General Physics (calculus based) Class Notes

Dr. Rakesh Kapoor, M.Sc., Ph.D.

Former Faculty-University of Alabama at Birmingham, Birmingham, AL 35294

Center of Mass and Linear Momentum


In this chapter we will learn the following concepts:

Center of mass (com) for a system of particles.

How to calculate center of mass of a continuous object.

How to apply Newton’s Second law to a system of particles.

Linear momentum for a single particle and a system of particles.

Impulse and its relation to change in momentum.

The principle of conservation of linear momentum.

We will use the conservation of linear momentum to study collisions in one and two dimensions.

Center of Mass

The center of mass (com) of a system of particles is the point that moves as though (1) all of the system’s mass were concentrated there and (2) all external forces were applied there.

Consider a system of two particles of masses CenterOfMass_LinearMomentum_1.gif and CenterOfMass_LinearMomentum_2.gif at positions CenterOfMass_LinearMomentum_3.gif and CenterOfMass_LinearMomentum_4.gif respectively. Position of center of mass CenterOfMass_LinearMomentum_5.gif of this system is defined as



Center of mass for many particles

If there are n particles, the center of mass of these n particles will be


Total mass M of n particles is


Therefore center of mass of n particles can also be written as


How to calculate Center of Mass in 3D?

We can calculate three position coordinates CenterOfMass_LinearMomentum_11.gif separately in three dimensions.


Here CenterOfMass_LinearMomentum_13.gif are the position coordinates for i th particle.


Similarly position vector CenterOfMass_LinearMomentum_15.gif of center of mass can be written as



Example 1: Center of Mass

Following figure shows a three-particle system, with masses CenterOfMass_LinearMomentum_18.gif, CenterOfMass_LinearMomentum_19.gif and CenterOfMass_LinearMomentum_20.gif. The scales on the axes are set by CenterOfMass_LinearMomentum_21.gif and CenterOfMass_LinearMomentum_22.gif.
What are (a) the x coordinate and (b) the y coordinate of the system’s center of mass? (c) If CenterOfMass_LinearMomentum_23.gif is gradually increased, does the center of mass of the system shift toward or away from that particle, or does it remain stationary?



How to calculate Center of Mass in Solid Objects?

We can assume that solid object is made up of n small elements of mass Δm each.


Let us consider, CenterOfMass_LinearMomentum_27.gif is the position vector of ith element of the object.

Coordinates of center of mass of such a solid object can be written as



When we make these elements extremely small such that n, in that case Δm0 and summation becomes integral.


When the density ρ of the object is uniform, the mass and volume at all the positions in the object can be given as  


Where dV is the volume of small element of mass dm and V is total volume of the object.



Symmetric Objects

These integrals can be avoided if the object has a point, a line or a plane of symmetry.

The center of mass of such an object then lies at that point, on that line, or in that plane.


The center of mass of a uniform sphere (which has a point of symmetry) is at the center of the sphere.

The center of mass of a uniform cuboid (which has planes of symmetry) is at the center of the intersection of two diagonals as shown.

CenterOfMass_LinearMomentum_34.gif CenterOfMass_LinearMomentum_35.gif

The center of mass of a uniform rectangular object lies at the intersection of the diagonals.


The center of mass of a banana (which has a plane of symmetry that splits it into two equal parts) lies somewhere in that plane.

The center of mass of an object need not lie within the object. There is no dough at the com of a dough-nut, and no iron at the com of a horseshoe.


Checkpoint-1: Center of Mass

The figure shows a uniform square plate from which four identical squares at the corners will be removed.
(a) Where is the center of mass of the plate originally?
Where is it after the removal of (b) square 1; (c) squares 1 and 2; (d) squares 1 and 3; (e) squares 1, 2, and 3; (f) all four squares? Answer in terms of quadrants, axes, or points (without calculation, of course).

Newton’s Second Law for a system of Particles

Consider multi-particle system comprises of n particles.

We are not interested in the individual motions of these particles.

We are only interested in the motion of the center of mass CenterOfMass_LinearMomentum_39.gif of this multi-particle system.

According to Newton’s second law


CenterOfMass_LinearMomentum_41.gif is the net external force acting on this multi-particle system.

M is the total mass of the system and CenterOfMass_LinearMomentum_42.gif is the acceleration of the center of mass of the system.

Components of net external force are related to components of acceleration of center of mass.


Checkpoint-2: Newton’s Second law

Two skaters on frictionless ice hold opposite ends of a pole of negligible mass. An axis runs along the pole, and the origin of the axis is at the center of mass of the two-skater system. One skater, Fred, weighs twice as much as the other skater, Ethel.
Where do the skaters meet if
(a) Fred pulls hand over hand along the pole so as to draw himself to Ethel,
(b) Ethel pulls hand over hand to draw herself to Fred, and
(c) both skaters pull hand over hand?

