1.9 Collision of projectiles  (Page 2/4)

Problem : Two projectiles “A” and “B” are projected simultaneously as shown in the figure. If they collide after 0.5 s, then determine (i) angle of projection “θ” and (ii) the distance “s”.

Relative motion

Solution : We see here that projectile “A” is approaching towards projectile “B” in horizontal direction. Their movement in two component directions should be synchronized so that they are at the same position at a particular given time. There is no separation in vertical direction at the start of motion. As such, relative velocity in y-direction should be zero for collision to occur.

Relative motion

$$v_{ABy}=u_{Ay}-u_{By}=0$$

$$\Rightarrow 20\sqrt{2}\sin \theta =20$$

$$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}=\sin {45}^0$$

$$\Rightarrow \theta ={45}^0$$

In the x-direction, the relative velocity is :

$$v_{ABx}=u_{Ax}-u_{Bx}=20\sqrt{2}\cos {45}^0-0=20m/s$$

The distance “s” covered with the relative velocity in 0.5 second is :

$$s=v_{ABx}Xt=20X0.5=10m$$

Problem : Two projectiles “A” and “B” are thrown simultaneously in opposite directions as shown in the figure. If they happen to collide in the mid air, then find the time when collision takes place.

Two projectiles

Solution : There is no separation in vertical direction at the start of motion. As such, relative velocity in y-direction should be zero for collision to occur.

$$v_{ABy}=u_{Ay}-u_{By}=0$$

$$\Rightarrow u_A\sin {\theta }_A=u_B\sin {\theta }_B$$

Putting values,

$$\Rightarrow \sin {\theta }_B=\frac{60}{50}X\sin {30}^0$$

$$\Rightarrow \sin {\theta }_B=\frac{3}{5}$$

The projectiles move towards each other with the relative velocity in horizontal direction. The relative velocity in x-direction is :

$$v_{ABx}=u_{Ax}-u_{Bx}=60\cos {30}^0+50\cos {\theta }_B$$

In order to find relative velocity in x-direction, we need to know “ $\cos {\theta }_B$ ”. Using trigonometric relation, we have :

$$\cos {\theta }_B=\sqrt{\left(1-\sin {}^2{\theta }_B\right)}=\sqrt{\{1-{\left(\frac{3}{5}\right)}^2\}}=\frac{4}{5}$$

Hence,

$$\Rightarrow v_{ABx}=60X\frac{\sqrt{3}}{2}+50X\frac{4}{5}=30\sqrt{3}+40=91.98\approx 92m/s$$

We should now understand that projectiles move towards each other with a relative velocity of 92 m/s. We can interpret this as if projectile “B” is stationary and projectile “A” moves towards it with a velocity 92 m/s, covering the initial separation between two particles for collision to take place. The time of collision, therefore, is :

$$\Rightarrow t=92/92=1s$$