Question

Suppose two particles, 1 and 2, are projected in vertical plane simultaneously.

Their angles of projection are ${30}^{\circ }$ and $\theta$ respectively with the horizontal. If they collide
after a time $t$ in air, then

• A. $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$ and they will have same speed just before the collision.
• B. $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$ and they will have different speed just before the collision.
• C. $x<1280\sqrt{3}-960\text{m}$
• D. it is possible that the particle collide when both of them are at their highest point.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$ and they will have different speed just before the collision.
C $x<1280\sqrt{3}-960\text{m}$
D it is possible that the particle collide when both of them are at their highest point.

For a collision to take place between two projectiles, two must have different speeds. If they have same speed the distance between the two won't change and hence collision won't happen.
Hence option A is incorrect.
From the frame of reference of $2^{nd}$ body, body 1 will appear to be travelling towards body 2 and hence should have same vertical component of velocity.
Therefore, $160sin{30}^{\circ }=100sin\theta$
$\Rightarrow sin\theta =\frac{160}{2\times 100}=\frac{4}{5}$
$\Rightarrow \theta =si{n}^{-1}\left(\frac{4}{5}\right)$
Since for collision between the two bodies, velocity between the two must be different. Hence option B is correct.
Time for collision between the two particles must be less than the time of flight of individual particle.
Time of flight for particle 1, $T=\frac{2usin\theta }{g}=\frac{2\times 160\times sin{30}^{\circ }}{10}=16s$
Velocity of particle 1 w.r.t 2, $\overrightarrow{V_{1/2}}={\overrightarrow{V}}_1-{\overrightarrow{V}}_2=160cos{30}^{\circ }-100cos\theta =160cos{30}^{\circ }-100\times \left(\frac{3}{5}\right)=80\sqrt{3}-60$
Time after which collision occur must be less than time of flight.
Therefore, $\frac{x}{\overrightarrow{V_{1/2}}}<T$
$\frac{x}{80\sqrt{3}-60}<16$
$x<1280\sqrt{3}-960m$
Hence option C is correct.
Since time to reach the highest point is less than time of flight and also the maximum height reached by both the particle is same since both have equal vertical component of velocity. Hence it is possible that both particles collide at their highest point.