Question

Particles $P$ and $Q$ of masses $20g$ of $40g$ respectively are projected from the positions $A$ and $B$ on the ground. The initial velocities of $P$ and $Q$ make angles of ${45}^{\circ }$ and ${135}^{\circ }$ respectively with the horizontal as shown in the figure. Each particle has an initial speed of $49m/s$. The separation $AB$ is $245m$. Both particles travel in the same vertical plane and undergo a collision. After the collision $P$ retraces its path. What is the distance of $Q$ (w.r.t its initial position) where it hits the ground?

Answer 1

Time $t$ which they collide is $t=\frac{245}{2\times 49\times \cos 45}=\frac{5}{\sqrt{2}}$ sec

Heught at time $t=\frac{5}{\sqrt{2}}$ sec

$h=49\sin 45t-\frac{1}{2}g{t}^2$

$=49\times \frac{1}{\sqrt{2}}\times \frac{5}{\sqrt{2}}-\frac{1}{2}\times 10\frac{25}{2}$

$=\frac{249}{2}-\frac{125}{2}=\frac{130}{2}=65$m

Applying collision $P$ retrace its path,it means horizontal velocity goes opposite.

$V_A=-49\cos 45=\frac{-49}{\sqrt{2}}$

By momentum conservation,

$=20\times 49\cos 45-40\times 49\cos 45=-20\times 49\cos 45+40v$

$=2\times 20\times \frac{49}{\sqrt{2}}-\frac{40\times 49}{\sqrt{2}}=40v$

$V=0$ it means $Q$ dropped after collision position of $Q$ before collision m in x-directon.

$x=49\cos 45\times t$

$=\frac{49}{\sqrt{2}}\times \frac{5}{\sqrt{2}}=\frac{245}{2}=122.5$m