Example 2: Newton’s Second law

Two skaters, one with mass 65 kg and the other with mass 40 kg, stand on an ice rink holding a pole of length 10 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet.
How far does the 40 kg skater move?


Hint :

Linear Momentum

The linear momentum of a particle CenterOfMass_LinearMomentum_45.gif is defined as


m is the mass of the particle and CenterOfMass_LinearMomentum_47.gif is its velocity and CenterOfMass_LinearMomentum_48.gif is the speed of light.

For slow moving objects (v<<c), the approximate expression for  momentum is given as


Since m is a scalar quantity and velocity CenterOfMass_LinearMomentum_50.gif is a vector quantity, therefore linear momentum CenterOfMass_LinearMomentum_51.gif is a vector.

As m is always positive, therefore CenterOfMass_LinearMomentum_52.gif and CenterOfMass_LinearMomentum_53.gif have the same direction.

The SI unit for linear momentum is the kilogram-meter per second (kg · m/s).


Relation between Linear Momentum and Force.

As per Newton's Law we know that


When mass of a particle remains constant we can rewrite it as


The time rate of change of the momentum of a particle is equal to the net external force acting on the particle and is in the direction of that force.

This relation between net force and momentum is always valid  (whether the mass is constant or changes with time).


Differentiation by parts give us


This is the complete expression of "Newton's second law".


If mass m of the particle is constant with time, then



Checkpoint-3: Linear Momentum

The figure gives the magnitude p of the linear momentum versus time t for a particle moving along an axis. A force directed along the axis acts on the particle.
(a) Rank the four regions indicated according to the magnitude of the force, greatest first.
(b) In which region is the particle slowing?


Hint :

The Linear Momentum of a system of Particles

Consider a system of n particles, each with its own mass, velocity, and linear momentum.

The system as a whole has a total linear momentum CenterOfMass_LinearMomentum_63.gif , which is the vector sum of linear momentum of the each individual particle.


We can also write total linear momentum CenterOfMass_LinearMomentum_65.gif as


Total Linear Momentum and velocity of center of mass

Velocity of center of mass of a multi-particle system is written as


Where M is the total mass of the system.

If we compare the above two equations, total linear momentum CenterOfMass_LinearMomentum_68.gif of a multi-particle system can be written as


The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.

Example 3: Linear Momentum

A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical wall and rebounds with speed 2.0 m/s.
What is the magnitude of the change in its linear momentum?

Solution :

Initial momentum CenterOfMass_LinearMomentum_70.gif is


Final momentum CenterOfMass_LinearMomentum_72.gif is


Therefore change in momentum is





When a constant acceleration is applied to an object from time CenterOfMass_LinearMomentum_77.gif to CenterOfMass_LinearMomentum_78.gif, as per kinematic equations the relation between its initial CenterOfMass_LinearMomentum_79.gif and final CenterOfMass_LinearMomentum_80.gif velocity is given as


Multiply both sides by mass m of the object.


Left hand side is change in momentum CenterOfMass_LinearMomentum_83.gif, of the object and right hand side is the product of force and time duration Δt of its action.


CenterOfMass_LinearMomentum_85.gif is also known as impulse. Symbol CenterOfMass_LinearMomentum_86.gif is used for impulse.


What if force is not constant?

We can divide time in small interval and calculate impulse in each interval, sum of impulses will give us net change in momentum.



When Δt0, above summation gets converted into integral from initial time CenterOfMass_LinearMomentum_90.gif to final time CenterOfMass_LinearMomentum_91.gif.


The right side of above equation is the impulse.



Computation of average force from change in momentum.

In most of the practical situations force is not constant during its time of action.

Variation of force with time is also not known.

In such situations, it is not easy to calculate impulse as an integral.

Example :

Consider a collision of a "baseball bat" and "ball".

Here bat exerts force on the ball and ball exerts force on the bat.

It is easy to measure start time CenterOfMass_LinearMomentum_96.gif when the ball touches the bat and ends time CenterOfMass_LinearMomentum_97.gif when the two object separate.

The momentum of the ball is changing during this time interval.

Suppose force F(t) at the time of collision, acting on the ball, is known as a function of time t. We can compute the impulse j along acting on the ball.



This indicates, impulse j is the area under the curve shown in Fig. 1.

In majority of the cases we do not know functional form of force F(t), therefore we can not compute impulse by this method.

But we can always measure initial momentum CenterOfMass_LinearMomentum_100.gif and final momentum  CenterOfMass_LinearMomentum_101.gif the ball.

Therefore change in momentum CenterOfMass_LinearMomentum_102.gif is given as


Impulse can be estimated from the change in momentum Δp.

Can we compute average force CenterOfMass_LinearMomentum_104.gif acting on the ball from the measured impulse?

Answer is YES.

We can assume a constant average force CenterOfMass_LinearMomentum_105.gif acting on the ball during collision interval CenterOfMass_LinearMomentum_106.gif.

