Impulse During An Impact Or Collision

Impulse During An Impact Or Collision

The impulsive force acting on the body produces a change in momentum of the body on which it acts. We know, Ft = mv - mu, therefore maximum force needed to produce a given impulse depends upon time. If time is short, the force required in a given  impulse or the change in momentum is large and vice - versa.

NEWTON’S THIRD LAW OF MOTION

When a force is applied to stop a moving body, we ourselves experience some force from the body being stopped. When a cricketer used his hands to stop a moving ball, his hands also experience some force from the ball and sometimes the force is unbearable. When we jump on a cemented road from some height, our feet get injured by the impact of the road.

From above examples we find that whenever one body exerts a force on another body, the second body exerts an equal and opposite force on the first body. The force exerted by the first body on the second body is called ‘action’ and the force exerted by the second body on the first body is called ‘reaction’

Statement:

The law states the “To every action there is an equal and opposite reaction”. Moreover, action and reaction act on different bodies.

Demonstration:

Two similar spring balances A and B joined by hook as shown in the figure, The other of the spring balance B is attached to a hook rigidly fixed in a rigid wall.

Demonstration - Newton’s third law of motion.

The other end of the spring balance A is pulled out to the left. Both the balances show the same reading (20 N) for the force.

The pulled balance A exerts a force of 20N on the balance B. It acts as action, B pulls the balance A in opposite direction with a force of 20 N. This force is known as reaction.

We conclude that action-reaction forces are equal and opposite and act on two different bodies.

Explanation :

If may be noted that action and reaction occur simultaneously. Action and reaction never act on same body. Had this been the case, there would have been no (accelerated) motion, Since action and reaction occur in pairs and act on two different bodies, it is impossible to have single isolated force.

Examples of Newton’s Third Law of Motion:

Some of our day-to-day observations can be explained in terms of Newton’s third law of motion as follows :

• Walking : To walk on the ground, we push the ground backwards with our foot. As a reaction, the ground pushes our foot forward with the same force. It is this forward reaction (force of the ground) that enables us to walk forward.

Walking becomes difficult when the ground is slippery or it is covered with snow or sand. This is because we can exert much smaller force in the form of backward action on the ground. The forward reaction of the ground will reduce accordingly.

• Swimming : While trying to swim, a swimmer pushes the water backwards with his hands and feet. This is the force of action. The water pushes the swimmer forward with the same force (of reaction).
• Recoiling of gun : When a bullet is fired from a gun, the gun recoils, i.e., the gun moves backwards through a small distance, giving jerk to the shoulder of the gunman. This is because on firing, the gun exerts some force on the bullet (i.e., action) in forward direction. In turn, the bullet exerts an equal force on the gun (i.e., reaction) in the backward direction. The distance moved by the gun is small because gun is much heavier than the bullet.

• Man and boat : When a man jumps from the boat to the river bank, the boat is pushed away from the bank. This happens due to equal forces of action and reaction.

• The flying of rockets and jet planes : In a rocket and jet plane the fuel burnt appears in the form of hot and highly compressed gases. These are made to escape from jet planes in the downward direction. As a reaction, the rocket moves upwards with the same force. The forces of action and reaction are shown in figure.

• The case of hose pipe : To put out fire, the firemen direct a powerful stream of water on fire, from a hosepipe. While doing so, the hosepipe is to be held strongly because of its tendency to move backwards. As the water rushes out at a great speed from the hose pipe in the forward direction (action), the hosepipe tends to move backwards, due to an equal force of reaction.

LAW OF CONSERVATION OF MOMENTUM:

According to this law, the total momentum of a system or a body remains constant if no net external force acts on the system. In other words, momentum is never created or destroyed. It has been deduced from Newton’s third law of motion. 

Suppose two balls A and B are moving in the same direction along a straight line with different velocities.

Let mA = mass of ball A

mB = mass of ball B

uA = initial velocity of A

uB = initial velocity of B

If uA > uB, the two balls collide with each other as shown in the given figure. Let this collision last for a short time t. During collision, suppose

FAB = force exerted by A on B, and

FBA = force exerted by B on A.

