Question

A block of $2\text{kg}$ is moving on a frictionless non-metal surface with a velocity of $1{\text{ms}}^{-1}$ towards another block of equal mass with a fixed spring is kept at rest as shown in figure. The Force constant of the spring is $100{\text{Nm}}^{-1}$. Find the maximum compression of the spring.

• A. $8\text{cm}$
• B. $3\text{cm}$
• C. $10\text{cm}$
• D. $1\text{cm}$

Answer 1

The correct option is C $10\text{cm}$

When compression in the spring is maximum, there is no relative motion between the blocks i.e., speed of both the block will be equal.


In collision, only internal force is involved. Hence, linear momentum is conserved.

$P_i=P_f$

$(2\times 1)+0=(2+2)v$

$\therefore v=\frac{1}{2}=0.5{\text{ms}}^{-1}$

Let, maximum compression in the spring be $x.$ As there is no non-conservative force acts on the system. Thus, mechanical energy of the system is conserved.

$E_i=E_f$

$\frac{1}{2}\times 2\times (1{)}^2+0=\frac{1}{2}k{x}^2+\frac{1}{2}\times (2+2)\times {\left(\frac{1}{2}\right)}^2$

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

$\Rightarrow 1=k{x}^2$

$\Rightarrow x=\frac{1}{\sqrt{k}}[\because k=100{\text{Nm}}^{-1}]$

$\therefore x=0.1\text{m}=10\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question ? Key concept : At the time of maximum compression in the spring, velocity of both the blocks are same. After this moment velocity of second block increase and velocity of first block decrease, this lead to compression in the spring be smaller.