When is the total linear momentum of a system constant?

The total linear momentum of a system of particles remains constant if the net external force acting on the system is zero.

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Conservation of Linear Momentum for a System

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Conservation of Linear Momentum for a System

Conservation of linear momentum for a system is a fundamental principle in physics stating that the total linear momentum of a system remains constant if no external forces act on it. This means that when objects within a closed system interact—such as during collisions or explosions—their combined momentum before and after the interaction stays the same. Widely applied in mechanics, engineering, and astrophysics, this law helps predict motion and behavior in systems ranging from subatomic particles to large celestial bodies. The conservation of momentum is crucial for understanding real-world phenomena where internal forces dominate and external influences are negligible.

1.0Definition of Linear Momentum of a System

The linear momentum of a system is the vector sum of the momenta of all particles within the system. It represents the total motion of the system and is given by

$P_{system} = \sum_{i = 1}^{n} m_{i} v_{i}$

Where $m_{i}$ and $v_{i}$ are the mass and velocity of the ith particle.

2.0Definition of Conservation of Linear Momentum for a System

The total linear momentum of a system of particles remains constant if the net external force acting on the system is zero.
If no external forces (or only balanced external forces) act on a system, the total momentum of all the objects in that system stays the same, even if the individual objects within the system interact (e.g., collide or push/pull each other).
$P_{t o t a l} = constant$

3.0COM Displacement from Particle Motion

$r_{c m} = \frac{m _{1} r _{1} + m _{2} r _{2} + \dots}{m _{1} + m _{2} + \dots}$

$d r_{c m} = \frac{m _{1} d r _{1} + m _{2} d r _{2} + \dots}{m _{1} + m _{2} + \dots}$

Displacement of COM,

$Δ r_{c m} = \frac{m _{1} Δ r _{1} + m _{2} Δ r _{2} + \dots}{m _{1} + m _{2} + \dots}$

4.0Velocity of COM

$M v_{c m} = m_{1} \frac{d r _{1}}{d t} + m_{2} \frac{d r _{2}}{d t} + m_{3} \frac{d r _{3}}{d t} + \dots + m_{n} \frac{d r _{n}}{d t}$

$M v_{c m} = m_{1} v_{1} + m_{2} v_{2} + m_{3} v_{3} + \dots + m_{n} v_{n}$

Here the right hand side term is the total momentum of the system, summation of momentum of the individual components(particle) of the system.

Hence velocity of center of mass of the system is the ratio of momentum of the system to the mass of the system.

∴ $P_{system} = M v_{c m}$

5.0Acceleration of COM

$M a_{c m} = m_{1} \frac{d v _{1}}{d t} + m_{2} \frac{d v _{2}}{d t} + m_{3} \frac{d v _{3}}{d t} + \dots + m_{n} \frac{d v _{n}}{d t}$ $= m_{1} a_{1} + m_{2} a_{2} + m_{3} a_{3} + \dots + m_{n} a_{n}$

$a_{c m} = \frac{Net force on system}{M} = \frac{Net external force + Net internal force}{M}$

(∵ Action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on system is zero)

$∴ F_{ext} = M a_{c m}$

where $F_{ext}$ is the sum of the 'external' forces acting on the system. The internal forces which the particles exert on one another play absolutely no role in the motion of the center of mass.

If no external force is acting on a system of particles, the acceleration of the center of mass of the system will be zero. If $a_{c m} = 0$ , it implies that $v_{c m}$ must be a constant and if vcm is a constant, it implies that the total momentum of the system must remain constant. It leads to the principle of conservation of momentum in absence of external forces.

$If F_{ext} = 0 then v_{c m} = constant$

“If resultant external force is zero on the system, then the net momentum of the system must remain constant”.

6.0Motion of COM in a Moving System

Conservation of Momentum at rest :

If $F_{e x t} = 0 an d V_{c m} = 0$ , then COM remains at rest. Individual components of the system may move and have non-zero momentum due to mutual forces (internal), but the net momentum of the system remains zero.

All the particles of the system are at rest.
Particles are moving such that their net momentum is zero.

