# Linear momentum of the system of particles, Newton’s second law, practice problems, FAQs

When a firecracker projected from the ground explodes in mid-air, the fragments fly off in different directions, describing their own parabolic paths. But the centre of mass of the cracker will still continue to move along the same parabolic path as shown.

Ever wondered what might be the reason behind the motion of the centre of mass even after the explosion?

## Linear Momentum

The product of the mass and velocity of the particle is known as linear momentum () of a particle. The direction of linear momentum is the same as the direction of the velocity.

The momentum of an n-particle system is the vector sum of the individual momenta of

n particles.

Or

Where, M is the sum of individual masses of particles in the system ().

## Newton’s Second Law of Motion

Newton’s second law of motion states that the rate of change of momentum of a system is equal to the net external force acting on the system and it is in the direction of that force.

We know that the acceleration of the centre of mass of a system is given as follows:

Also, the external force on the system is given by,

Equating equation and ,

By Newton’s second law of motion, we get the following:

Point to Remember

## COM at Rest

If a system is initially at rest and there is no net external force acting on it, then there will be no shift in position of the COM of the system.

## Practice problems

Q1. In a boat of mass 4M and length l on a frictionless water surface, two men and of mass and 2M, respectively, are standing at two opposite ends. Now, A travels a distance of , i.e., relative to the boat towards its centre, and moves a distance of and meets A. Find the distance travelled by the boat on the water when they meet.

Solution: Given,

Mass of the boat, m3 = 4M

Length of the boat = l

Mass of man A, m1 = M

Mass of man B, m2 = 2M

Distance travelled by A, x1 =

Distance travelled by B, x2 =

When A is walking rightwards, he pushes the boat leftwards, and when is walking leftwards, he pushes the boat rightwards.let boat displaced in right direction . we can assume any direction if the assumed direction is wrong then your answer comes negative.

Let the distance travelled by the boat when they meet be x.

Consider the boat and the two people together as a system.

As there is no net external force acting on the system along the x-direction,

Since the system was initially at rest,we have,

Q2. For the arrangement shown in the image, mA = 2 kg and mB = 1 kg. The string is light and inextensible. Find the acceleration of COM of both the blocks. Neglect the friction everywhere.

Mass of block A, mA = 2 kg

Mass of block B, mB = 1 kg

Consider both the blocks together as a single system.

(Recreatethe image)

Acceleration of the blocks is as follows:

Since block A is heavier, it moves down and block B moves upwards as shown in the figure. Consider the upward direction as negative.

Now, the acceleration of the centre of mass of the system is given as follows:

Q3. Two identical buggies move one after the other due to inertia with the same velocity . A man of mass is riding the rear buggy. At a certain moment, the man jumps into the front buggy with velocity relative to his buggy. Knowing that the mass of each buggy is equal to , find the velocities with which the buggies will move after the man jumps.

Initial speed of the two buggies =

Mass of each buggy =

Mass of the man =

When the man jumps from one buggy to another, the velocity of the first buggy decreases to as he pushes it backwards.
Upon his landing on the second buggy, let the velocity of that buggy be .

By taking the man and the first buggy as a system during the jump, we get the following:

The net external force acting on the system along the x-direction is zero and thus, the momentum is conserved along the x-direction. Using principle of conservation of momentum for this system before and during the jumping instance,

Now, consider the man and second buggy as a system,

Conserving momentum along the x-direction during and after the jumping instance,

Q4. A block of mass is placed on a triangular block of mass , which in turn is placed on a horizontal surface as shown in figure. Assuming a frictionless surface, find the velocity of the triangular block when the smaller block reaches the bottom end.

(Recreate the image)

Solution: Given,

Mass of the small block =

Mass of the larger triangular block =

Initial height of the smaller block =

Let be the final velocity of the small block with respect to the larger block.

Let be the velocity of the larger block when the small block reaches the bottom position.

(Recreate the image)

As the net external force along the x-direction is zero, the momentum is conserved along the x-direction.

Also, the resultant velocity of the smaller block of mass is as follows:

Conserving the mechanical energy of the system (both the blocks together),

Putting value of from equation , we get

## FAQs

Q1. An astronaut in space is fixing the spaceship using a wrench. If the astronaut wants to move towards the spaceship, in which direction should he throw the wrench?
Answer:  Astronauts should throw the wrench opposite to the direction in which he/she wants to move. Due to absence of external forces, the momentum remains conserved and thus the position of the centre remains the same.

Q2. If the centre of mass of three particles is at rest, then choose the incorrect option

A. There is a possibility that two of them are at rest.
B. It is possible when two of them are moving or all three are at rest.
C. It is possible when at least two of them are moving or all three are at rest.
D. It is possible when all three are moving or all three are at rest.

If centre of mass of three particle is at rest, then

But, if only two of them have zero velocity, let

Then,

cannot be zero if only two of them are at rest.

Therefore at least two of them must have non zero velocity or all three should have zero velocity.

Q3. Two persons A and B having masses are sitting at the opposite ends of a stationary boat. If a person walks towards other & sits with him, does the centre of mass of the system shift?
Answer: Initially both persons and the boat are at rest. Therefore the C.O.M of the system is also at rest.

As there is no external force acting on the system, therefore, the C.O.M of the system will continue to remain at rest and it will not shift.

Q4. A small block of mass is placed on a body of mass at rest. If the block is set in horizontal motion with velocity as shown. Does the block of mass M start moving?

(Assume that the small block does not leave surface of and all surfaces are smooth).
Answer: Yes, the block will start moving right, due to the normal reaction exerted on it by .

As there is no external force acting on the system, net momentum is conserved. Hence as the velocity of the mass decreases that for mass starts increasing until reaches point A.