We shall assume that no other external unbalanced forces are acting on the balls.

After collision for t seconds, the balls move separately. Let

vA = velocity of A after collision,

vB = velocity of B after collision.

Change in momentum of A = momentum of A after collision – momentum of A before collision

 

Rate of change of momentum of

 

This is equal to force exerted by B on A, i.e., FBA (as per Newton’s second law of motion).

∴ …(i)

Similarly, rate of change of momentum of

This must be equal to force exerted by A on B, i.e., FAB.

∴ …(ii)

According to Newton’s third law of motion, the force FAB exerted by ball A on ball B (say, action) and the force FBA exerted by ball B on ball A, (the reaction) must be equal and opposite to each other. Therefore, 

Using equations (i) and (ii), we get

or

…(iii)

Now, total momentum of the two balls after collision, and 

total momentum of the two balls before collision.

Therefore, equation (iii) shows that total momentum of the two balls remains unchanged on collision, i.e., total linear momentum is conserved and is not affected by the mutual action and reaction of the balls. This is the law of conservation of linear momentum. 

Applications of the Law of Conservation of Momentum:

As the law of conservation of (linear) momentum has been deduced from Newton’s third law of motion, therefore, all the applications/examples of Newton’s third law of motion can be explained in terms of the law of conservation of momentum.

(a) For example, flight of jet planes and rockets can be understood in terms of the law of conservation of momentum. Before firing, the momentum of the rocket is zero. On firing, the burnt gases rush out through the nozzle in the downward direction. The rocket moves such that

momentum of rocket + momentum of escaping gases = 0.

∴ momentum of rocket = – momentum of escaping gases.

Negative sign shows that rocket moves upwards (in a direction opposite to the direction of escaping gases), with momentum equal to the momentum of the escaping gases.

(b) Similarly the case of the gun can also be understood

Let

m₁ = mass of bullet,

m₂ = mass of gun,

v₁ = velocity of bullet,

v₂ = velocity of gun,

Total momentum of bullet and gun on firing = m₁v₁ + m₂v₂

Before firing, both the gun and the bullet are at rest, therefore total momentum of bullet and gun = 0.

As no external forces are involved in the process, therefore, applying the law of conservation of momentum,

m₁v₁ + m₂v₂ = 0

            m₂v₂ = –m₁v₁

             

Negative sign in equation shows that v₂ is in a direction opposite to v₁, i.e., the gun recoils or moves backwards when the bullet moves forward. Further, as , therefore, i.e., as the gun is much heavier than the bullet, the recoil velocity of the gun is much smaller than the velocity of the bullet.

Question: A field gun a mass 1.5 t fires a shell of mass 15 kg with a velocity of 150 m/s. Calculate the velocity of the recoil of the gun.

Solution: Mass of gun = 1.5 t = 1.5 × 1000 kg = 1500 kg

Mass of shell = 15 kg

Velocity of shell = 150 m/s.

Velocity of recoil of the gun = ?

Momentum of gun = Mass of gun × velocity of recoil of the gun = 1500 V kg m.s

Momentum of shell = Mass of shell × velocity of shell = 15 × 10 kg m/s.

By the law of conservation of momentum :

Momentum of gun = Momentum of shell

1500 V = 15 × 150 or V =

The recoil velocity of gun = 1.5 m/sec.

Question2. A hunter of 45 kg is standing on ice fires a bullet on 100 gram with a velocity of 500 ms-1 by a gun of 5 kg. Find the recoil velocity of the hunter.

Solution: The initial momentum of the system, P₁ = Momentum of hunter + momentum of gun  + momentum of bullet

or P₁ + 45 × 0 + 5 × 0 + 0.1 × 0 = 0 ... (1)

Final momentum of the system, P₁ = Momentum of hunter + Momentum of gun + momentum of bullet 

P₂ = 45 V + 5 V + 0.1 × 500 (Here V is the recoil velocity of gun with hunter).

P₂ = 50 V + 50 ...(2)

By the conservation of momentum

P₁ = P₂

0 = 50 V + 50 or V = -1 m/s.

The recoil velocity of gun with hunter is 1 m/s.