Question

Particle P and Q of masses 20 g and 40 g, respectively are projected from the position A and B on the ground. The initial velocities of P and Q makes angle of ${45}^o$ and ${135}^o$, respectively with the horizontal as shown in the figure. Each particle has an initial speed of 49ms${}^{-1}$. The separation AB is 245m. Both undergo a collision. After the collision, P retraces its path. The position of Q when it hits the ground is :

• A. $245m$
• B. $\frac{245}{3}m$
• C. $\frac{245}{2}m$
• D. $\frac{245}{\sqrt{2}}m$

Answer 1

The correct option is C $\frac{245}{2}m$

The horizontal component of velocity $(v\cos {45}^{\circ }=\frac{49}{\sqrt{2}}m/s)$ of both particles is same and in opposite directions.
If we look at the projectile motion equation,
The maximum height that can be attended by each particle is
$H=\frac{u^2{\sin }^2{45}^{\circ }}{g}=61.25m$
Also, the horizontal range is given by
$R=\frac{u^2\sin 2\times {45}^{\circ }}{g}=245m$
Thus, both the particles will collide in the middle and at the highest point. Hence, will have only a horizontal component of velocity.
Using conservation of momentum,
$m_P{v}_P+m_Q{v}_Q=m_P{u}_P+m_Q{u}_Q$
Since P retraces path $finalvelocity=-initialvelocity$
$v_P=-u_P=-\frac{49}{\sqrt{2}}m/s$
$-20\times \frac{49}{\sqrt{2}}+40\times v_Q=20\times \frac{49}{\sqrt{2}}-40\times \frac{49}{\sqrt{2}}$
$v_Q=0$
Thus, particle Q comes to rest after collision and since the collision is in the middle of th epath the particle vertically drops down. Hence position is $\frac{245}{2}m$.