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# Collision of projectiles

Module by: Sunil Kumar Singh. E-mail the author

Collision between two projectiles is a rare eventuality. The precise requirement of collision is that the two projectiles are at the same position (having same x and y coordinates) at a given time instant. This is the condition for two point objects to collide or for a collision to occur. This is possible only rarely. Consider projections of two projectiles as shown in the figure.

We see that paths of two projectiles cross twice. Thus, there are two positions (or corresponding time instants) when it is possible that projectiles occupy same position and thus may collide with each other. But, there are infinite possibilities that they would not. Projectile “A” may rise to the required height but “B” may be at lower or higher position. Similarly, even if two projectiles are at same height they may be horizontally separated. To top these, the projectiles may have time difference at the start of their motion. However, if two projectiles collide then we can make lot many simplifying assumptions resulting from the requirement of collision that two projectiles are at the same position at the same time. In this module, we shall examine these simplifying aspects of projectile motion under collision.

## Analysis of motion

Two projectiles need to approach towards each other for collision to take place. If we look at this requirement in component form, then projectiles should approach towards each other both vertically and horizontally. It is possible that projectiles have different projection times. We can, however, extend the analysis even when projectiles are projected at different times by accounting motion for the additional time available to one of projectiles. For the sake of simplicity however, we consider that two projectiles are initiated at the same time instant.

The most important aspect of analysis of projectile motion involving collision is that we can interpret condition of collision in terms of relative velocity. We actually use the fact that components of relative velocity in either x or y direction is uniform motion - not accelerated one. This simplifies the analysis a great deal.

### Relative motion in x-direction

If “ x 0 x 0 ” is the initial separation between projectiles, then for collision they should cover this separation with x-component of relative velocity. Since two projectiles are initiated at the same time instant, the time when collision occurs is given by :

t = x 0 v A B x t = x 0 v A B x

where v A B x v A B x is the relative speed of approach of A with respect B. Note that time expression evaluates to same value whether we compute it with v A B x v A B x or v B A x v B A x . On the other hand, if there is no initial separation in x – direction, the projectiles should cover same horizontal distance for all time intervals. It is so because projectiles have to reach same horizontal i..e x-position at the point of collision in two dimensional space. This means that relative velocity of projectiles in x-direction is zero for the condition of collision. Mathematically,

v A B x = v A x v B x = 0 v A B x = v A x v B x = 0

### Relative motion in y-direction

If “ y 0 y 0 ” is the initial separation between projectiles, then for collision they should cover this separation with y-component of relative velocity. Since two projectiles are initiated at the same time instant, the time when collision occurs is given by :

t = y 0 v A B y t = y 0 v A B y

where v A B y v A B y is the relative speed of approach of A with respect B. Note that time expression evaluates to same value whether we compute it with v A B y v A B y or v B A y v B A y . On the other hand, if there is no initial separation in y – direction, the projectiles should cover same vertical distance for all time intervals. It is so because projectiles have to reach same vertical i..e y-position at the point of collision in two dimensional space. This means that relative velocity of projectiles in y-direction is zero for the condition of collision. Mathematically,

v A B y = v A y v B y = 0 v A B y = v A y v B y = 0

## Collision of projectiles initiated without vertical separation

In this case, there is no initial vertical separation. Such is the case, when projectiles are projected from same horizontal level. Both projectiles should rise to same height for all time. Clearly, relative velocity in vertical i.e y-direction is zero :

v A B y = v A y v B y = 0 v A B y = v A y v B y = 0

On the other hand, time of collision is obtained by considering relative motion in x-direction :

t = x 0 v A B x t = x 0 v A B x

There are different cases for projection from the same horizontal level. Some important cases are : (i) one projectile is projected at certain angle to the horizontal while the other projectile is projected vertically and (ii) Both projectiles are projected at certain angles to the horizontal. Here, we shall work out examples for each of these cases.

### Example 1

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”.

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.

v A B y = u A y u B y = 0 v A B y = u A y u B y = 0

20 2 sin θ = 20 20 2 sin θ = 20

sin θ = 1 2 = sin 45 0 sin θ = 1 2 = sin 45 0

θ = 45 0 θ = 45 0

In the x-direction, the relative velocity is :

v A B x = u A x u B x = 20 2 cos 45 0 - 0 = 20 m / s v A B x = u A x u B x = 20 2 cos 45 0 - 0 = 20 m / s

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

s = v A B x X t = 20 X 0.5 = 10 m s = v A B x X t = 20 X 0.5 = 10 m

### Example 2

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.

