Simple Harmonic Motion Ques 44
- A small block is connected to one end of a massless spring of unstretched 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 $\omega=\frac{\pi}{3} rad / s$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure.
Point $P$ is at a horizontal distance of $10 m$ from $O$. If the pebble hits the block at $t=1 s$, the value of $v$ is
(take, $g=10 m / s^{2}$ )
(2012)
(a) $\sqrt{50} m / s$
(b) $\sqrt{51} m / s$
(c) $\sqrt{52} m / s$
(d) $\sqrt{53} m / s$
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Answer:
Correct Answer: 44.(a)
Solution:
Formula:
- $t$ =time of flight of projectile
$$ \begin{aligned} & =\frac{2 v \sin \theta}{g} \quad\left(\theta=45^{\circ}\right) \\ \therefore \quad v=\frac{g t}{2 \sin \theta} & =\frac{10 \times 1}{2 \times 1 / \sqrt{2}} \\ & =\sqrt{50} m / s \end{aligned} $$
BETA
