1. Centre of Mass:

Particle: A point mass with negligible size when compared to the distances covered.

Mass Moment: $\vec{M} = m \vec{r}$

Centre of mass: For a system of particles centre of mass is that point at which its total mass is supposed to be concentrated.

(i) Centre of mass of the system of discrete particles:

Total mass of the body: $M = m_{1} + m_{2} + \dots + m_{n}$ , then

${\vec{R}}_{cm} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + m_{3} {\vec{r}}_{3} + \dots}{m_{1} + m_{2} + m_{3} + \dots} = \frac{1}{M} \sum_{i = 1}^{n} m_{i} z_{i}$

Coordinates of the centre of mass:

$x_{cm} = \frac{1}{M} Σ m_{i} x_{i}, y_{cm} = \frac{1}{M} Σ m_{i} y_{i}$ and $z_{cm} = \frac{1}{M} Σ m_{i} z_{i}$

(ii) Centre of mass of continuous distribution of particles:

${\vec{R}}_{cm} = \frac{1}{M} \int \vec{r} d m$

$x_{cm} = \frac{1}{M} \int x d m, y_{cm} = \frac{1}{M} \int y d m$ and $z_{cm} = \frac{1}{M} \int z d m$

$x, y and z$ are the coordinates of the COM of the $d m$ mass.

2. Centre of Mass of Some Common Systems:

(i) A system of two-point masses, $m_{1} r_{1} = m_{2} r_{2}$ .

(ii) Rectangular plate (By symmetry)		$x_{cm} = \frac{b}{2}, y_{cm} = \frac{L}{2}$
(iii) A triangular plate (By qualitative argument)		at the centroid: $y_{cm} = \frac{h}{3}$
(iv) A semi-circular ring		$y_{cm} = \frac{2 R}{π}, x_{cm} = 0$
(v) A semi-circular disc		$y_{cm} = \frac{4 R}{3 π}, x_{cm} = 0$
(vi) A hemispherical shell		$y_{cm} = \frac{R}{2}, x_{cm} = 0$
(vii) A solid hemisphere		$y_{cm} = \frac{3 R}{8}, x_{cm} = 0$
(viii) A circular cone solid		$y_{cm} = \frac{h}{4}$
(ix) A circular cone (hollow)		$y_{cm} = \frac{h}{3}$

Note: The centre of mass lies closer to the heavier mass.

(ii) The centre of mass after removal of a part of a body:

Original mass $(M) -$ mass of the removed part $(m) =$ Mass of the remaining body.

The formula changes to:

$x_{cm} = \frac{M x - m x'}{M - m}; y_{cm} = \frac{M y - m y'}{M - m}; z_{cm} = \frac{M z - m z'}{M - m}$

3. Motion of Centre of Mass:

(i) Velocity of centre of mass of system:

${\vec{v}}_{cm} = \frac{m_{1} \frac{d {\vec{r}}_{1}}{d t} + m_{2} \frac{d {\vec{r}}_{2}}{d t} + m_{3} \frac{d {\vec{r}}_{3}}{d t} \dots . . + m_{n} \frac{d {\vec{r}}_{n}}{d t}}{M}$ $= \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3} \dots . . + m_{n} {\vec{v}}_{n}}{M} = \frac{Σ {\vec{p}}_{i}}{M} = \frac{{\vec{p}}_{system}}{M}$

$\Rightarrow {\vec{p}}_{system} = M {\vec{v}}_{cm}$

(ii) Acceleration of centre of mass of system:

${\vec{a}}_{cm} = \frac{m_{1} \frac{d {\vec{v}}_{1}}{d t} + m_{2} \frac{d {\vec{v}}_{2}}{d t} + m_{3} \frac{d {\vec{v}}_{3}}{d t} \dots . . + m_{n} \frac{d {\vec{v}}_{n}}{d t}}{M}$

$\Rightarrow {\vec{a}}_{cm} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2} + m_{3} {\vec{a}}_{3} \dots \dots \dots + m_{n} {\vec{a}}_{n}}{M}$

$\Rightarrow {\vec{a}}_{cm} = \frac{Net force on system}{M} = \frac{Net External Force + Net internal Force}{M}$

$\Rightarrow {\vec{a}}_{cm} = \frac{Net External Force}{M}$

$\Rightarrow {\vec{F}}_{e x t} = M {\vec{a}}_{cm}$

(iii) If $F_{ext} = 0$ and $v_{cm} = 0$ , then the centre of mass 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.