Motion of COM in a Moving System : COM at rest

A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions, then, since the explosive forces are internal and there is no external force on the system for explosion, therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero.
Two men standing on a frictionless platform, push each other, then their net momentum remains zero because the push forces are internal for the two men system.
A boat floating in a lake also has net momentum zero if the people on it changes their position, because the friction force required to move the people is internal in the boat system.
Objects initially at rest, if moved under mutual forces (electrostatic or gravitation) also have net momentum zero.
A light spring of spring constant k kept compressed between two blocks of masses m1 and m2 on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions such that the net momentum is zero.
In a fan, all particles are moving but COM is at rest

Conservation of Momentum moving with uniform velocity :

If $F_{e x t} = 0, t h e n V_{c m}$ remains constant, therefore, net momentum of the system also remains conserved. Individual components of the system may have variable velocities and momentum due to mutual forces (internal), but the net momentum of the system remains constant and COM continues to move with the initial velocity.

All the particles of the system are moving with the same velocity. E.g.: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity.

Conservation Of Momentum moving with uniform velocity

Internal explosions / breaking does not change the motion of COM and net momentum remains conserved. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragments fly such that their net momentum remains conserved.
Man jumping from cart or buggy also exert internal forces therefore net momentum of the system does not change and hence, motion of COM remains conserved.
Two moving blocks connected by a light spring on a smooth horizontal surface. If the acting forces are only due to spring then COM will remain in its motion and momentum will remain conserved.
Particles colliding in absence of external impulsive forces also have their momentum conserved.

Conservation of Momentum moving with acceleration :

If an external force is present then COM continues its original motion as if the external force is acting on it, irrespective of internal forces.

Example:

Projectile motion : An axe thrown in air at an angle with the horizontal will perform a complicated motion of rotation as well as parabolic motion under the effect of gravitation.

7.0Application of Conservation of Linear Momentum of a System

Firing a Bullet from a Gun :

Application of Conservation of Linear Momentum : Firing a Bullet from a Gun

If the bullet and gun is the system, then the force exerted by the trigger will be internal.

Now, as initially both bullet and gun are at rest so, $p_{G} + p_{B} = 0$ . From this is evident that: $p_{G} = - p_{B}$ , i.e., if a bullet acquires forward momentum, the gun will acquire equal and opposite (backward) momentum.

$m v + M V = 0 \Rightarrow V = - \frac{m}{M} v$

If the bullet moves forward the gun recoils or kicks backwards. Heavier the gun lesser will be the recoil velocity V

Kinetic Energy $K = \frac{p ^{2}}{2 m}$ and $∣ p_{B} ∣ = ∣ p_{G} ∣$

Kinetic Energy of gun $K_{G} = \frac{p ^{2}}{2 M}$

Kinetic Energy of bullet $K_{B} = \frac{p ^{2}}{2 m}$

$\frac{K _{G}}{K _{B}} = \frac{m}{M} < 1 (since M ≫ m)$

Thus, the kinetic energy of a gun is less than that of a bullet i.e., the kinetic energy of bullet and gun will not be equal. Initial kinetic energy of the system is zero as both are at rest. Final kinetic energy of the system is greater than zero. So, here kinetic energy of the system is not constant but increases. However, energy is always conserved. Here chemical energy of gun powder is converted into KE.

Block-Bullet System

When bullet remains embedded in the block

When bullet remains embedded in the block

Conserving momentum of bullet and block

$m v + 0 = (M + m) V$

Velocity of block $V = \frac{m v}{M + m}$

By conservation of mechanical energy

$\frac{1}{2} (M + m) V^{2} = (M + m) g h \Rightarrow V = 2 g h$

Maximum Height gained by block $h = \frac{V ^{2}}{2 g} = \frac{m ^{2} v ^{2}}{2 g ( M + m ) ^{2}}$

If bullet emerges out of the block

Block-Bullet System - f bullet emerges out of the block

Conserving momentum $m v + 0 = m v_{1} + M v_{2}$

$m (v - v_{1}) = M v_{2}$ ………..(1)

Conserving energy

$\frac{1}{2} M v_{2}^{2} = M g h \Rightarrow v_{2} = 2 g h$ ……..(2)

From equation(1) and (2)

$m (v - v_{1}) = M 2 g h \Rightarrow \frac{m ^{2} ( v - v _{1} ) ^{2}}{2 g M ^{2}}$

Spring Block System

Consider two blocks, resting on a frictionless surface and connected by a massless spring as shown in figure. If the spring is stretched (or compressed) and then released from rest,

