Collision usually occurs in the absence of any external force or negligible external force. Two bodies collide because a body comes in the line of motion of other body. The forces during collision between colliding bodies are essentially internal forces. The internal force is a variable force of relatively high magnitude, which operates for a very small period.

Figure 2: The internal force is a variable force of relatively high magnitude, which operates for a very small period.

Force during collision

The figure here captures the attributes of the collision force. It is important to note that the plot has time in milli-seconds.

The collision forces between colliding objects appear in pair and follow Newton's third law. The force pairs are equal and opposite. This has important implications. Irrespective of the masses or sizes of the colliding bodies, the pair (two) forces are always equal at any time during the collision. What it means that force-time plot of either pair force is exact replica of force-time plot of the other collision force.

Figure 3: The force pairs are equal and opposite.

Forces during collision

As pointed out earlier, it is not very useful studying contact force as it is a variable force and secondly, we can not measure the same with any accuracy for the short interval it operates. We can, however, relate the collision force to the change in linear momentum. Here is a word of caution. If the collision forces are internal forces, then how can there be a change in linear momentum? We are actually referring change in linear momentum of one of the colliding bodies, which changes in such a manner so that linear momentum of the system of colliding bodies remain constant. Now, using Newton's second law of motion, the collision force operating on one of the colliding objects is :

F ( t ) = ⅆ p ⅆ t F ( t ) = ⅆ p ⅆ t

⇒ ⅆ p = F ( t ) ⅆ t ⇒ ⅆ p = F ( t ) ⅆ t

Integrating for the period of collision,

⇒ ∫ ⅆ p = ∫ F ( t ) ⅆ t ⇒ ∫ ⅆ p = ∫ F ( t ) ⅆ t

⇒ Δ p = ∫ F ( t ) ⅆ t ⇒ Δ p = ∫ F ( t ) ⅆ t

This is an important result. This enables us to measure the product of two immeasurable quantities (i) variable force of relatively high magnitude, F(t) and (ii) very small time, dt, during the collision in terms of measurable change in momentum, Δp, before and after the collision. The integral of the product of force and time equals change in momentum and is known as impulse denoted by the symbol "J" :

Like linear momentum, impulse is a vector quantity. Evidently, it has the same dimension and unit as that of linear momentum. It may be noted here that impulse is a measure of the product of force and time of operation, but not either of them individually. Further, it must also be understood that impulse is rarely evaluated using its defining integral; rather it is evaluated as the change in linear momentum - most of the time.

We shall learn that a collision may be elastic (where no loss of kinetic energy of the system takes place) or inelastic (where certain loss of kinetic energy of the system is involved).

Example 1

Problem : A ball of 0.1 kg in horizontal flight moves with a speed 50 m/s and is hit by a swinging bat in the opposite direction. As a result, the ball reverses its direction and moves with the same speed as before. Find the impulse that acts on the ball during its contact with the bat.

Solution : We calculate the impulse on the ball by measuring the change in linear momentum. Here, vectors involved are one dimensional. Therefore, we can us scalar representation of quantities with appropriate sign. Let us consider that the ball is moving in positive x-direction. Then,

Figure 4: The ball reverses its direction with same speed when hit by the bat.

Collision

J = Δ p = p f - p i = m v f - m v i J = Δ p = p f - p i = m v f - m v i

⇒ J = m ( v f - v i ) = 0.1 x ( - 50 - 50 ) = - 10 kg x m s ⇒ J = m ( v f - v i ) = 0.1 x ( - 50 - 50 ) = - 10 kg x m s

Negative sign means that the impulse is acting in the "-x" direction, in which the ball finally moves.

Component form of impulse

Like linear momentum, impulse is a vector quantity. Impulse vector on a body is directed in the direction of Δp. This direction may not be the direction of either initial or final linear momentum of the body under investigation. Like stated before in the course, the direction of a vector and that of "change in vector" may not be same. The component of impulse vector along any of the coordinate direction is equal to the difference of the components of linear momentum in that direction. Mathematically,

J x = Δ p x = p fx - p ix J y = Δ p y = p fy - p iy J z = Δ p z = p fz - p iz J x = Δ p x = p fx - p ix J y = Δ p y = p fy - p iy J z = Δ p z = p fz - p iz

(2)

Example 2

Problem : A ball of 0.1 kg in horizontal flight moves with a speed 50 m/s and is hit by a swinging bat as shown in the figure at an angle with horizontal. As a result, the ball is reflected in a direction, making 45° with the horizontal. Find the impulse that acts on the ball during its contact with the bat.

Figure 5: The ball is deflected at an angle with horizontal.

