The center of mass (CM) represents the system's total mass and moves with constant velocity or stays at rest without external forces. It simplifies motion analysis and helps apply momentum conservation in collisions.
In a perfectly elastic collision, the total kinetic energy and momentum are conserved because no energy is lost to heat, sound, or deformation. This is because the internal forces between colliding particles do not dissipate energy; they only change the direction of motion without altering the total kinetic energy. Momentum conservation holds due to Newton's Third Law, which states that the forces between the colliding particles are equal and opposite.
Momentum is conserved in collisions due to Newton’s Third Law of motion, which states that the forces between the colliding objects are equal and opposite. The internal forces during the deformation of objects do not affect the total momentum of the system. The deformation may convert kinetic energy into potential energy (such as in elastic deformation), but the total momentum of the system remains constant if no external forces are acting.
In a collision, the normal velocity components change according to the coefficient of restitution, while tangential components remain unchanged if there's no friction. Analyzing these components separately simplifies understanding and calculations of post-collision behavior.
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Centre of Mass And Collision
The Center of Mass is the point where the entire mass of a system is concentrated and plays a crucial role in analyzing motion, especially during collisions. Collisions are interactions where objects exert forces on each other for a short time. They are classified as elastic (both momentum and kinetic energy conserved), inelastic (momentum conserved, but not kinetic energy), and perfectly inelastic (objects stick together). The key principle in all collisions is the conservation of linear momentum, and in elastic collisions, kinetic energy is also conserved. Analyzing collisions in the center of mass frame simplifies the problem by making the total momentum zero.
1.0Definition of Centre of Mass
It is the point in a system where the total mass can be considered concentrated, and it moves as if all external forces act at that point.
COM of two particles divides internally the line joining two particles in the inverse ratio of their masses.
xcm=m1+m2m1x1+m2x2⇒0=m1+m2−m1d1+m2d2⇒m1d1=m2d2⇒d2d1=m1m2d1∝m1d1=d2=12ifm1=m2The center of mass lies at the midpoint between two particles of equal mass.d1>d2ifm1<m2andd1<d2ifm1>m2 COM is always nearer to the particle having larger mass.
(1) If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and vice-versa.
(2) The center of mass of uniform bodies lies on their geometric centres.
Types of mass distributions
Linear mass density
λ=LM
Surface mass density
σ=AM
Volume mass density
ρ=VM
Centre of Mass of Geometrical Bodies
Bodies
Geometrical Diagram
Location of COM
Rod
2L,0
Circular Arc
[0,θ2Rsin2θ]
Semi-circular arc
[Half-ring]
[0,π2R]
Disc
[0,3θ4Rsin22θ]
Semi-disc
[0,3π4R]
Hollow sphere
[0,2R]
Solid sphere
[0,83R]
Hollow cone
[0,3H]
Solid cone
[0,4H]
3.0Motion of Center of Mass and Conservation of Linear Momentum
Displacement of COM due to Displacement of Particles of SystemΔrcm=m1+m2+⋯m1Δr1+m2Δr2+⋯Velocity of Center of Mass of systemMvcm=m1dtdr1+m2dtdr2+m3dtdr3+⋯+mndtdrnMvcm=m1v1+m2v2+m3v3+⋯+mnvn∴Psystem=MvcmAcceleration of Center of Mass of systemMacm=m1dtdv1+m2dtdv2+m3dtdv3+⋯+mndtdvnMacm=m1a1+m2a2+m3a3+⋯+mnanFext=MacmIf Fext=0 then vcm=ConstantIf the net external force on a system is zero, its total momentum remains constant.
Motion of the Center of Mass in a Moving System
COM at Rest
COM Moving With Uniform Velocity
COM Moving With Acceleration
If Fext=0andVcm=0 the center of mass remains at rest. While individual components may have non-zero momentum due to internal forces, the system's net momentum stays zero.
If Fext=0 the velocity of the center of mass remains constant, and the system's net momentum is conserved. While individual components may change velocity due to internal forces, the total momentum stays constant, and the COM moves with its initial velocity.
If an external force acts, the COM moves as if that force is solely acting on it, regardless of internal forces.
