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 a System of 'N' Discrete Particles

${\vec{r}}_{\text{cm}}=\frac{1}{M}{\sum }_{i=1}^n{m}_i{\vec{r}}_i$

In cartesian coordinate system

$$\begin{gathered} x_{\text{cm}}=\frac{\sum m_i{x}_i}{\sum m_i}, \\ {y}_{\text{cm}}=\frac{\sum m_i{y}_i}{\sum m_i}, \\ {z}_{\text{cm}}=\frac{\sum m_i{z}_i}{\sum m_i} \end{gathered}$$

Position of COM of Two Particles

• COM of two particles divides internally the line joining two particles in the inverse ratio of their masses.

$$\begin{gathered} x_{\text{cm}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\Rightarrow 0=\frac{-m_1{d}_1+m_2{d}_2}{m_1+m_2} \\ \Rightarrow m_1{d}_1=m_2{d}_2\Rightarrow \frac{d_1}{d_2}=\frac{m_2}{m_1} \\ {d}_1\propto \frac{1}{m} \\ {d}_1=d_2=1_{2}if{m}_1=m_2\text{The center of mass lies at the midpoint between two particles of equal mass.} \\ {d}_1>d_{2}if{m}_1<m_{2}and{d}_1<d_{2}if{m}_1>m_2\text{ COM is always nearer to the particle having larger mass.} \end{gathered}$$

2.0COM of a Continuous Mass Distribution

$$\begin{gathered} x_{\text{cm}}=\frac{\int xdm}{\int dm}, \\ {y}_{\text{cm}}=\frac{\int ydm}{\int dm}, \\ {z}_{\text{cm}}=\frac{\int zdm}{\int dm}, \\ {\vec{r}}_{\text{cm}}=\frac{1}{M}\int \vec{r}dm \end{gathered}$$

Note:

(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 $\lambda =\frac{M}{L}$

Surface mass density $\sigma =\frac{M}{A}$

Volume mass density $\rho =\frac{M}{V}$

Centre of Mass of  Geometrical Bodies

Bodies	Geometrical Diagram	Location of COM
Rod		$\frac{L}{2},0$
Circular Arc		$[0,\frac{2R}{\theta }\sin \frac{\theta }{2}]$
Semi-circular arc [Half-ring]		$[0,\frac{2R}{\pi }]$
Disc		$[0,\frac{4R}{3\theta }{\sin }^2\frac{\theta }{2}]$
Semi-disc		$[0,\frac{4R}{3\pi }]$
Hollow sphere		$[0,\frac{R}{2}]$
Solid sphere		$[0,\frac{3R}{8}]$
Hollow cone		$[0,\frac{H}{3}]$
Solid cone		$[0,\frac{H}{4}]$

3.0Motion of Center of Mass and Conservation of Linear Momentum

$$\begin{gathered} \text{Displacement of COM due to Displacement of Particles of System} \\ {\overrightarrow{\Delta r}}_{\text{cm}}=\frac{m_1{\overrightarrow{\Delta r}}_1+m_2{\overrightarrow{\Delta r}}_2+\cdots }{m_1+m_2+\cdots } \\ \text{Velocity of Center of Mass of system} \\ M{\vec{v}}_{\text{cm}}=m_1\frac{d{\vec{r}}_1}{dt}+m_2\frac{d{\vec{r}}_2}{dt}+m_3\frac{d{\vec{r}}_3}{dt}+\cdots +m_n\frac{d{\vec{r}}_n}{dt} \\ M{\vec{v}}_{\text{cm}}=m_1{\vec{v}}_1+m_2{\vec{v}}_2+m_3{\vec{v}}_3+\cdots +m_n{\vec{v}}_n \\ \therefore {\vec{P}}_{\text{system}}=M{\vec{v}}_{\text{cm}} \\ \text{Acceleration of Center of Mass of system} \\ M{\vec{a}}_{\text{cm}}=m_1\frac{d{\vec{v}}_1}{dt}+m_2\frac{d{\vec{v}}_2}{dt}+m_3\frac{d{\vec{v}}_3}{dt}+\cdots +m_n\frac{d{\vec{v}}_n}{dt} \\ M{\vec{a}}_{\text{cm}}=m_1{\vec{a}}_1+m_2{\vec{a}}_2+m_3{\vec{a}}_3+\cdots +m_n{\vec{a}}_n \\ {\vec{F}}_{\text{ext}}=M{\vec{a}}_{\text{cm}} \\ \text{If }{\vec{F}}_{\text{ext}}=0\text{ then }{\vec{v}}_{\text{cm}}=\text{Constant} \\ \text{If the net external force on a system is zero, its total momentum remains constant.} \end{gathered}$$

