Question

A small block is connected to one end of a massless spring of un-stretched length $4.9$m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2$m and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega =\pi /3$ rad/s. Simultaneously at $t=0$, a small pebble is projected with speed v form point P at an angle of ${45}^o$ as shown in the figure. Point P is at a horizontal distance of $10m$ from O. If the pebble hits the block at $t=1$s, the value of v is (take $g=10m/s^2$)

• A. $\sqrt{50}$ m/s
• B. $\sqrt{51}$ m/s
• C. $\sqrt{52}$ m/s
• D. $\sqrt{53}$ m/s

Answer 1

Answer: (A)

$\sqrt{50}$ m/s

The block starts its oscillation from its positive extremum.

$\therefore x_{block}\left(t\right)=4.9+0.2\cos \omega t$

$\therefore x_{block}(t=1s)=4.9+0.2\cos (\frac{\pi }{3}\times 1)=5$ $m$

$\therefore$ Range of pebble $=5m$

So, $\frac{v^2\sin 2\theta }{g}=5$

$⟹v=\sqrt{50}$ $m/s$