Then $F_{ext} = 0$ so $p_{s} = p_{1} + p_{2} = constant$

However, initially both the blocks were at rest so, $p_{1} + p_{2} = 0$

$p_{2} = - p_{1}$ i.e. at any instant the two blocks will have momentum equal in magnitude but opposite in direction(though they have different values of momentum at different positions)
As momentum, $p = m_{1} v_{1} + m_{2} v_{2} = 0 \Rightarrow v_{2} = - (\frac{m _{1}}{m _{2}}) v_{1}$ the two blocks always move in opposite directions with lighter blocks moving faster.
Kinetic Energy $K E = \frac{p ^{2}}{2 m}$ and $∣ p_{1} ∣ = ∣ p_{2} ∣, \frac{K E _{1}}{K E _{2}} = \frac{m _{2}}{m _{1}}$ or the kinetic energy of two blocks will not be equal but in the inverse ratio of their masses and so lighter blocks will have greater kinetic energy.
Initially kinetic energy of the blocks is zero (as both are at rest) but after some time kinetic energy of the blocks is not zero (as both are in motion). So, kinetic energy is not constant but changes. Here during the motion of the blocks KE is converted into elastic potential energy of the spring and vice– versa but total mechanical energy of the system remains constant.
Kinetic energy + Potential energy = Mechanical Energy = Constant

8.0Key Points Linear Momentum of a System

For an isolated system, the initial momentum of the system is equal to the final momentum of the system. If the system consists of n bodies having momenta

$p_{1}, p_{2}, p_{3}, \dots, p_{n} then p_{1} + p_{2} + p_{3} + \dots + p_{n} = constant$

As linear momentum depends on frame of reference, observers in different frames would find different values of linear momenta of a given system but each would agree that his own value of linear momentum does not change with time. But the system should be isolated and closed, i.e., law of conservation of linear momentum is independent of frame of reference though linear momentum depends on the frame of reference.
Conservation of linear momentum is equivalent to Newton's III law of motion for a system of two particles. In the absence of external force from law of conservation of linear momentum,

$\Rightarrow$ $p_{1} + p_{2} = constant i.e. m_{1} v_{1} + m_{2} v_{2} = constant$

Differentiating the above expression with respect to time

$m_{1} \frac{d v _{1}}{d t} + m_{2} \frac{d v _{2}}{d t} = 0 [as m is constant]$

$m_{1} a_{1} + m_{2} a_{2} = 0 [∵ \frac{d v _{1}}{d t} = a_{1}]$

$\Rightarrow F_{1} + F_{2} = 0 [∵ F = m a]$

$\Rightarrow F_{1} = - F_{2}$

This law is universal, i.e., it applies to macroscopic as well as microscopic systems.
The total linear momentum of the system in the centroidal frame is zero.

It implies $\sum_{n} m_{n} v_{n} = 0 or m_{1} v_{1} + m_{2} v_{2} + \dots + m_{n} v_{n} = 0$

9.0Sample Questions on Conservation of Linear Momentum

Q-1. If man walks from A to B find displacement of man and plank.

Solution: Initial momentum of system man and plank is zero. Net ext. force on this system is zero.

Thus $p_{sys} = 0 \Rightarrow v_{c m} = 0 \Rightarrow Δ s_{c m} = 0$

If the man moves towards right, the plank will move towards left by a distance (X) (taking the two blocks together as the system).

The horizontal position of CM remains same $m_{1} (l - x) = m_{2} x$

$x = \frac{- m _{1}}{m _{1} + m _{2}} l (left)$

Q-2. A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil ?

Solution: Consider the situation shown in figure. Suppose the man moves at a speed w towards right and the platform recoils at a speed V towards the left, both relative to the ice. Hence, the speed of the man relative to the platform is V + w. By the question,

V + w=v or w=v-V……(1)

Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially both the man and the platform were at rest.

Thus,

$0 = M V - m w$

$M V = m (v - V)$ using (1)

$V = \frac{m v}{M + m}$

Q-3.A flat car of mass M is at rest on a frictionless floor with a child of mass m standing at its edge. If a child jumps off from the car towards right with an initial velocity u, with respect to the car, find the velocity of the car after its jump.

Solution: Let the car attains a velocity v, and the net velocity of the child with respect to earth will be u – v, as u is its velocity with respect to the car.

Initially, the system was at rest, thus according to momentum conservation, momentum after jump must be zero, as

mu-v=Mv

$V = \frac{m v}{M + m}$

Q-4.A shell is fired from a cannon with a speed of 100 m/s at an angle 60° with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of explosion.

