A small block is connected to one end of a massless spring of un-stretched length $4.9m$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2m$ 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 $P$ is at 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 $t = 1s,$ the value of $v$ is (take $g = 10m/s^{2}$) [image:14_image3.emf]

Question

Dr. Abubacker · Accepted Answer

(a) [image:14_image55.emf]
The block is released from $A$
$x = 4.9m + (0.2m)\sin\left( \omega t + \frac{\pi}{2} \right)$
at$\ t = 1s;x = 5m$
so range of projectile will be $5m$
Now$\ 5 = \frac{v^{2}\sin{90{^\circ}}}{g} \Rightarrow v^{2} = 50 \Rightarrow v = \sqrt{50}$

Explanation

Reviewed by Dr. Abubacker

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