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 A B y = u A y u B y = 0 v A B y = u A y u B y = 0

u A sin θ A = u B sin θ B u A sin θ A = u B sin θ B

Putting values,

sin θ B = 60 50 X sin 30 0 sin θ B = 60 50 X sin 30 0

sin θ B = 3 5 sin θ B = 3 5

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

v A B x = u A x u B x = 60 cos 30 0 + 50 cos θ B v A B x = u A x u B x = 60 cos 30 0 + 50 cos θ B

In order to find relative velocity in x-direction, we need to know “ cos θ B cos θ B ”. Using trigonometric relation, we have :

cos θ B = 1 sin 2 θ B = { 1 3 5 2 } = 4 5 cos θ B = 1 sin 2 θ B = { 1 3 5 2 } = 4 5

Hence,

v A B x = 60 X 3 2 + 50 X 4 5 = 30 3 + 40 = 91.98 92 m / s v A B x = 60 X 3 2 + 50 X 4 5 = 30 3 + 40 = 91.98 92 m / 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 :

t = 92 / 92 = 1 s t = 92 / 92 = 1 s

## Collision of projectiles initiated without horizontal separation

In this case, there is no initial horizontal separation. Projectiles are thrown from different levels. Both projectiles travel same horizontal distnace for all time. Clearly, relative velocity in horizontal i.e x-direction is zero :

v A B x = v A x v B x = 0 v A B x = v A x v B x = 0

On the other hand, time of collision is obtained by considering relative motion in y-direction :

t = y 0 v A B x t = y 0 v A B x

### Example 3

Problem : A fighter plane is flying horizontally at a speed of 360 km/hr and is exactly above an aircraft gun at a given moment. At that instant, the aircraft gun is fired to hit the plane, which is at a vertical height of 1 km. If the speed of the shell is 720 km/hr, then at what angle should the gun be aimed to hit the plane (Neglect resistance due to air)?

Solution : This is a case of collision between fighter plane and shell fired from the gun. Since plane is overhead, it is required that horizontal component of velocity of the shell be equal to that of the plane as it is flying in horizontal direction. This will ensure that shell will hit the plane whenever it rises to the height of the plane.

Now the horizontal and vertical components of shell velocity are :

u x = 720 cos θ u x = 720 cos θ

u y = 720 sin θ u y = 720 sin θ

For collision,

u x = 720 cos θ = 360 u x = 720 cos θ = 360

cos θ = 1 2 = cos 60 0 cos θ = 1 2 = cos 60 0

θ = 60 0 θ = 60 0

However, we need to check that shell is capable to rise to the height of the plane. Here,

720 k m / h r = 720 X 5 18 = 200 m / s 720 k m / h r = 720 X 5 18 = 200 m / s

The maximum height achieved by the shell is obtained as here :

H = u 2 sin 2 θ 2 g = 200 2 3 2 2 2 X 10 = 1500 m H = u 2 sin 2 θ 2 g = 200 2 3 2 2 2 X 10 = 1500 m

The shell indeed rises to the height of the plane (1000 m) and hence will hit it.

### Example 4

Problem : Two projectiles are projected simultaneously from a point on the ground “O” and an elevated position “A” respectively as shown in the figure. If collision occurs at the point of return of two projectiles on the horizontal surface, then find the height of “A” above the ground and the angle at which the projectile "O" at the ground should be projected.