Solution: As we know in absence of external force the motion of centre of mass of

a body remains unaffected. Thus, here the centre of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is

$v_{M} = u cos θ = 100 \times cos 6 0^{\circ} = 50 m/s$

Sample Question - conservation of momentum

Let $v_{1}$ be the speed of the fragment which moves along the negative x-direction and the other fragment has speed $v_{2}$ Which must be along a positive x-direction. Now from momentum conservation, we have

$m v = - \frac{m}{2} v_{1} + \frac{m}{2} v_{2}$

$2 v = v_{2} - v_{1}$

$v_{2} = 2 v + v_{1} = (2 \times 50) + 50 = 150 m/s$

Q-5.Each of the blocks shown in figure has mass 1 kg. The rear block moves with a speed of 2 m/s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant 50 N/m. Find the maximum compression of the spring. (Assume surface is frictionless)

Sample Question 5 on conservation of momentum

Solution: Maximum compression will take place when the blocks move with equal velocity. As no net external horizontal force acts on the system of the two blocks, the total linear momentum will remain constant. If V is the common speed at maximum compression, we have,

$(1 kg) (2 m/s) = (1 kg) V + (1 kg) V or V = 1 m/s$

Initial kinetic energy = $\frac{1}{2} (1 kg) (2 m/s)^{2} = 2 J$

Final kinetic energy = $\frac{1}{2} (1 kg) (1 m/s)^{2} + \frac{1}{2} (1 kg) (1 m/s)^{2} = 1 J$

The kinetic energy lost is stored as the elastic energy in the spring

Hence , $\frac{1}{2} (50 N/m) x^{2} = 2 J - 1 J = 1 J$

Conservation of Linear Momentum for a System

1.0Definition of Linear Momentum of a System

The linear momentum of a system is the vector sum of the momenta of all particles within the system. It represents the total motion of the system and is given by

$P_{system} = \sum_{i = 1}^{n} m_{i} v_{i}$

Where $m_{i}$ and $v_{i}$ are the mass and velocity of the ith particle.

2.0Definition of Conservation of Linear Momentum for a System

The total linear momentum of a system of particles remains constant if the net external force acting on the system is zero.
If no external forces (or only balanced external forces) act on a system, the total momentum of all the objects in that system stays the same, even if the individual objects within the system interact (e.g., collide or push/pull each other).
$P_{t o t a l} = constant$

3.0COM Displacement from Particle Motion

$r_{c m} = \frac{m _{1} r _{1} + m _{2} r _{2} + \dots}{m _{1} + m _{2} + \dots}$

$d r_{c m} = \frac{m _{1} d r _{1} + m _{2} d r _{2} + \dots}{m _{1} + m _{2} + \dots}$

Displacement of COM,

$Δ r_{c m} = \frac{m _{1} Δ r _{1} + m _{2} Δ r _{2} + \dots}{m _{1} + m _{2} + \dots}$

4.0Velocity of COM

$M v_{c m} = m_{1} \frac{d r _{1}}{d t} + m_{2} \frac{d r _{2}}{d t} + m_{3} \frac{d r _{3}}{d t} + \dots + m_{n} \frac{d r _{n}}{d t}$

$M v_{c m} = m_{1} v_{1} + m_{2} v_{2} + m_{3} v_{3} + \dots + m_{n} v_{n}$

Here the right hand side term is the total momentum of the system, summation of momentum of the individual components(particle) of the system.

Hence velocity of center of mass of the system is the ratio of momentum of the system to the mass of the system.

∴ $P_{system} = M v_{c m}$

5.0Acceleration of COM

$a_{c m} = \frac{Net force on system}{M} = \frac{Net external force + Net internal force}{M}$

(∵ Action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on system is zero)

$∴ F_{ext} = M a_{c m}$

where $F_{ext}$ is the sum of the 'external' forces acting on the system. The internal forces which the particles exert on one another play absolutely no role in the motion of the center of mass.

$If F_{ext} = 0 then v_{c m} = constant$

“If resultant external force is zero on the system, then the net momentum of the system must remain constant”.

6.0Motion of COM in a Moving System

Conservation of Momentum at rest :

All the particles of the system are at rest.
Particles are moving such that their net momentum is zero.

A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions, then, since the explosive forces are internal and there is no external force on the system for explosion, therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero.
Two men standing on a frictionless platform, push each other, then their net momentum remains zero because the push forces are internal for the two men system.
A boat floating in a lake also has net momentum zero if the people on it changes their position, because the friction force required to move the people is internal in the boat system.
Objects initially at rest, if moved under mutual forces (electrostatic or gravitation) also have net momentum zero.
A light spring of spring constant k kept compressed between two blocks of masses m1 and m2 on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions such that the net momentum is zero.
In a fan, all particles are moving but COM is at rest

Conservation of Momentum moving with uniform velocity :

All the particles of the system are moving with the same velocity. E.g.: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity.