In terms of constant average force CenterOfMass_LinearMomentum_107.gif , the magnitude of impulse CenterOfMass_LinearMomentum_108.gif can be written as


Therefore average force acting on the ball can be estimated by dividing change in momentum Δp with time interval Δt.


In graphical terms, value of the magnitude of average force CenterOfMass_LinearMomentum_111.gif will be such that area under the curve in Fig. 2 is same as in Fig. 1.


Checkpoint-4: Collisions and Impulse

A paratrooper whose chute fails to open lands in snow; he is hurt slightly. Had he landed on bare ground, the stopping time would have been 10 times shorter and the collision lethal.
Does the presence of the snow increase, decrease, or leave unchanged the values of
(a) the paratrooper’s change in momentum,
(b) the impulse stopping the paratrooper, and
(c) the force stopping the paratrooper?

Checkpoint-5: Collisions and Impulse

The figure shows an overhead view of a ball bouncing from a vertical wall without any change in its speed. Consider the change  CenterOfMass_LinearMomentum_113.gif in the ball’s linear momentum.
(a) Is CenterOfMass_LinearMomentum_114.gif positive, negative, or zero?
(b) Is CenterOfMass_LinearMomentum_115.gif positive, negative, or zero?
(c) What is the direction of CenterOfMass_LinearMomentum_116.gif ?  


Example 3: (Collisions and Impulse)

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 70 kg, and the collision on the floor lasts 0.082 s.
What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Solution :

First we need to calculate velocity of the person just before the collision


Initial kinetic energy is zero, if v is the final velocity then


Therefore magnitude of v is


Now initial momentum CenterOfMass_LinearMomentum_121.gif is


Since final momentum CenterOfMass_LinearMomentum_123.gif, therefore


(a) Magnitude of impulse CenterOfMass_LinearMomentum_125.gif.




(a) Average force CenterOfMass_LinearMomentum_129.gif.





Conservation of Linear Momentum

We know that in the presence of net external force CenterOfMass_LinearMomentum_134.gif, the time rate of change of momentum is given as   


If no external force is present, CenterOfMass_LinearMomentum_136.gif, in that case




This means final momentum will be equal to the initial momentum.

If no net external force acts on a system of particles, the total linear momentum of the system cannot change.

This result is called the law of conservation of linear momentum.  

If only one component of the external force is zero, say  CenterOfMass_LinearMomentum_139.gif, in that case only x component CenterOfMass_LinearMomentum_140.gif of the momentum will be constant.




If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

Example 5: Conservation of Linear Momentum

A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s.
What speed does the man acquire as a result?

Solution :

Checkpoint-6: Conservation of linear Momentum

An initially stationary device lying on a frictionless floor explodes into two pieces, which then slide across the floor. One piece slides in the positive direction of an x axis. (a) What is the sum of the momenta of the two pieces after the explosion? (b) Can the second piece move at an angle to the x axis? (c) What is the direction of the momentum of the second piece?

Different types of collisions

In the absence of any external force, linear momentum is always conserved regardless of the collision type.

Elastic Collision:

When two or more objects collide in the absence of any external force, linear momentum of the system is conserved and collision is elastic if kinetic energy of the system is also conserved.


In an elastic collision, the kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change.

Inelastic Collision:

When two or more objects collide in the absence of any external force, linear momentum of the system is conserved and collision is inelastic if kinetic energy of the system is not conserved.


Checkpoint-6: Inelastic Collision

Body 1 and body 2 are in a completely inelastic one-dimensional collision. What is their final momentum if their initial momenta are, respectively, (a) 10 kg·m/s and 0; (b) 10 kg·m/s and 4 kg·m/s; (c) 10 kg·m/s and -4 kg·m/s

Checkpoint-7: Elastic Collision

What is the final linear momentum of the target (green sphere) in following figure, if the initial linear momentum of the projectile (orange sphere) is 6 kg·m/s and the final linear momentum of the projectile is (a) 2 kg·m/s and (b) −2 kg·m/s? (c) What is the final kinetic energy of the target if the initial and final kinetic energies of the projectile are, respectively, 5 J and 2 J?



Checkpoint-8: Two dimensional Collision

In following figure, suppose that the projectile has an initial momentum of 6 kg·m/s, a final x component of momentum of 4 kg·m/s, and a final y-component of momentum of −3 kg·m/s. For the target, what then are (a) the final x component of momentum and (b) the final y component of momentum?


Examples of Elastic and Inelastic Collisions :

For elastic collision Kinetic energy of the system is conserved.



Momentum is also conserved.



When we solve above two equations we can get the values of final velocities CenterOfMass_LinearMomentum_154.gif of both particles in terms of initial velocities CenterOfMass_LinearMomentum_155.gif and masses CenterOfMass_LinearMomentum_156.gif.