Motion of the Center of Mass in a Moving System

4.0Impulse

Impulse of a force F exerting on a body for the time period $T=t_{1}tot=t_2$ is defined as :

$$\begin{gathered} \vec{I}={\int }_{t_1}^{t_2}\vec{F}dt\Rightarrow \vec{I}=\int \vec{F}dt=\int m\frac{d\vec{v}}{dt}dt=\int md\vec{v} \\ \vec{I}=m({\vec{v}}_2-{\vec{v}}_1)=\Delta \vec{P}=\text{change in momentum due to force} \\ {\vec{I}}_{\text{res}}={\int }_{t_1}^{t_2}{\vec{F}}_{\text{res}}dt=\Delta \vec{P}\text{(impulse-momentum theorem)} \end{gathered}$$

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.

$\vec{I}=\int \vec{F}dt=\Delta \vec{P}={\vec{P}}_f-{\vec{P}}_i$

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

1. Gravitational and spring forces are always non-impulsive.
2. Normal, tension, and friction forces are case-dependent.
3. 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

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

• $$\begin{gathered} e=\frac{\text{impulse of recovery}}{\text{impulse of deformation}}=\frac{\int Rdt}{\int Bdt} \\ e=\frac{\text{velocity of separation along line of impact}}{\text{velocity of approach along line of impact}}=\frac{|v_B^{\prime }-v_A^{\prime }|}{|u_A-u_B|} \end{gathered}$$
• 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

• In an elastic collision, if the surface is rough, then Kf< Ki due to impulsive friction. 

$K_f<K_i$

• 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):

1. Draw the line of impact (normal at contact point).
2. Resolve velocity components of both contact points along this line before and after collision.
3. Use:

$e=\frac{\text{velocity of separation along line of impact}}{\text{velocity of approach along line of impact}}$

$e=\frac{V_{2x}-V_{1x}}{u_{1x}-u_{2x}}$

Collision in one dimension (Head on)

$$\begin{gathered} e=\frac{v_2-v_1}{u_1-u_2},(u_1-u_2)e=(v_2-v_1) \\ {m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2 \\ {v}_2=v_1+e(u_1-u_2)and{v}_1=\frac{m_1{u}_1+m_2{u}_2-m_{2}e(u_1-u_2)}{m_1+m_2} \\ {v}_2=\frac{m_1{u}_1+m_2{u}_2-m_{2}e(u_1-u_2)}{m_1+m_2} \end{gathered}$$

Collision in two dimension (oblique)

1. Impulses act along the common normal, changing the linear momentum and velocity in that direction.
2. No impulse acts along the tangent, so momentum and velocity stay unchanged in the tangential direction.
3. Net impulse is zero, so the total momentum of the system is conserved in all directions.
4. 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:

1. The speed of the ball after reflection, and
2. 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:

$v=e{v}_{0}cos\alpha$

After collision components of velocity v' are v0 sin and $v=e{v}_{0}cos\alpha$
$|v^{\prime }|=\sqrt{(v_0\sin \alpha {)}^2+(e{v}_0\cos \alpha {)}^2}$

$\tan \beta =\frac{v_0\sin \alpha }{e{v}_0\cos \alpha }=\frac{\tan \alpha }{e}$

Note: For elastic collision $(e=1)v^{\prime }=v_0,and\beta =\alpha$

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 $\mu \frac{kg}{s}$ kg/s and relative velocity ${\vec{v}}_{\text{rel}}$ then the force exerted by this mass on the system has magnitude $\mu \left|{\vec{v}}_{\text{rel}}\right|$

${\vec{F}}_t={\vec{v}}_{\text{rel}}\left(\frac{dm}{dt}\right)$

Steps to Solve Variable Mass Problems

1.List all forces acting on the object and apply them to the mass.

