5.9 Collision of projectiles  (Page 4/4)

According to question, collision occurs in the mid air. The necessary condition for collision is that direction of relative velocity and initial displacement between projectiles should be same.

In x-direction,

$$v_{ABx}=u_{Ax}-u_{Bx}=10\sqrt{2}\cos {45}^0-\left(-10\right)=10\sqrt{2}X\frac{1}{\sqrt{2}}+10=20m/s$$

In y-direction,

$$v_{ABy}=u_{Ay}-u_{By}=10\sqrt{2}\cos {45}^0-0=10\sqrt{2}X\frac{1}{\sqrt{2}}=10m/s$$

The angle that relative velocity makes with horizontal is :

$$\Rightarrow \tan \theta =\frac{v_{ABy}}{v_{ABx}}=\frac{10}{20}=\frac{1}{2}$$

This is also the slope of displacement AB,

$$\Rightarrow \tan \theta =\frac{1}{2}=\frac{y}{x}$$

$$\Rightarrow \frac{x}{y}=2$$

Hence, option (c) is correct.

It is given that the projectiles collide. Now the initial separation in vertical direction is zero. It follows then that component of relative velocity between two projectiles in vertical direction should be zero for collision to take place. This is possible if the vertical component of velocity of projectile form "O" is equal to the speed of projectile form "A" (Remember that acceleration due to gravity applies to both projectiles and relative acceleration in vertical direction is zero).

$$u_1\sin {60}^0=u_2$$

$$\Rightarrow \frac{u_1}{u_2}=\frac{1}{\sin {60}^0}=\frac{2}{\sqrt{3}}$$

Hence, option (a) is correct.

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

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

$$10\sin \theta =5\sqrt{2}X\sin {45}^0=5\sqrt{2}X\frac{1}{\sqrt{2}}=5$$

$$\sin \theta =\frac{1}{2}=\sin {30}^0$$

$$\theta ={30}^0$$

In the x-direction, the initial separation is 15 m and the relative velocity is :

$$v_{ABx}=u_{Ax}-u_{Bx}=10\cos {30}^0+5\sqrt{2}x\cos {45}^0$$

$$v_{ABx}=10X\sqrt{3}\frac{}{2}+5\sqrt{2}X\sqrt{2}=5\sqrt{3}+5=5\left(\sqrt{3}+1\right)$$

The time after which collision may occur is :

$$\Rightarrow t=\frac{20}{5\left(\sqrt{3}+1\right)}=\frac{4}{\left(\sqrt{3}+1\right)}=1.464s$$

We should, however, check whether projectiles stay that long in the air in the first place? Now, we have seen that the projectiles should have same vertical component of velocity for collision to occur. As time of flight is a function of vertical component of velocity, the time of the flight of the projectiles are equal. For projectile “A” :

$$T=\frac{2u_A\sin {30}^0}{g}=\frac{2X10X1}{10X2}=1s$$

This means that projectiles would not stay that long in the air even though the necessary conditions (but not sufficient conditions) for the collision are fulfilled. Clearly, collision does not occur in this case. We should clearly understand that "analysis of motion with respect to collision" and "requirements of collision" are not same. In this module, we have studied the "consequence of collision" - not the "conditions for collision".