Internal explosions / breaking does not change the motion of COM and net momentum remains conserved. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragments fly such that their net momentum remains conserved.
Man jumping from cart or buggy also exert internal forces therefore net momentum of the system does not change and hence, motion of COM remains conserved.
Two moving blocks connected by a light spring on a smooth horizontal surface. If the acting forces are only due to spring then COM will remain in its motion and momentum will remain conserved.
Particles colliding in absence of external impulsive forces also have their momentum conserved.

Conservation of Momentum moving with acceleration :

If an external force is present then COM continues its original motion as if the external force is acting on it, irrespective of internal forces.

Example:

Projectile motion : An axe thrown in air at an angle with the horizontal will perform a complicated motion of rotation as well as parabolic motion under the effect of gravitation.

7.0Application of Conservation of Linear Momentum of a System

Firing a Bullet from a Gun :

If the bullet and gun is the system, then the force exerted by the trigger will be internal.

$m v + M V = 0 \Rightarrow V = - \frac{m}{M} v$

If the bullet moves forward the gun recoils or kicks backwards. Heavier the gun lesser will be the recoil velocity V

Kinetic Energy $K = \frac{p ^{2}}{2 m}$ and $∣ p_{B} ∣ = ∣ p_{G} ∣$

Kinetic Energy of gun $K_{G} = \frac{p ^{2}}{2 M}$

Kinetic Energy of bullet $K_{B} = \frac{p ^{2}}{2 m}$

$\frac{K _{G}}{K _{B}} = \frac{m}{M} < 1 (since M ≫ m)$

Block-Bullet System

When bullet remains embedded in the block

Conserving momentum of bullet and block

$m v + 0 = (M + m) V$

Velocity of block $V = \frac{m v}{M + m}$

By conservation of mechanical energy

$\frac{1}{2} (M + m) V^{2} = (M + m) g h \Rightarrow V = 2 g h$

Maximum Height gained by block $h = \frac{V ^{2}}{2 g} = \frac{m ^{2} v ^{2}}{2 g ( M + m ) ^{2}}$

If bullet emerges out of the block

Conserving momentum $m v + 0 = m v_{1} + M v_{2}$

$m (v - v_{1}) = M v_{2}$ ………..(1)

Conserving energy

$\frac{1}{2} M v_{2}^{2} = M g h \Rightarrow v_{2} = 2 g h$ ……..(2)

From equation(1) and (2)

$m (v - v_{1}) = M 2 g h \Rightarrow \frac{m ^{2} ( v - v _{1} ) ^{2}}{2 g M ^{2}}$

Spring Block System

Consider two blocks, resting on a frictionless surface and connected by a massless spring as shown in figure. If the spring is stretched (or compressed) and then released from rest,

Then $F_{ext} = 0$ so $p_{s} = p_{1} + p_{2} = constant$

However, initially both the blocks were at rest so, $p_{1} + p_{2} = 0$

$p_{2} = - p_{1}$ i.e. at any instant the two blocks will have momentum equal in magnitude but opposite in direction(though they have different values of momentum at different positions)
As momentum, $p = m_{1} v_{1} + m_{2} v_{2} = 0 \Rightarrow v_{2} = - (\frac{m _{1}}{m _{2}}) v_{1}$ the two blocks always move in opposite directions with lighter blocks moving faster.
Kinetic Energy $K E = \frac{p ^{2}}{2 m}$ and $∣ p_{1} ∣ = ∣ p_{2} ∣, \frac{K E _{1}}{K E _{2}} = \frac{m _{2}}{m _{1}}$ or the kinetic energy of two blocks will not be equal but in the inverse ratio of their masses and so lighter blocks will have greater kinetic energy.
Initially kinetic energy of the blocks is zero (as both are at rest) but after some time kinetic energy of the blocks is not zero (as both are in motion). So, kinetic energy is not constant but changes. Here during the motion of the blocks KE is converted into elastic potential energy of the spring and vice– versa but total mechanical energy of the system remains constant.
Kinetic energy + Potential energy = Mechanical Energy = Constant

8.0Key Points Linear Momentum of a System

For an isolated system, the initial momentum of the system is equal to the final momentum of the system. If the system consists of n bodies having momenta

$p_{1}, p_{2}, p_{3}, \dots, p_{n} then p_{1} + p_{2} + p_{3} + \dots + p_{n} = constant$