(iv) If $F_{ext} = 0$ and $v_{cm}$ remains constant, therefore, the net momentum of the system also remains conserved. Individual components of the system may have variable velocity and momentum due to the internal forces, but the net momentum of the system remains constant and the centre of mass continues to move with the constant velocity.

(v) If an external force is present then centre of mass continues its original motion as if the external force is acting on it, irrespective of internal forces.

4. Linear Momentum of a System of Particles and Conservation of Linear Momentum:

The total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass, ${\vec{p}}_{system} = M {\vec{v}}_{cm}$

According to Newton’s second law of motion, ${\vec{F}}_{ext} = M {\vec{a}}_{cm}$ and $\frac{d {\vec{p}}_{system}}{d t} = {\vec{F}}_{ext}$

Suppose the net external force acting on the system of particles is zero then, $\frac{d {\vec{p}}_{system}}{d t} = 0$ and ${\vec{p}}_{system} =$ constant.

Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum of a system of particles.

5. Impulse:

Impulse of a force $\vec{F}$ acting on a body is defined as:-

$\vec{J} = \int_{t_{i}}^{t_{f}} \vec{F} d t =$ Area under the force-time graph

$\vec{J} = Δ \vec{p}$ (impulse-momentum theorem)

(i) Units:

(a) SI Unit: $(N s)$ or $(kg m s^{- 1})$

(b) CGS Unit: dyne-second

(ii) Dimensions: ${MLT}^{- 1}$

Important points:

(i) Gravitational force and spring force are always non-impulsive.

(ii) An impulsive force can only be balanced by another impulsive force.

6. Collision of Bodies:

The event or the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart and affect each other's motion (velocity, momentum, energy or direction of motion) is defined as a collision.

In collision, the particles come closer before collision and after collision they either stick together or move away from each other. The particles need not come in contact with each other for a collision.

The law of conservation of linear momentum is necessarily applicable in a collision if the system is isolated(no external force), whereas the law of conservation of mechanical energy is not.

The line passing through the common normal to the surfaces in contact during collision is called line of impact(LOI). It is also the line along which the normal contact force acts during the collision.

(i) On the basis of line of impact, collisions are classified as:
(a) Head-on collision: If the velocities of the colliding bodies before and after collision are along the line of impact.
(b) Oblique or $2 D$ collision: If the velocities of the colliding bodies before and after collision are not along the line of impact.

(ii) On the basis of energy, collisions are classified as:
(a) Elastic collision, (b) Inelastic collision and (c) Perfectly inelastic collision

7. Coefficient of Restitution $(e)$ :

$e = \frac{I m p u l s e o f r e f o r m a t i o n}{I m p u l s e o f d e f o r m a t i o n} = \frac{\int {\vec{F}}_{r} d t}{\int {\vec{F}}_{d} d t}$ $= \frac{V e l o c i t y o f s e p a r a t i o n a l o n g l i n e o f i m p a c t}{V e l o c i t y o f a p p r o a c h a l o n g l i n e o f i m p a c t}$

(i) Elastic Collision

e = 1

$\Rightarrow$ Impulse of Reformation $=$ Impulse of Deformation

$\Rightarrow$ Velocity of separation $=$ Velocity of approach

$\Rightarrow$ Kinetic Energy is conserved.

(ii) Perfectly Inelastic Collision

e = 0

$\Rightarrow$ Impulse of Reformation $= 0$

$\Rightarrow$ Velocity of separation $= 0$

$\Rightarrow$ Particles moves together after collision.

$\Rightarrow$ Kinetic Energy is not conserved.

(iii) Inelastic Collision

0 < e < 1

$\Rightarrow$ Impulse of Reformation $<$ Impulse of Deformation

$\Rightarrow$ Velocity of separation $<$ Velocity of approach

$\Rightarrow$ Kinetic Energy is not conserved.

8. Head-on Collision:

Consider the Head-on inelastic collision of two particles,

Let the coefficient of restitution for collision is $e$ ,

(i) Momentum is conserved, $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \dots (i)$

(ii) Kinetic energy is not conserved.

