Newton's Third Law of Motion, Law of Conservation of Linear Momentum, and Applications

Newton's Third Law of Motion

Newton's third law of motion is a fundamental principle that describes the nature of forces between interacting objects. It states that for every action, there is an equal and opposite reaction. This means that forces always occur in pairs. If object A exerts a force on object B, then object B simultaneously exerts a force of equal magnitude and opposite direction on object A.

Let's break this down. Imagine two bodies, Body 1 and Body 2. If Body 1 exerts a force on Body 2, denoted as F₁₂, then Body 2 simultaneously exerts a force on Body 1, denoted as F₂₁. According to Newton's third law:

F₁₂ = -F₂₁

The negative sign indicates that the forces are in opposite directions. It's crucial to understand that these two forces act on *different* bodies. The action force acts on one body, and the reaction force acts on the other. They are equal in magnitude, opposite in direction, and act simultaneously.

Example 1: Walking When you walk, your foot pushes backward on the ground (action). The ground, in turn, pushes forward on your foot with an equal and opposite force (reaction), propelling you forward.

Example 2: Rocket Propulsion A rocket expels hot gases downwards at high speed (action). These gases exert an equal and opposite upward force on the rocket (reaction), which is what lifts the rocket into space.

Example 3: A book on a table A book placed on a table exerts a downward force (due to gravity) on the table. The table exerts an equal and upward force (normal force) on the book. These are action-reaction forces. Note that the gravitational force exerted by the Earth on the book (weight) is an action force, and the reaction force is the gravitational force exerted by the book on the Earth. The normal force from the table on the book is the reaction to the force the book exerts on the table. It's important to distinguish between these pairs.

Example 4: Swimming A swimmer pushes water backward (action). The water pushes the swimmer forward (reaction).

Newton's third law helps us understand interactions involving contact forces (like pushing, pulling, friction, normal force) and non-contact forces (like gravitational force, electrostatic force).

Law of Conservation of Linear Momentum

The law of conservation of linear momentum is a direct consequence of Newton's laws of motion, particularly the third law. Momentum (denoted by p) is defined as the product of an object's mass (m) and its velocity (v):

p = mv

Momentum is a vector quantity, meaning it has both magnitude and direction.

The law of conservation of linear momentum states that if the net external force acting on a system is zero, then the total linear momentum of the system remains constant. In simpler terms, in the absence of external forces, the total momentum before an event (like a collision or explosion) is equal to the total momentum after the event.

Consider a system of two interacting particles, Particle 1 and Particle 2, with masses m₁ and m₂, and initial velocities u₁ and u₂, respectively. If they interact, their velocities change to v₁ and v₂.

Initial total momentum of the system, P_initial = m₁u₁ + m₂u₂

Final total momentum of the system, P_final = m₁v₁ + m₂v₂

If there are no external forces acting on the system, then according to the law of conservation of linear momentum:

P_initial = P_final

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

This law holds true for any system of particles, whether it's a collision between two billiard balls, an explosion of a bomb, or the recoil of a gun.

Let's see how this relates to Newton's third law. Consider two bodies interacting. Let F₁₂ be the force exerted by body 1 on body 2, and F₂₁ be the force exerted by body 2 on body 1. By Newton's third law, F₁₂ = -F₂₁.

From Newton's second law, force is the rate of change of momentum: F = dp/dt. So, F₁₂ = dp₂/dt and F₂₁ = dp₁/dt, where p₁ and p₂ are the momenta of body 1 and body 2, respectively.

Substituting these into Newton's third law:

dp₂/dt = - dp₁/dt

dp₂/dt + dp₁/dt = 0

d(p₁ + p₂)/dt = 0

This means that the rate of change of the total momentum (p₁ + p₂) of the system is zero. If the rate of change of momentum is zero, then the momentum itself is constant. This proves that the total momentum of the system is conserved, provided that F₁₂ and F₂₁ are the *only* forces acting (i.e., no external forces).

Applications of Newton's Third Law and Conservation of Linear Momentum

These principles have wide-ranging applications in physics and engineering.

1. Recoil of a Gun

When a bullet is fired from a gun, the gun and the bullet form a system. Initially, both are at rest, so the total momentum is zero. When the gun is fired, the bullet moves forward with a certain momentum. To conserve the total momentum (which must remain zero), the gun recoils backward with an equal and opposite momentum.

Let m_g be the mass of the gun, v_g be its recoil velocity, m_b be the mass of the bullet, and v_b be its velocity.

Initial momentum = 0

Final momentum = m_gv_g + m_bv_b

By conservation of momentum:

m_gv_g + m_bv_b = 0

m_gv_g = - m_bv_b

The negative sign indicates that the recoil velocity of the gun is in the opposite direction to the velocity of the bullet. Since the mass of the gun is much larger than the mass of the bullet, the recoil velocity of the gun is much smaller than the velocity of the bullet.

2. Explosion of a Bomb

Consider a bomb at rest. Its initial momentum is zero. When it explodes, it breaks into several fragments moving in various directions. The vector sum of the momenta of all these fragments must be zero to conserve the total momentum of the system.