Solution : There is no initial separation between two projectiles in x-direction. For collision to occur, the relative motion in x-direction should be zero. In other words, the component velocities in x-direction should be equal so that two projectiles cover equal horizontal distance at any given time. Hence,

u O x = u A x u O x = u A x

u O cos θ = u A u O cos θ = u A

cos θ = u A u O = 5 10 = 1 2 = cos 60 0 cos θ = u A u O = 5 10 = 1 2 = cos 60 0

θ = 60 0 θ = 60 0

We should ensure that collision does occur at the point of return. It means that by the time projectiles travel horizontal distances required, they should also cover vertical distances so that both projectiles are at “C” at the same time. In the nutshell, their times of flight should be equal. For projectile from "O",

T = 2 u O sin θ g T = 2 u O sin θ g

For projectile from "A",

T = 2 H g T = 2 H g

For projectile from "A",

T = 2 u O sin θ g = 2 H g T = 2 u O sin θ g = 2 H g

Squaring both sides and putting values,

H = 4 u O 2 sin 2 θ 2 g H = 4 u O 2 sin 2 θ 2 g

H = 4 X 10 2 sin 2 60 0 2 X 10 H = 4 X 10 2 sin 2 60 0 2 X 10

H = 20 3 2 2 = 15 m H = 20 3 2 2 = 15 m

We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding at the end of their flights and hence their times of flight should be equal. We can, however, proceed to analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles in the vertical direction is “H”. This separation is covered with the component of relative velocity in vertical direction.

v O A y = u O y u A y = u O sin 60 0 0 = 10 X 3 2 = 5 3 m / s v O A y = u O y u A y = u O sin 60 0 0 = 10 X 3 2 = 5 3 m / s

Now, time of flight of projectile from ground is :

T = 2 u O sin θ g = 2 x 10 x sin 60 0 10 = 3 T = 2 u O sin θ g = 2 x 10 x sin 60 0 10 = 3

Hence, the vertical displacement of projectile from "A" before collision is :

H = v O A y X T = 5 3 x 3 = 15 m / s H = v O A y X T = 5 3 x 3 = 15 m / s

## Collision of projectiles initiated from different horizontal and vertical levels

In this case, there are finite initial horizontal and vertical separations. For collision to occur, projectiles need to cover these separations simultaneously. Clearly, component relative velocity in x and y directions are finitie. We can obtain time of collision by consideration of relative motion in either direction,

t = x 0 v A B x t = x 0 v A B x

or

t = y 0 v A B x t = y 0 v A B x

### Example 5

Problem : Two projectiles are projected simultaneously from two towers as shown in the figure. If the projectiles collide in the air, then find the distance “s” between the towers.

Solution : We see here that projectiles are approaching both horizontally and vertically. Their movement in two component directions should be synchronized so that they are at the same position at a particular given time. For collision, the necessary requirement is that relative velocity and displacement should be in the same direction.

It is given that collision does occur. It means that two projectiles should cover the displacement with relative velocity in each of the component directions.

In x-direction,

v A B x = u A x u B x = 10 2 cos 45 0 10 = 10 2 x 1 2 + 10 = 20 m / s v A B x = u A x u B x = 10 2 cos 45 0 10 = 10 2 x 1 2 + 10 = 20 m / s

If “t” is time after which collision occurs, then

s = v A B x X t = 20 t s = v A B x X t = 20 t

Clearly, we need to know “t” to find “s”. The component of relative velocity in y-direction is :

v A B y = u A y u B y v A B y = u A y u B y

v A B y = u sin 45 0 0 = 10 2 X 1 2 = 10 m / s v A B y = u sin 45 0 0 = 10 2 X 1 2 = 10 m / s

The initial vertical distance between points of projection is 30-10 = 20 m. This vertical distance is covered with component of relative velocity in vertical direction. Hence, time taken to collide, “t”, is :

t = 20 10 = 2 t = 20 10 = 2

Putting this value in the earlier equation for “s”, we have :

s = 20 t = 20 x 2 = 40 m s = 20 t = 20 x 2 = 40 m

## Exercises

### Exercise 1

Two projectiles are projected simultaneously from two towers as shown in the figure. If collision takes place in the air, then what should be the ratio "x/y" :

a 1 2 b 1 c 2 d 3 a 1 2 b 1 c 2 d 3

### Exercise 2

Two balls are projected simultaneously with speeds " u 1 u 1 ” and “" u 2 u 2 ” from two points "O" and "A" respectively as shown in the figure. If the balls collide, then find the ratio " u 1 u 2 u 1 u 2 ” (consider g = 10 m / s 2 m / s 2 ) :

a 2 3 b 3 2 c 1 / 3 d 3 a 2 3 b 3 2 c 1 / 3 d 3

### Exercise 3

Two projectiles “A” and “B” are projected simultaneously towards each other as shown in the figure. Determine if they collide.

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