Question

A block of mass 2.0 kg is moving on a friction less horizontal surface with a velocity of 1.0 m/s below figure towards another block of equal mass kept at rest.The spring constant of the spring fixed at one end is 100 N/m.Find the maximum compression of the spring.

Answer 1

Mass of each block $M_B=2kg$

Initial velocity of the 1st block, $v=1\frac{m}{sec}$

$V_a=1\frac{m}{sec}$

$V_b=0\frac{m}{sec}$

Spring constant of the spring=$100\frac{N}{m}$

The block A strikes the spring with a velocity $1\frac{m}{sec}$

After the collision,it's velocity decreases continuosly and at an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity.

Let that velocity be V.

$\left(\frac{1}{2}\right)M_A{V}_A^2+\left(\frac{1}{2}\right)M_B{V}_B^2=\left(\frac{1}{2}\right)M_A{V}^2+\left(\frac{1}{2}\right)M_B{V}^2+\left(\frac{1}{2}\right)k{x}^2\left(\frac{1}{2}\right)\times 2(1{)}^2+0$

$=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\times v^2+\left(\frac{1}{2}\right)x^2\times 100$

(when x=maximum compression of spring).

$\Rightarrow 1-2v^2=50x^2...\left(i\right)$

As there is no external force in the horizontal direction,the momentum should be conserved.

$\Rightarrow M_A{V}_A+M_B{V}_B=(M_A+M_B)V$

$\Rightarrow 2\times 1=4\times V$

$\Rightarrow V=\left(\frac{1}{2}\right)\frac{m}{sec}...\left(ii\right)$

Putting in equation (i)

$1=2\times \left(\frac{1}{4}\right)+50x^2$

$\Rightarrow \left(\frac{1}{4}\right)=50x^2$

$\Rightarrow x^2=\frac{1}{100}{m}^2$

$\Rightarrow x=\left(\frac{1}{10}\right)m$

$\Rightarrow =0.1m=10cm$