Question

Initially distance between two small blocks A and B is 1 m and natural length of spring is 2 m. Now the system is released . If B hits the vertical wall elastically then : (All surfaces are smooth)

• A. Maximum extension in spring during complete motion is 20 cm.
• B. Maximum speed of block B during whole complete is $\frac{2}{\sqrt{3}}$ m/sec.
• C. At maximum extension speed of block A is $\frac{4\sqrt{3}}{5}$ m/sec.
• D. Impulse by vertical wall on block B is $4\sqrt{3}N.sec$.
• E. Blank

Answer 1

The correct option is E Blank

Let the respective displacements of the block from the initial position be $x_1,x_2$ in opposite directions.

Thus $x_1+x_2=2-1=1$

Also since no external force acts on the system ,the center of mass of the system remains at the same place.

Thus $3x_1=2x_2$

$⟹x_1=\frac{2}{5}m,x_2=\frac{3}{5}m$

By conservation of momentum and energy,

$\frac{1}{2}\times 10\times 1^2=\frac{1}{2}\times 3\times v_1^2+\frac{1}{2}\times 2\times v_2^2$

$⟹v_1=\frac{2}{\sqrt{3}}m/s,v_2=\sqrt{3}m/s$

By elastic collision,

Impulse=$2m_B{v}_1=4\sqrt{3}Ns$
On maximum extension, both blocks move with same velocity,

Let that velocity be $v$.

Thus by conservation of energy,

$\frac{1}{2}\times 2\times {\sqrt{3}}^2+\frac{1}{2}\times 3\times (\frac{2}{\sqrt{3}}{)}^2=\frac{1}{2}\times 10\times x_{max}^2+\frac{1}{2}\times 5\times v^2$

By momentum conservation,

$2\sqrt{3}+2\sqrt{3}=5v$

$⟹v=\frac{4\sqrt{3}}{5}m/s$

$⟹x=\frac{1}{5}m=20cm$