As linear momentum depends on frame of reference, observers in different frames would find different values of linear momenta of a given system but each would agree that his own value of linear momentum does not change with time. But the system should be isolated and closed, i.e., law of conservation of linear momentum is independent of frame of reference though linear momentum depends on the frame of reference.
Conservation of linear momentum is equivalent to Newton's III law of motion for a system of two particles. In the absence of external force from law of conservation of linear momentum,

$\Rightarrow$ $p_{1} + p_{2} = constant i.e. m_{1} v_{1} + m_{2} v_{2} = constant$

Differentiating the above expression with respect to time

$m_{1} \frac{d v _{1}}{d t} + m_{2} \frac{d v _{2}}{d t} = 0 [as m is constant]$

$m_{1} a_{1} + m_{2} a_{2} = 0 [∵ \frac{d v _{1}}{d t} = a_{1}]$

$\Rightarrow F_{1} + F_{2} = 0 [∵ F = m a]$

$\Rightarrow F_{1} = - F_{2}$

This law is universal, i.e., it applies to macroscopic as well as microscopic systems.
The total linear momentum of the system in the centroidal frame is zero.

It implies $\sum_{n} m_{n} v_{n} = 0 or m_{1} v_{1} + m_{2} v_{2} + \dots + m_{n} v_{n} = 0$

9.0Sample Questions on Conservation of Linear Momentum

Q-1. If man walks from A to B find displacement of man and plank.

Solution: Initial momentum of system man and plank is zero. Net ext. force on this system is zero.

Thus $p_{sys} = 0 \Rightarrow v_{c m} = 0 \Rightarrow Δ s_{c m} = 0$

If the man moves towards right, the plank will move towards left by a distance (X) (taking the two blocks together as the system).

The horizontal position of CM remains same $m_{1} (l - x) = m_{2} x$

$x = \frac{- m _{1}}{m _{1} + m _{2}} l (left)$

V + w=v or w=v-V……(1)

Thus,

$0 = M V - m w$

$M V = m (v - V)$ using (1)

$V = \frac{m v}{M + m}$

Solution: Let the car attains a velocity v, and the net velocity of the child with respect to earth will be u – v, as u is its velocity with respect to the car.

Initially, the system was at rest, thus according to momentum conservation, momentum after jump must be zero, as

mu-v=Mv

$V = \frac{m v}{M + m}$

Solution: As we know in absence of external force the motion of centre of mass of

a body remains unaffected. Thus, here the centre of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is

$v_{M} = u cos θ = 100 \times cos 6 0^{\circ} = 50 m/s$

$m v = - \frac{m}{2} v_{1} + \frac{m}{2} v_{2}$

$2 v = v_{2} - v_{1}$

$v_{2} = 2 v + v_{1} = (2 \times 50) + 50 = 150 m/s$

$(1 kg) (2 m/s) = (1 kg) V + (1 kg) V or V = 1 m/s$

Initial kinetic energy = $\frac{1}{2} (1 kg) (2 m/s)^{2} = 2 J$

Final kinetic energy = $\frac{1}{2} (1 kg) (1 m/s)^{2} + \frac{1}{2} (1 kg) (1 m/s)^{2} = 1 J$

The kinetic energy lost is stored as the elastic energy in the spring

Hence , $\frac{1}{2} (50 N/m) x^{2} = 2 J - 1 J = 1 J$

Frequently Asked Questions

What is the linear momentum of a system?

When is the total linear momentum of a system constant?

Frequently Asked Questions

What is the linear momentum of a system?

When is the total linear momentum of a system constant?

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Conservation of Linear Momentum for a System

1.0Definition of Linear Momentum of a System

2.0Definition of Conservation of Linear Momentum for a System

3.0COM Displacement from Particle Motion

4.0Velocity of COM

5.0Acceleration of COM

6.0Motion of COM in a Moving System

7.0Application of Conservation of Linear Momentum of a System

8.0Key Points Linear Momentum of a System

9.0Sample Questions on Conservation of Linear Momentum

Table of Contents

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Conservation of Linear Momentum for a System

1.0Definition of Linear Momentum of a System

2.0Definition of Conservation of Linear Momentum for a System

3.0COM Displacement from Particle Motion

4.0Velocity of COM

5.0Acceleration of COM

6.0Motion of COM in a Moving System

7.0Application of Conservation of Linear Momentum of a System

8.0Key Points Linear Momentum of a System

9.0Sample Questions on Conservation of Linear Momentum

Table of Contents