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For every action, there is an equal and opposite reaction is Newton’s third law of motion. This tells us that all forces in nature acts in pairs. These actions and reactions help in understanding the motion of bodies on which forces act. The law also helps in resolving issues in several applications of forces, namely when two bodies collide. The momentum of the bodies before collision and after collision can be worked out using the third law.

If we consider bodies moving along a straight path, the momentum they possess is called linear momentum. If two spheres in a linear motion collide, their momentum before and after the collision can be related using Newton’s third law of motion.

Law of Conservation of Momentum is derived from Newton's third law of motion using the mathematical expression of force, which is derived from Newton’s second law of motion, enunciates that in the absence of external forces, if two bodies collide, the total momentum of the bodies before the collision and after the collision remains the same. This is the law of conservation of linear momentum. After collision if the two bodies stick together, their common speed or velocity can be calculated by using the law of conservation of linear momentum.

**Conservation of Momentum**

There are very few laws in physics that are known to be valid in all situations. The “law of conservation of momentum” is one such very important law for which no exception has been found so far. According to the law of conservation of momentum, when no external unbalanced force is acting, the sum of the momenta of a system of particles is constant. This law is applied for a collision between two bodies, According to the law of conservation of momentum the total momentum of the colliding bodies before collision is equal to the total momentum after collision.

We can apply this law for different situations. For example, consider a boy standing on a boat at rest. When he jumps from the boat on to the bank, the boat will no longer be at rest. The subsequent motion of the boat and its velocity can be explained using the law of conservation of momentum.

When a bullet is fired from a gun, the gun gets some velocity in the opposite direction. The velocity of the gun, commonly called recoil velocity, can be calculated by applying the law of conservation of momentum.

• When one object exerts a force on another object, the second object instantaneously exerts a force back on the first.

• These two forces are always equal in magnitude but opposite in direction.

• These forces act on different objects but never on the same object.

• Every action has an equal and opposite reaction.

There are several applications of Newton’s third law of motion; the launching of satellites is one among them.

**Derivation of law of conservation of momentum from Newton's third law:**

Law of conservation of momentum statesthat if no external force acts on a system of particles, the algebraic sum of the linear momenta of the particles remains conserved."

Consider two particles A and B, Which collide head on as shown in figure The particles move in a straight line before and after collision.

Let particle A have initial velocity u_{1} and particle B has initial velocity u_{2} . The two particles will collide, if u_{1} > u_{2} . Let after collision the final velocities of A and B becomes v_{1} and v_{2} respectively . The two particles will seperate after collision, if v_{2} > v_{1} . Let the two particles A and B have m_{1} and m_{2} respectively.

Initial momentum of particle A is m_{1}u_{1}

Initial momentum of particle B is m_{2}u_{2}

Final momentum of particle A is m_{1}v_{1}

Final momentum of particle B is m_{2}v_{2}

Rate of change of momentum of particle A = $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$ ........(1)

Rate of change of momentum of particle B = $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$ ........(2)

Let $\overrightarrow{{\text{F}}_{\text{AB}}}$ and $\overrightarrow{{\text{F}}_{\text{BA}}}$ , be the forces exerted by particle B on A and particle A on B respectively.

According to Newton's second law of motion

$\overrightarrow{{\text{F}}_{\text{AB}}}$ = $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$ and $\overrightarrow{{\text{F}}_{\text{BA}}}$ = $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$

Then, by Newton's third law of motion, we have

F_{AB }+ F_{BA }= 0

F_{AB }= - F_{BA}

ie, m_{1}v_{1}-m_{1}u_{1} = -(m_{2}v_{2}-m_{2}u_{2})

⇒ m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}

or initial momentum of the system = final momentum of the system, which is the law of conversation of momentum.

For every action, there is an equal and opposite reaction is Newton’s third law of motion. This tells us that all forces in nature acts in pairs. These actions and reactions help in understanding the motion of bodies on which forces act. The law also helps in resolving issues in several applications of forces, namely when two bodies collide. The momentum of the bodies before collision and after collision can be worked out using the third law.

If we consider bodies moving along a straight path, the momentum they possess is called linear momentum. If two spheres in a linear motion collide, their momentum before and after the collision can be related using Newton’s third law of motion.

Law of Conservation of Momentum is derived from Newton's third law of motion using the mathematical expression of force, which is derived from Newton’s second law of motion, enunciates that in the absence of external forces, if two bodies collide, the total momentum of the bodies before the collision and after the collision remains the same. This is the law of conservation of linear momentum. After collision if the two bodies stick together, their common speed or velocity can be calculated by using the law of conservation of linear momentum.

**Conservation of Momentum**

There are very few laws in physics that are known to be valid in all situations. The “law of conservation of momentum” is one such very important law for which no exception has been found so far. According to the law of conservation of momentum, when no external unbalanced force is acting, the sum of the momenta of a system of particles is constant. This law is applied for a collision between two bodies, According to the law of conservation of momentum the total momentum of the colliding bodies before collision is equal to the total momentum after collision.

We can apply this law for different situations. For example, consider a boy standing on a boat at rest. When he jumps from the boat on to the bank, the boat will no longer be at rest. The subsequent motion of the boat and its velocity can be explained using the law of conservation of momentum.

When a bullet is fired from a gun, the gun gets some velocity in the opposite direction. The velocity of the gun, commonly called recoil velocity, can be calculated by applying the law of conservation of momentum.

• When one object exerts a force on another object, the second object instantaneously exerts a force back on the first.

• These two forces are always equal in magnitude but opposite in direction.

• These forces act on different objects but never on the same object.

• Every action has an equal and opposite reaction.

There are several applications of Newton’s third law of motion; the launching of satellites is one among them.

**Derivation of law of conservation of momentum from Newton's third law:**

Law of conservation of momentum statesthat if no external force acts on a system of particles, the algebraic sum of the linear momenta of the particles remains conserved."

Consider two particles A and B, Which collide head on as shown in figure The particles move in a straight line before and after collision.

Let particle A have initial velocity u_{1} and particle B has initial velocity u_{2} . The two particles will collide, if u_{1} > u_{2} . Let after collision the final velocities of A and B becomes v_{1} and v_{2} respectively . The two particles will seperate after collision, if v_{2} > v_{1} . Let the two particles A and B have m_{1} and m_{2} respectively.

Initial momentum of particle A is m_{1}u_{1}

Initial momentum of particle B is m_{2}u_{2}

Final momentum of particle A is m_{1}v_{1}

Final momentum of particle B is m_{2}v_{2}

Rate of change of momentum of particle A = $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$ ........(1)

Rate of change of momentum of particle B = $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$ ........(2)

Let $\overrightarrow{{\text{F}}_{\text{AB}}}$ and $\overrightarrow{{\text{F}}_{\text{BA}}}$ , be the forces exerted by particle B on A and particle A on B respectively.

According to Newton's second law of motion

$\overrightarrow{{\text{F}}_{\text{AB}}}$ = $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$ and $\overrightarrow{{\text{F}}_{\text{BA}}}$ = $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$ = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$

Then, by Newton's third law of motion, we have

F_{AB }+ F_{BA }= 0

F_{AB }= - F_{BA}

ie, m_{1}v_{1}-m_{1}u_{1} = -(m_{2}v_{2}-m_{2}u_{2})

⇒ m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}

or initial momentum of the system = final momentum of the system, which is the law of conversation of momentum.

**Activity 1**

**Amrita.olabs.co.in** has developed an interactive online simulation to simulates the Newton's third law of motion. Using this simulation Newton's third law of motion can be understood as there a reaction for anaction.

**Go to Activity**