(iii) According to Newton's law $e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}} \dots (i i)$

By solving equations (i) and (ii):

$v_{1} = (\frac{m_{1} - e m_{2}}{m_{1} + m_{2}}) u_{1} + (\frac{(1 + e) m_{2}}{m_{1} + m_{2}}) u_{2} = \frac{m_{1} u_{1} + m_{2} u_{2} - m_{2} e (u_{1} - u_{2})}{m_{1} + m_{2}}$

$v_{2} = (\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}) u_{2} + (\frac{(1 + e) m_{1}}{m_{1} + m_{2}}) u_{1} = \frac{m_{1} u_{1} + m_{2} u_{2} - m_{1} e (u_{2} - u_{1})}{m_{1} + m_{2}}$

9. Elastic Collision $(e = 1)$ :

(i) If the two bodies are of equal masses: $m_{1} = m_{2} = m$ , then $v_{1} = u_{2}$ and $v_{2} = u_{1}$

Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the bodies will exchange their velocities.

(ii) If the mass of a body is negligible as compared to other.

(a) If $m_{1} ≫ m_{2}$ and $u_{2} = 0$ then $v_{1} = u_{1}, v_{2} = 2 u_{1}$ .

When a heavy body $A$ collides against a light body $B$ at rest, the body $A$ should keep on moving with the same velocity and the body $B$ will move with velocity double that of $A$ .

(b) If $m_{2} ≫ m_{1}$ and $u_{2} = 0$ then $v_{2} = 0, v_{1} = - u_{1}$ .

When a light body $A$ collides against a heavy body $B$ at rest, the body $A$ should start moving with the same velocity just in opposite direction while the body $B$ should practically remain at rest.

10. Inelastic Collision:

Loss in kinetic energy in an inelastic collision,

$Δ K = \frac{m_{1} m_{2}}{2 (m_{1} + m_{2})} (1 - e^{2}) {(u_{1} - u_{2})}^{2}$

11. Oblique Collision:

Conserving the momentum of the system in directions along normal ( $x$ axis in our case) and tangential ( $y$ axis in our case),

$m_{1} u_{1} \cos α_{1} + m_{2} u_{2} \cos α_{2} = m_{1} v_{1} \cos β_{1} + m_{2} v_{2} \cos β_{2}$ and

$m_{2} u_{2} \sin α_{2} - m_{1} u_{1} \sin α_{1} = m_{2} v_{2} \sin β_{2} - m_{1} v_{1} \sin β_{1}$

Since no force is acting on $m_{1}$ and $m_{2}$ along the tangent (i.e. $y$ -axis) the individual momentum of $m_{1}$ and $m_{2}$ remains conserved. Therefore,

$m_{1} u_{1} \sin α_{1} = m_{1} v_{1} \sin β_{1}$ and $m_{2} u_{2} \sin α_{2} = m_{2} v_{2} \sin β_{2}$

By using Newton's experimental law along the line of impact, $e = \frac{v_{2} \cos β_{2} - v_{1} \cos β_{1}}{u_{1} \cos α_{1} - u_{2} \cos α_{2}}$

12. Variable Mass System:

If a mass is added or ejected from a system, at rate $μ kg s^{- 1}$ and relative velocity ${\vec{v}}_{rel}$ (w.r.t. the system), then the force exerted by this mass on the system has magnitude $μ |{\vec{v}}_{rel}|$ .

Thrust Force, ${\vec{F}}_{t} = {\vec{v}}_{rel} (\frac{d m}{d t})$

13. Rocket Propulsion:

At $t = 0$

$v = u$

$m = m_{0}$

At $t = t$

$v = v$

$m = m$

exhaust velocity $= v_{r}$

Thrust force on the rocket, $F_{r} = v_{r} (- \frac{d m}{d t})$

Velocity of the rocket at any instant, $v = u - g t + v_{r} l n (\frac{m_{0}}{m})$

If gravity is ignored and the initial velocity of the rocket, $u = 0$ , then, $v = v_{r} l n (\frac{m_{0}}{m})$

Two particles of mass m and 2m have their position vectors as a function of time as r1t=t i^-t3 j^+2t2 k^ and r1t=t i^-t3 j^-t2 k^ respectively (where t is the time). Which one of the following graphs represents the path of the centre of mass?

Important Questions on Centre of Mass, Momentum and Collisions

Learn and Prepare for any exam you want !