Question

A particle is suspended from a fixed point by a string of length $5m$. It is projected from the equilibrium position with such a velocity that the string slackens after the particle has reached a height $8m$ above the lowest point. Find the velocity of the particle, just before the string slackens. Find also, to what height the particle can rise further?

• A. $2.71m{s}^{-1},0.48m$
• B. $5.42m{s}^{-1},0.96m$
• C. $2.71m{s}^{-1},0.96m$
• D. $5.42m{s}^{-1},0.48m$

Answer 1

The correct option is B $5.42m{s}^{-1},0.96m$

Let the velocity when string slackens be $v$.

As particle is $8m$ above the lowest point, it makes angle, $\theta =co{s}^{-1}(\frac{8-5}{5})={53}^0$, with the veritcal

Now, string slackens means the tension in the string becomes zero.

So, $mgcos{53}^0=\frac{m{v}^2}{5}⟹v=\sqrt{5gcos{53}^0}$

$⟹v=\sqrt{3g}=\sqrt{3\times 9.8}=5.42m/s$

Now, the particle won't perform circular motion. The component of velocity in vertical direction is, $v_y=vsin{53}^0=5.42\times 0.8=4.336m/s$

So, further height it can reach is,

$h=\frac{v_y^2}{2g}=0.96m$

Option B is correct.