2.Apply thrust force ${\vec{F}}_t$ with magnitude $(\vec{v}\frac{dm}{dt})$ directed along ${\vec{v}}_r$ if mass is increasing or $-{\vec{v}}_r$ if mass is decreasing.

3.Find the net force: ${\vec{F}}_{\text{net}}=m\frac{d\vec{v}}{dt}$ where m is the mass at a particular instant.

4.Integrate with appropriate limits to find the velocity at time t.

Rocket propulsion

$$\begin{gathered} \text{Thrust force on the rocket},F_t=v_r\left(-\frac{dm}{dt}\right)\text{(upwards)} \\ \text{Weight of the rocket,W=mg downwards} \\ \text{Net force on the rocket},F_{\text{net}}=F_t-W \\ {F}_{\text{net}}=v_r\left(-\frac{dm}{dt}\right)-mg \\ \text{Net acceleration of the rocket},a=\frac{F}{m} \\ v=-gt+v_r\ln \left(\frac{m_0}{m}\right)\dots \dots ..(1) \end{gathered}$$

Note:

1.$$\begin{gathered} F_t=v_r\left(-\frac{dm}{dt}\right)\text{is upwards, as}{v}_r\text{is downwards and}\frac{dm}{dt}\text{is negative.} \\ 2.\text{If gravity is ignored and initial velocity of the rocket u=0 equation (1) reduces to}v=v_r\ln \left(\frac{m_0}{m}\right) \end{gathered}$$

8.0C- Frame

1. A C-frame of reference is attached to the center of mass (CM) of a system, where the CM is at rest $v_{CM}=0$
2. In the C-frame, the total momentum of the system is always zero: $\vec{P}=\sum {\vec{P}}_i$. The C-frame is sometimes called the zero momentum frame.
3. The C-frame moves with velocity $v_{cm}$ relative to the ground frame.
4. When no external forces act on the system, the C-frame can be considered inertial.
5. Analyzing problems in the C-frame is often simpler than in the ground frame.

A system of two particles :

• Given two particles with masses $m_{1}and{m}_2$ and velocities $v_{1}and{v}_2$
• in the K- reference Frame we will derive the expressions for their momentum and the total kinetic energy in the C-Frame.

$$\begin{gathered} \text{The momentum of the first particle in the C-system is} \\ {\vec{P}}_{1/c}=m_1{\vec{v}}_{1/c}=m_1\left({\vec{v}}_1-{\vec{V}}_c\right) \\ {\vec{V}}_c\text{ is the velocity of the centre of mass (of the C system) in the K reference frame.} \\ {\vec{V}}_c=\frac{1}{m}\sum m_i{\vec{v}}_i=\frac{1}{m}\sum {\vec{p}}_i \\ {\vec{p}}_i=\mu ({\vec{v}}_1-{\vec{v}}_2) \\ \mu =\frac{m_1{m}_2}{m_1+m_2}=\text{reduced mass} \\ \text{The momentum of the second particle in the C frame is} \\ {\vec{P}}_{2/c}=\mu ({\vec{v}}_2-{\vec{v}}_1) \\ \text{In the C-frame, the momenta of the two particles are equal in magnitude but opposite in direction; the magnitude of each particle’s momentum is.} \\ {\vec{P}}_{1/c}=\mu {\vec{v}}_{\text{rel}} \\ \text{The total kinetic energy of the two particles in the C-frame is} \\ {K}_{s/c}=K_1+K_2=\frac{P^2}{2m_1}+\frac{P^2}{2m_2} \\ \mu =\frac{m_1{m}_2}{m_1+m_2},\frac{1}{m_1+m_2}=\frac{1}{m_1}+\frac{1}{m_2}=\frac{1}{\mu } \\ {K}_{s/c}=\frac{P^2}{2\mu }=\frac{\mu v_{\text{rel}}^2}{2} \\ \text{If the particles interact, then total mechanical energy in the C frame is E=T+U Where U is the potential energy of interaction of the given particles.} \end{gathered}$$