If a bomb of mass M at rest explodes into two fragments of masses m₁ and m₂ (M = m₁ + m₂), with velocities v₁ and v₂, respectively.

Initial momentum = 0

Final momentum = m₁v₁ + m₂v₂

Conservation of momentum:

m₁v₁ + m₂v₂ = 0

m₁v₁ = - m₂v₂

This means the two fragments move in exactly opposite directions with momenta of equal magnitude.

3. Rocket Propulsion (Revisited)

A rocket expels mass (hot gases) downwards. The rocket and the expelled gases form a system. Initially, the system is at rest (or moving with some velocity). When the gases are expelled downwards with a certain velocity, they gain downward momentum. To conserve momentum, the rocket gains an equal amount of upward momentum.

The force exerted by the rocket on the gases is the action, and the force exerted by the gases on the rocket is the reaction, which provides the thrust. This thrust is what accelerates the rocket upwards.

Let M be the mass of the rocket and v its velocity. Let dm be the mass of fuel expelled in time dt with velocity v_e relative to the rocket. The velocity of the expelled gas relative to the ground is (v - v_e).

The momentum of the rocket at time t is Mv. At time t+dt, the mass of the rocket is M-dm, and its velocity is v+dv. Its momentum is (M-dm)(v+dv). The momentum of the expelled gas is dm(v - v_e).

Total momentum at t+dt = (M-dm)(v+dv) + dm(v - v_e)

By conservation of momentum (assuming no external forces):

Mv = (M-dm)(v+dv) + dm(v - v_e)

Mv = Mv + M dv - dm v - dm dv + dm v - dm v_e

Neglecting the product of infinitesimals (dm dv):

0 = M dv - dm v_e

M dv = dm v_e

If we consider the change in momentum of the expelled mass, let dm be positive. The momentum change of the rocket is M dv. The momentum of the expelled gas relative to the ground is dm * (v - v_e). The change in momentum of the system is the sum of the change in momentum of the rocket and the momentum of the expelled gases: d(Mv) + dm(v - v_e) = 0 (if we consider dm as the mass *added* to the exhaust, so it's removed from the rocket). Let's use a simpler approach: The thrust on the rocket is due to the reaction force from the expelled gases. The rate of change of momentum of the expelled gases is (dm/dt) * v_e (relative to the rocket). This force acts on the rocket.

The thrust force (F) on the rocket is given by:

F = - (dm/dt) v_e

Here, dm/dt is the rate at which mass is expelled. The negative sign indicates that if v_e is downwards, the force F is upwards. This force causes the rocket to accelerate according to Newton's second law: F = Ma, where M is the instantaneous mass of the rocket.

4. Collision of Objects

When two objects collide, they exert forces on each other (action-reaction). These forces are internal to the system formed by the two objects. If we neglect external forces like friction or air resistance during the brief moment of collision, the total momentum of the system is conserved.

Example: Billiard Balls When a cue ball strikes another billiard ball, the force the cue ball exerts on the second ball (action) is equal and opposite to the force the second ball exerts on the cue ball (reaction). The total momentum of the two balls before the collision is equal to their total momentum after the collision.

Example: Car Crash In a car crash, the two cars exert large forces on each other. If we consider the system of two cars, their total momentum just before the crash is equal to their total momentum just after the crash, assuming external forces are negligible. This principle is used in accident reconstruction.

5. Jet Engine

A jet engine works on a similar principle to rockets. It takes in air, compresses it, mixes it with fuel, and ignites it. The resulting hot gases are expelled at high speed from the rear of the engine. This expulsion of gases (action) creates an equal and opposite force (reaction) that propels the aircraft forward.

6. Jumping from a Boat

If you jump forward from a stationary boat, you exert a force on the boat backward (action). The boat exerts an equal and opposite force on you, propelling you forward. Simultaneously, the boat moves backward. This is to conserve the total momentum of the system (you + boat). If you jump forward with momentum p, the boat moves backward with momentum -p.

Example: A person on ice If a person standing on frictionless ice throws a heavy ball, they will move backward. If they catch a ball thrown towards them, they will move slightly in the direction of the ball's motion (depending on their initial state).

7. Firing a Cannonball from a Moving Cannon

Consider a cannon on a moving railway carriage. If the cannon fires a ball forward, the cannon and carriage will recoil backward. The conservation of momentum applies to the entire system (cannon + carriage + ball).

Let M be the mass of the cannon and carriage, and V be their initial velocity. Let m be the mass of the cannonball and v its velocity relative to the cannon. Let the mass of the cannon be m_c and the mass of the carriage be m_car.

If the cannon fires the ball with velocity v relative to the cannon, and the cannon recoils with velocity u relative to the ground, then the velocity of the ball relative to the ground is u+v.

Initial momentum = (m_c + m_car)V

Final momentum = (m_c + m_car)u + mv

By conservation of momentum:

(m_c + m_car)V = (m_c + m_car)u + mv

This equation can be used to find the recoil velocity u if V, m, v, and masses are known.