JEE Physics Notes: Conservation of Linear Momentum

Physics Notes on Conservation of Linear Momentum

linear momentum of a particle

Linear momentum is a vector quantity because it equals the product of a scalar quantity m and a vector quantity v. Its direction is along v, it has dimensions ML/T, and its SI unit is kg.m/s . If a particle is moving in an arbitrary direction, p must have three components, and Equation 9.1 is equivalent to the component equations Linear momentum component equations

As you can see from its definition, the concept of momentum provides a quantitative distinction between heavy and light particles moving at the same velocity. For example, the momentum of a bowling ball moving at 10 m/s is much greater than that of a tennis ball moving at the same speed. Newton called the product mv quantity of motion; this is perhaps a more graphic description than our present-day word momentum, which comes from the Latin word for movement.
Using Newton’s second law of motion, we can relate the linear momentum of a particle to the resultant force acting on the particle: The time rate of change of the linear momentum of a particle is equal to the net force acting on the particletime rate of change of the linear momentum

Conservation of Momentum for a Two-Particle System
Consider two particles 1 and 2 that can interact with each other but are isolated from their surroundings (Fig. 9.1). That is, the particles may exert a force on each other, but no external forces are present. It is important to note the impact of Newton’s third law on this analysis. If an internal force from particle 1 (for example, a gravitational force) acts on particle 2, then there must be a second internal force—equal in magnitude but opposite in direction—that particle 2 exerts on particle 1.

Momentum for a Two-Particle System

Figure 9.1 

At some instant, the momentum of particle 1 is p1 = m1v1 and the momentum of particle 2 is p2 = m2v2 .

Note that F12 = -F21 . The total momentum of the system ptot is equal to the vector sum p1 + p2 .

Suppose that at some instant, the momentum of particle 1 is p1 and that of particle 2 is p2 . Applying Newton’s second law to each particle, we can write Momentum for a Two-Particle Systemwhere F21 is the force exerted by particle 2 on particle 1 and F12 is the force exerted by particle 1 on particle 2.

Newton’s third law tells us that F12 and F21 are equal in magnitude and opposite in direction.

That is, they form an action–reaction pair F12 = –F21 . We can express this condition as Momentum for a Two-Particle SystemBecause the time derivative of the total momentum ptot = p1 + p2 is zero, we conclude that the total momentum of the system must remain constant: Momentum for a Two-Particle Systemor, equivalently, Momentum for a Two-Particle Systemwhere p1i and p2i are the initial values and p1f and p2f the final values of the momentum during the time interval dt over which the reaction pair interacts. Equation 9.5 in component form demonstrates that the total momenta in the x, y, and z directions are all independently conserved: Momentum for a Two-Particle System

This result, known as the law of conservation of linear momentum, can be extended to any number of particles in an isolated system. We can state it as follows:
Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.

This law tells us that the total momentum of an isolated system at all times equals its initial momentum.
Notice that we have made no statement concerning the nature of the forces acting on the particles of the system. The only requirement is that the forces must be internal to the system.

Applications of Conservation of Momentum

Following few examples with illustrate the law of conservation of momentum.

(a) Recoil of Gun

A gun and a bullet constitute one isolated system. On firing the gun, bullet moves out with a very high velocity \vec{v}. The gun experiences a recoil. It moves in the opposite direction as shown in the below figure. Velocity ‘\vec{V}’ of the recoil gun can be calculated by the application of law of conservation of momentum.

Before Firing After Firing 
Momentum of bullet = 0

Momentum of gun = 0

Total momentum of the system = 0

Momentum of bullet = m\vec{v}

Momentum of gun = M\vec{V}

Total momentum of the system = M\vec{V}+m\vec{v}

Here ‘m’ and ‘M’ are the masses of bullet and gun respectively. According to the law of conservation of momentum, momentum before collision and after collision must be same.

M\vec{V}+m\vec{v} = 0

or, M\vec{V}=-m\vec{v}

or,

\vec{V}=-\frac{m\vec{v}}{M}

Negative sign indicates that direction of motion of gun is in opposite direction.

(b) Rocket and Jet Plane

Fuel and oxygen is burnt in the ignition chamber. As hot gases escape from a rear opening, with some momentum, the rocket moves in the forward direction with the same momentum.

(c)  Explosion of a Bomb

Momentum of a bomb before explosion is zero. After explosion different fragments fly in various directions. It will be observed that their momenta, when represented by the slide of a polygon, from a closed polygon, indicating that net momentum after explosion is also zero. Thus, if the bomb exploded into two fragments, they must move in opposite directions.

(d) A man Jumping from a Boat

When a man jumps from the boat to the shore, the boat is pushed backward. It can, exactly, be explained as in the case of recoil of gun.

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