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 \(\omega = \dfrac{\pi}{3}\text{rad/s}\). Simultaneously at \(t = 0\), a small pebble is projected with speed \(v\) from point \(P\) at an angle of \(45^o\) 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 \(g = 10 ~m/s^2)\)


45 _————— o 10 m P

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

Answer 1

The correct option is C $\sqrt{50}m/s$

For the block, $T=\frac{2\pi }{\omega }=6\text{seconds}$.
For block, the position in $1\text{seconds}$ will be $x=0.2\cos \omega t$
$\Rightarrow$ At $t=1s$, block will be at a distance $4.9M+0.1m=5m$ from wall.
$\therefore$ Range of pebble is also $5m$
Now, for pebble, $v\cos {45}^o\times \left(1s\right)=5m$