Question

Particles P and Q of mass 20 g and 40 g respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q makes ${45}^{\circ }$ and ${135}^{\circ }$ angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49 m/s. The separation AB is 245 m. Both particle travel in the same vertical plane and undergo a collision. After the collision, P retraces its path. How much time after the collision does the particle Q take to reach the ground? Take $g=9.8m/s^2$

• A. 3.5 s
• B. 4.5 s
• C. 5.5 s
• D. 2.5 s

Answer 1

The correct option is A 3.5 s

Each particle has the same initial speed of 49 m/s. Horizontal velocity of each particle is same being
$49cos{45}^{\circ }=\frac{49}{\sqrt{2}}m/s$
Both the particles are projected simultaneously. Hence they undergo a collision in the middle of range AB. This will be at the highest point.

R = Range of each particle $=\frac{u^{2}sin2\theta }{g}$

$R=\frac{(49{)}^2\times sin{90}^{\circ }}{9.8}$; R = 245 m

$\because$ Linear momentum is conserved.

$20\left(ucos{45}^{\circ }\right)-40\left(ucos{45}^{\circ }\right)$

$=40v-20\left(ucos{45}^{\circ }\right)$

$\therefore 0=40v$ or v = 0 = horizontal component of velocity of particle Q.

Q thus comes to rest while P retraces its path. Hence Q will fall freely under gravity. Since the collision occurs midway between A and B, Q hits the ground just midway between A and B.

Time to fall will also be equal to time to reach maximum height:

$t=\frac{u\sin \theta }{g}=3.53s$