4.0Impulse
Impulse of a force F exerting on a body for the time period T=t1tot=t2 is defined as :
I=∫t1t2Fdt⇒I=∫Fdt=∫mdtdvdt=∫mdvI=m(v2−v1)=ΔP=change in momentum due to forceIres=∫t1t2Fresdt=ΔP(impulse-momentum theorem)
Note: Impulse applied to an object over a time interval can be determined from the area under the force-time (F-t) graph for that interval.
Instantaneous Impulse
It describes scenarios where a force acts briefly, like a bat hitting a ball. Although the force's magnitude and duration may be unknown, impulse (force × time) can be calculated by measuring the initial and final momenta.
I=∫Fdt=ΔP=Pf−Pi
Impulsive force
An impulsive force is a force of high magnitude that acts over a very short time, causing a significant change in momentum. Impulsive forces are often associated with collisions, where the application time is so brief that the particle's motion is minimal. The distinction between impulsive and non-impulsive forces is relative, with no clear boundary.
Key Points
Gravitational and spring forces are always non-impulsive.
Normal, tension, and friction forces are case-dependent.
An impulsive force can only be countered by another impulsive force.
Collision or Impact
It is an event where an impulsive force acts briefly between bodies, causing a change in their velocities.
Note on Collisions: (a) Physical contact between particles is not necessary. (b) Collision duration Δt is very short compared to typical motion times. (c) External non-impulsive forces (e.g., gravity) are neglected, as impulsive forces during collision are much larger.
Note:A collision is essentially a redistribution of total momentum between particles, making the law of conservation of linear momentum essential for its analysis.
Line of Impact
The line of impact, along which the collision force acts on both bodies, passes through the common normal to the surfaces in contact. Its direction is determined by:
(a) The geometry of the colliding objects
(b) The direction of momentum changes.
If one particle is stationary before the collision, the line of impact aligns with its motion after the collision.
5.0Classification of Collisions
Basis
Type
Description
Line of Impact
(i) Head-on Collision
Velocities are along the same line before and after collision.
(ii) Oblique Collision
Velocities are along different lines before and/or after collision.
Energy Consideration
(i) Elastic Collision
- Shape fully restored after collision. - Momentum and KE conserved.
(ii) Inelastic Collision
- Partial shape loss. - Only momentum conserved, some KE lost.
(iii) Perfectly Inelastic
- Particles stick together. - Max KE lost, momentum still conserved.
Note:Collisions between real objects are neither perfectly elastic nor perfectly inelastic; they are generally inelastic.
Examples of Line of Impact and Related Collisions
(1) Two balls A and B are approaching each other such that their centers are moving along line CD.
Head on Collision (If line of motion and line of impact is same)
(2)Two balls A and B are approaching each other such that their centers are moving along dotted lines as shown in figure.
Oblique Collision
(3) The ball is falling on a stationary wedge.
6.0Coefficient of Restitution
e=impulse of deformationimpulse of recovery=∫Bdt∫Rdte=velocity of approach along line of impactvelocity of separation along line of impact=∣uA−uB∣∣vB′−vA′∣
The ratio of magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e.
Note:The coefficient of restitution (e) depends on the material, not the shape or mass of the object, and is constant for a pair of materials. In contact collisions, e is always less than one.
0 ≤ e ≤ 1
Case
Value of e
Impulse
Velocity Relation
Kinetic Energy
Collision Type
Elastic Collision
e=1
Reformation = Deformation
Separation = Approach
KE conserved
Elastic
Inelastic Collision
0<e<1
Reformation < Deformation
Separation < Approach
KE partially lost
Inelastic
Perfectly Inelastic
e=0
Reformation = 0
Separation = 0 (objects stick together)
Maximum KE loss
Perfectly Inelastic
In an elastic collision, if the surface is rough, then Kf< Ki due to impulsive friction.
Kf<Ki
K refers to kinetic energy; friction causes partial loss even in elastic collisions.
When a particle B collides with a moving rod at point C, analyze motion using the line of impact.
To find the coefficient of restitution (e):
Draw the line of impact (normal at contact point).
Resolve velocity components of both contact points along this line before and after collision.
Use:
e=velocity of approach along line of impactvelocity of separation along line of impact
v1=v2 Both bodies move with the same velocity after collision.
2
e=1,m1=m2=m
v1=u2,v2=u1 Velocities are exchanged (head-on elastic).
3
m1>>m2
v2=u1+e(u1−u2)
Collision in two dimension (oblique)
Impulses act along the common normal, changing the linear momentum and velocity in that direction.