Collision

Solution : We calculate the impulse on the ball by measuring the change in linear momentum in horizontal (x-direction) and vertical (y-direction) directions. Let us consider that the ball is initially moving in positive x-direction. Then,

J x = Δ p x = p fx - p ix = m ( v fx - v ix ) ⇒ J x = 0.1 x ( - 50 cos 45 0 - 50 ) = - 8.54 kg x m s J x = Δ p x = p fx - p ix = m ( v fx - v ix ) ⇒ J x = 0.1 x ( - 50 cos 45 0 - 50 ) = - 8.54 kg x m s

Similarly, component of impulse in y-direction is :

J y = Δ p y = p fy - p iy = m ( v fy - v iy ) ⇒ J y = 0.1 x ( 50 sin 45 0 - 0 ) = - 3.54 kg x m s J y = Δ p y = p fy - p iy = m ( v fy - v iy ) ⇒ J y = 0.1 x ( 50 sin 45 0 - 0 ) = - 3.54 kg x m s

The impulse on the ball is :

J = J x i + J y j ⇒ J = ( - 8.54 i + 3.54 j ) kg x m s J = J x i + J y j ⇒ J = ( - 8.54 i + 3.54 j ) kg x m s

This example and the one given earlier illustrate that the impulse and its evaluation in terms of linear momentum is independent of whether the collision is head on or not. No doubt the result is different in two cases. The impulse force is greater for head on collision than when the colliding object glances the target at an angle. It is expected also. The change in linear momentum when the striking object reverses it motion is maximum and as such impulse imparted is also maximum when collision is head on.

We may not be able to relate impulse with the instantaneous force during the collision, but we can estimate the average force if we have accurate estimate of the time involved during collision. For the sake of simplicity, we consider one-dimensional contact force. Looking at the expression of impulse i.e the integral of force over the time of collision is graphically equal to the area under the curve.

Figure 6: The integral of force over the time of collision is graphically equal to the area under the curve.

Force - time plot during collision

J = ∫ F ( t ) ⅆ t = Area J = ∫ F ( t ) ⅆ t = Area

It emerges, therefore, that average force is equal to the ordinate of the rectangle of area equal to that of area under the actual force - time curve as shown here.

Figure 7: Average force is equal to the ordinate of the rectangle of area equal to that of area under the force - time actual curve.

Average collision force

Mathematically,

Example 3

Problem : A ball of 0.1 kg in horizontal flight moves with a speed 50 m/s and is hit by a swinging bat in the opposite direction. As a result, the ball reverses its direction and moves with the same speed as before. If the time of contact is 1 milli-second, then find the average force that acts on the ball during its contact with the bat.

Solution : We calculate the impulse on the ball by measuring the change in its linear momentum. Let us consider that the ball is moving in positive x-direction. Then,

Figure 8: The ball reverses its direction with same speed when hit by the bat.

Collision

J = Δ p = p f - p i = m v f - m v i J = Δ p = p f - p i = m v f - m v i

⇒ J = m ( v f - m v i ) = 0.1 x ( - 50 - 50 ) = - 10 kg x m s ⇒ J = m ( v f - m v i ) = 0.1 x ( - 50 - 50 ) = - 10 kg x m s

The magnitude of average force is :

F avg = 10 1 x 10 - 3 = 10000 N F avg = 10 1 x 10 - 3 = 10000 N

A collision may take place when external force is operating like when a cricket ball strikes the racket in the presence of gravitational force. However, gravitational force is relatively small (mg = 0.1 x 10 = 1.5 N). This force is negligible in comparison to the collision force of 10000 N! It is, therefore, realistic to assume that collision takes place in the absence of external force.

This concept of average force allows us to estimate the average force on a target, which is bombarded with successive colliding masses like a stream of bullets or balls as shown in the figure below. Here, the average force is given as :

Figure 9: The ball hits the target in quick succession.

Series Collision

F avg = J Δ t = ∑ Δ p Δ t F avg = J Δ t = ∑ Δ p Δ t

If the projectiles have equal mass "m" and there are "n" numbers of the projectiles hitting the target, then

⇒ ∑ Δ p = n m Δ v ⇒ ∑ Δ p = n m Δ v

and

There are two types of collision - elastic and inelastic. In the elastic collision, the colliding bodies is restored to the normal shape and size when the bodies separate. The deformation of colliding bodies is temporary and is limited to the period of collision. This means that kinetic energy of the system is simply stored as elastic energy during collision and is released to the system when bodies regain their normal condition. On the other hand, there is loss of kinetic energy during inelastic collision as kinetic energy is converted to some other form of energy like sound energy or heat energy, which can not be released to the system as kinetic energy subsequent to the collision.

