $=\frac{1}{2}m{v}^2-\frac{1}{4}m{v}^2=\frac{1}{4}m{v}^2$

The difference will be converted to P.E of spring.

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

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