No impulse acts along the tangent, so momentum and velocity stay unchanged in the tangential direction.
Net impulse is zero, so the total momentum of the system is conserved in all directions.
Coefficient of restitution (e) is applied only along the common normal:
Relative speed of separation=e ✕ Relative speed of approach
Illustration-1:A ball of mass m strikes the floor with speed v0 at an angle to the vertical (normal). If the coefficient of restitution is e Determine:
The speed of the ball after reflection, and
The angle of reflection w.r.t the normal.
Solution:
The tangential component of velocity, v0 sin , remains unchanged after the collision.
Let v represent the normal component of the velocity after the collision.
Applying the coefficient of restitution e along the common normal direction:
This gives the normal component after the collision as:
v=ev0cosα
After collision components of velocity v' are v0 sin and v=ev0cosα ∣v′∣=(v0sinα)2+(ev0cosα)2
tanβ=ev0cosαv0sinα=etanα
Note: For elastic collision (e=1)v′=v0,andβ=α
Illustration-2:A ball of mass m collides elastically with another identical ball at rest. Prove that if the collision is oblique, the balls will move at orthogonal to each other after the collision.
Solution:
Head-on elastic collision: The two particles exchange velocities.
For Ball 1:
The normal component vcosθ becomes zero after the collision.
The tangential component remains unchanged.
For Ball 2:
The normal component vcosθ becomes zero after the collision.
The tangential component remains unchanged.
Note: When two identical bodies collide elastically at an angle, with one body at rest before the collision, the bodies will move in perpendicular directions afterward.
7.0Variable Mass System
If mass is added or ejected , at a rate of μskg kg/s and relative velocity vrel then the force exerted by this mass on the system has magnitude μ∣vrel∣
Ft=vrel(dtdm)
Steps to Solve Variable Mass Problems
1.List all forces acting on the object and apply them to the mass.
2.Apply thrust force Ft with magnitude (vdtdm) directed along vr if mass is increasing or −vr if mass is decreasing.
3.Find the net force: Fnet=mdtdv where m is the mass at a particular instant.
4.Integrate with appropriate limits to find the velocity at time t.
Rocket propulsion
Thrust force on the rocket,Ft=vr(−dtdm)(upwards)Weight of the rocket,W=mg downwardsNet force on the rocket,Fnet=Ft−WFnet=vr(−dtdm)−mgNet acceleration of the rocket,a=mFv=−gt+vrln(mm0)……..(1)
Note:
1.Ft=vr(−dtdm)is upwards, asvris downwards anddtdmis negative.2.If gravity is ignored and initial velocity of the rocket u=0 equation (1) reduces tov=vrln(mm0)
8.0C- Frame
A C-frame of reference is attached to the center of mass (CM) of a system, where the CM is at rest vCM=0
In the C-frame, the total momentum of the system is always zero: P=∑Pi. The C-frame is sometimes called the zero momentum frame.
The C-frame moves with velocity vcm relative to the ground frame.
When no external forces act on the system, the C-frame can be considered inertial.
Analyzing problems in the C-frame is often simpler than in the ground frame.
A system of two particles :
Given two particles with masses m1andm2and velocities v1andv2
in the K- reference Frame we will derive the expressions for their momentum and the total kinetic energy in the C-Frame.
The momentum of the first particle in the C-system isP1/c=m1v1/c=m1(v1−Vc)Vc is the velocity of the centre of mass (of the C system) in the K reference frame.Vc=m1∑mivi=m1∑pipi=μ(v1−v2)μ=m1+m2m1m2=reduced massThe momentum of the second particle in the C frame isP2/c=μ(v2−v1)In the C-frame, the momenta of the two particles are equal in magnitude but opposite in direction; the magnitude of each particle’s momentum is.P1/c=μvrelThe total kinetic energy of the two particles in the C-frame isKs/c=K1+K2=2m1P2+2m2P2μ=m1+m2m1m2,m1+m21=m11+m21=μ1Ks/c=2μP2=2μvrel2If the particles interact, then total mechanical energy in the C frame is E=T+U Where U is the potential energy of interaction of the given particles.
Table of Contents
1.0Definition of Centre of Mass
2.0COM of a Continuous Mass Distribution
3.0Motion of Center of Mass and Conservation of Linear Momentum
