Question

Initially distance between two small blocks A and B is $1m$ and natural length of spring is $2m$. 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 $20cm$
• B. maximum speed of block B during whole complete is $\frac{2}{\sqrt{3}}$
• 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

Answer 1

Answer: (B), (C), (D)

The correct options are
B maximum speed of block B during whole complete is $\frac{2}{\sqrt{3}}$
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

Since the centre of mass of the entire system remains at rest, so after considering mass ($m$) $2kg$ moving distance $x_1$ to left and mass $3kg$ ($M$) moving $x_2$ to right when released from compression, we can write. These distances are also the amplitudes of the two masses.
$2x_1=3x_2$

$\frac{x_1}{x_2}=\frac{3}{2}$

Adding $1$ to both sides:

$\frac{x}{x_2}=\frac{5}{2}$

$x_2=\frac{mx}{m+M}=\frac{2}{5}x$

$x=$ compressed length $=1m$

Therefore, $x_2=\frac{2}{5}$

Similarly, $x_1=\frac{Mx}{m+M}=\frac{3x}{5}$

Tension in the system, $T=kx=10N$

Now with respect to $3kg$ mass ($M$):

Acceleration, $a=\frac{T}{M}=\frac{k(m+M)x_2}{mM}$

Frequency, $\omega =\sqrt{\frac{a}{x_2}}=\sqrt{\frac{k(m+M)}{mM}}$

Maximum speed of block $B$ is:
$v_b=\omega x_2=\sqrt{\frac{k(m+M)}{mM}}\frac{2}{5}=\frac{2}{\sqrt{3}}$

Block $B$ collides with wall elasitically after covering $0.4m$ which is also its amplitude.

Impulse $=$ total change in momentum $=2m{v}_b$ $=2\times 3\times \frac{2}{\sqrt{3}}=4\sqrt{3}N/s$