Question

A small block is connected to one end of a massless spring of unstretched length \(4.9~\text{m}\). The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by \(0.2~\text{m}\) and then released. It then executes SHM with an angular frequency of \(\omega = \left(\dfrac{\pi}{3}\right) \text{rad s}^{-1}\). Simultaneoulsy, at \(t = 0\), a small pebble is projected with a speed \(v\) from the point \(P\) at an angle of \(45^{\circ}\) as shown in the figure. The point \(P\) is at a horizontal distance of \(10~\text{cm}\) from \(O\). If this pebble hits the block at \(t =1~\text{s}\), the value of \(v\) is

10m

• A. $\sqrt{53}{\text{ms}}^{-1}$
• B. $\sqrt{52}{\text{ms}}^{-1}$
• C. $\sqrt{51}{\text{ms}}^{-1}$
• D. $\sqrt{50}{\text{ms}}^{-1}$

Answer 1

The correct option is D $\sqrt{50}{\text{ms}}^{-1}$

In this problem, we don't need to consider the SHM part rather we will focus on projectile part only.
Since, the pebble hits the block after $1\text{s}$, we can easily calculate the speed of projection $\left(v\right)$ from this "Time of flight" data only.
So, we have
Time of flight $t=T=\frac{2u\sin \theta }{g}$, i.e., $1\text{s}=\frac{2u.\sin {45}^{\circ }}{10}$
$\therefore u=5\sqrt{2}{\text{ms}}^{-1}=\sqrt{50}{\text{ms}}^{-1}$
Hence, option $\left(A\right)$ is correct.