In reality most of the macroscopic collisions involving physical contact are inelastic as kinetic energy of the system is irrevocably converted to other form of energy. For example, when a tennis ball is released from a height, it does not bounce to the same height on the return journey. However, if the ground is plane and hard, then the ball may transverse the greater part of the height and we may say that the collision is approximately elastic.

There is an interesting set up, which explains various aspects of elastic collision. We will study this set up, which comprises of two blocks and a spring to get an insight into the collision process. We imagine two identical blocks "A" and "B" moving on a smooth floor as shown in the figure below. The block "A" moves right with speed v A v A towards block "B". On the other hand block "B" moves with speed v B v B such that v A > v B v A > v B . We also imagine that a "mass0less" spring is attached to the block "B" as shown in the figure. Let us also consider that the spring follows Hook's law for elongation (F = -kx).

Figure 10: The spring is compressed.

Collision

At the time when block "A" hits the spring, the end of the spring in contact with block "A" acquires the speed of the block "A" ( v A v A ). The end of the spring in contact with block "B", however, moves with speed v B v B . As the block "A" moves with relative speed towards "B", the spring is compressed. The spring force for any intermediate compression "x" is given by :

F ( t ) = - k x F ( t ) = - k x

The spring forces act on both the blocks but in opposite directions. The spring force is equivalent to the internal collision force. The spring force pushes the block "A" towards left and hence decelerates it. Let its speed at a given instant after hitting the spring be v A ' v A ' . On the other hand, spring force pushes the block "B" towards right and hence accelerates it. As such, block "B" acquires certain speed, say v B ' v B ' . If v A ' > v B ' v A ' > v B ' , then the spring is further compressed. The speed of block "A" further decreases and that of block "B" further increases. The spring force during this period keeps increasing. This process continues till the speeds of blocks become equal. Let us consider that common speed of each of the blocks be V.This situation corresponds to the maximum compression and maximum spring force i.e the maximum collision force.

The maximum spring force continues to decelerate block "A" and accelerates block "B". As such, block "B" ( v B '' v B '' ) begins to move faster than the block "A" ( v A '' v A '' ) . It results in elongation of spring. The spring force during this period keeps decreasing. The spring force, however, continues to decelerates block "A" and accelerates block "B". This process continues till the spring attains its normal length. At this moment, the block "A" looses contact with the spring. The block "B" moves with greater speed ( v B ''' v B ''' ) than block "A" ( v A ''' v A ''' ). Therefore, the separation between two blocks keeps increasing with time as they move with different speeds.

Figure 11: The spring is elongated.

Collision

The spring force - time plot during the contact with spring approximates the force curve during a collision :

Figure 12: The force - time plot.

Collision force

Since the spring is mass-less and elastic (follows Hooke's law), the kinetic energy of the system before the collision is temporarily converted to elastic potential energy i.e spring energy during the compression of the spring. The stored elastic potential energy is then released as kinetic energy during elongation of the spring. At the end of elongation when spring attains its normal length, the kinetic energy of the system is restored as before. In the nutshell,

During collision, however, total energy has component of elastic potential energy as well and is equal to the kinetic energy of the system before and after the collsion :

E = K + U = K before = K after E = K + U = K before = K after

In the same fashion, we can visualize inelastic collision. In this case, some of the energy is converted to a form of energy, which can not be regained as kinetic energy like when a rubber ball hits a ground and is unable to regain the height. Here,

In some cases, the colliding object may get embedded into the other object. In that case, all the kinetic energy of the striking object is lost to the surrounding/ target. Such collision is completely in-elastic.

1: Collision of objects is a brief phenomenon of interaction with force of relatively high magnitude.

2: The integral of the product of force and time equals change in momentum and is known as impulse denoted by the symbol "J" :

J = Δ p = ∫ F ( t ) ⅆ t J = Δ p = ∫ F ( t ) ⅆ t

3: The component of impulse vector along any of the coordinate direction is equal to the difference of the components of linear momentum in that direction. Mathematically,

J x = Δ p x = p fx - p ix J y = Δ p y = p fy - p iy J z = Δ p z = p fz - p iz J x = Δ p x = p fx - p ix J y = Δ p y = p fy - p iy J z = Δ p z = p fz - p iz

4: The average force during collision is by the ratio of impulse and time interval as :

F avg = ∫ F ( t ) ⅆ t Δ t = J Δ t = Δ p Δ t F avg = ∫ F ( t ) ⅆ t Δ t = J Δ t = Δ p Δ t

5: Elastics collision :

P before = P after P before = P after

and

K before = K after K before = K after

6: Inelastic collision :

P before = P after P before = P after

and

K before > K after K before > K after