Physics - W.E.P. Question with Solution | TestHub

The hanging ball can only receive a horizontal impulse (the thread is vertical and takes any vertical impulse), so for a head-on collision the projectile ball $M$ must be moving horizontally at the moment of impact, i.e. the collision takes place at the top (apex) of its projectile trajectory. Since the peg is $2\ell$ above the ground and the thread has length $\ell$, the ball $m$ hangs $\ell$ above the ground, so the maximum height of the projectile is $H=\ell$ and the collision happens at half the range: $x=R/2$.

Conserving horizontal momentum and using $e=1$ for the elastic collision,

$$M{v}_0\text{cos}\theta +0=Mv+mv\prime ,e=1=\frac{v\prime -v}{v_0\text{cos}\theta }\Rightarrow v\prime =\frac{2M{v}_0\text{cos}\theta }{M+m}$$

For the ball to complete the whole circle around the peg,

$$v\prime =\frac{2M{v}_0\text{cos}\theta }{M+m}\ge \sqrt{5g\ell }\Rightarrow v_0\text{cos}\theta \ge \frac{M+m}{2M}\sqrt{5g\ell }$$

From the projectile motion of $M$: $H=\ell =\frac{v_0^2{\text{sin}}^2\theta }{2g}$, so $v_0\text{sin}\theta =\sqrt{2g\ell }$, and

$$x=\frac{R}{2}=\frac{v_0^2\text{sin}\theta \text{cos}\theta }{g}=\frac{(v_0\text{cos}\theta )\sqrt{2g\ell }}{g}\ge \frac{(M+m)\sqrt{5g\ell }\cdot \sqrt{2g\ell }}{2Mg}=\frac{(M+m)\ell \sqrt{10}}{2M}$$

$$\text{Solving, }x\ge \frac{3\sqrt{10}}{4}=\frac{3\times 3.162}{4}\Rightarrow x\ge 2.37\text{m}$$

Of the given choices only 2.5 m satisfies this, so the answer is (D).