3.1 Motion of mass of the system

1.1 Position of centre of mass for a system of two particles

2.1 Rod
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies

3.1 Motion of mass of the system

5.1 Rocket Propulsion

6.1 Instantaneous impulse

9.1 For direct collision
9.2 For collision in two dimension
9.3 For oblique collision

10.1 Head on elastic collision
10.2 Head on inelastic collision

Position of centre of mass for a system of n particles is given by, $${\overrightarrow r _{COM}} = \frac{{{m_1}{{\overrightarrow r }_1} + {m_2}{{\overrightarrow r }_2} + ... + {m_n}{{\overrightarrow r }_n}}}{{{m_1} + {m_2} + ... + {m_n}}}$$ where,

$M = {m_1} + {m_2} + ... + {m_n}$, (Total mass of the system)

So, $$M{\overrightarrow r _{COM}} = {m_1}{\overrightarrow r _1} + {m_2}{\overrightarrow r _2} + ... + {\overrightarrow r _n}\quad ...(i)$$

Differentiating the above equation wrt $t$ we get, $$M\frac{{d{{\overrightarrow r }_{COM}}}}{{dt}} = {m_1}\frac{{d{{\overrightarrow r }_1}}}{{dt}} + {m_2}\frac{{d{{\overrightarrow r }_2}}}{{dt}} + ... + {m_n}\frac{{d{{\overrightarrow r }_n}}}{{dt}}$$ or $$M{\overrightarrow v _{COM}} = {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2} + ... + {m_n}{\overrightarrow v _n}\quad ...(ii)$$

Differentiating the equation $(ii)$ wrt $t$ we get, $$M\frac{{d{{\overrightarrow v }_{COM}}}}{{dt}} = {m_1}\frac{{d{{\overrightarrow v }_1}}}{{dt}} + {m_2}\frac{{d{{\overrightarrow v }_2}}}{{dt}} + ... + {m_n}\frac{{d{{\overrightarrow v }_n}}}{{dt}}$$ or $$M{\overrightarrow a _{COM}} = {m_1}{\overrightarrow a _1} + {m_2}{\overrightarrow a _2} + ... + {m_n}{\overrightarrow a _n}\quad ...(iii)$$

We can write equation $(ii)$ as, $m\overrightarrow v = \overrightarrow p $, (linear momentum),

We can write equation $(iii)$ as, $m\overrightarrow a = \overrightarrow F $, (linear momentum),

Summarizing all the equations related to the motion of centre of mass,

Note: If no external force is applied on the system, centre of mass $(COM)$ of the system will be at rest or moves with constant velocity.

Let us understand the above concept with the help of some illustrations.

Illustration 1: Consider a man of mass $M$ is standing on a plank of mass $m$ and length $l$. There is sufficient friction between the feet of the man and plank, so that man can easily walk from one end to the other on the plank. There is no friction between the plank and the surface on which it is kept.

Initial situation: The man is standing still at one corner of the plank and no external force acts on the system. So, the centre of mass of the system will be at rest.

Initial position of centre of mass of the systems is, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_{syste{m_i}}}}} = \frac{{{m_{Man}}{{\overrightarrow r }_{Man}} + {m_{Plank}}{{\overrightarrow r }_{Plank}}}}{{{m_{Man}} + {m_{Plank}}}} \\ {\overrightarrow r _{CO{M_{syste{m_i}}}}} = \frac{{M\left( {0\widehat i + y\widehat j} \right) + m\left( {\frac{l}{2}\widehat i + 0\widehat j} \right)}}{{M + m}} \\ {\overrightarrow r _{CO{M_{syste{m_i}}}}} = \frac{{My\widehat j + \frac{{ml}}{2}\widehat i}}{{M + m}}\quad ...(i) \\\end{aligned} \end{equation} $$

Final situation: Now, the man starts walking on plank from one end to another. Since no external force acts on the system. Therefore centre of mass will be at rest and will remain at the same position.

Also, let the plank is displaced by a distance $x$.

Final position of centre of mass of the system is, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_{syste{m_f}}}}} = \frac{{{m_{Man}}{{\overrightarrow r }_{Man}} + {m_{Plank}}{{\overrightarrow r }_{Plank}}}}{{{m_{Man}} + {m_{Plank}}}} \\ {\overrightarrow r _{CO{M_{syste{m_f}}}}} = \frac{{M\left[ {\left( {l + x} \right)\widehat i + y\widehat j} \right] + m\left[ {\left( {\frac{l}{2} + x} \right)\widehat i + 0\widehat j} \right]}}{{M + m}} \\ {\overrightarrow r _{CO{M_{syste{m_f}}}}} = \frac{{\left[ {M(l + x) + m\left( {\frac{l}{2} + x} \right)} \right]\widehat i + My\widehat j}}{{M + m}}\quad ...(ii) \\\end{aligned} \end{equation} $$

Since no external force acts on the system, so the centre of mass will be at rest. Therefore the position of centre of mass $(COM)$ will remain same.

$${\overrightarrow r _{CO{M_{syste{m_i}}}}} = {\overrightarrow r _{CO{M_{syste{m_f}}}}}$$ or $$\frac{{\frac{{ml}}{2}\widehat i + My\widehat j}}{{M + m}} = \frac{{\left[ {M(l + x) + m\left( {\frac{l}{2} + x} \right)} \right]\widehat i + My\widehat j}}{{M + m}}$$ On solving, $$x = - \frac{{Ml}}{{M + m}}\widehat i\quad ...(iii)$$

The plank will be displaced by a distance $x = \frac{{Ml}}{{M + m}}$ towards negative $x$ axis as shown in the figure.

Note: $-ve$ sign in equation $(iii)$ suggests that the displacement of plank is in direction opposite to what we have considered.

Question 10. Find the displacement of the wedge of mass $M$ when the small block of mass $m$ slides down from top to bottom.

Let the centre of mass of the wedge be at point $P$ i.e. $(a,b)$. The centre of mass of small block is at point $Q$ i.e. $(0,h)$.

Initial position of centre of mass of the systems is, $$\begin{equation} \begin{aligned} {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{{m_{Wedge}}{{\vec r}_{Wedge}} + {m_{Block}}{{\vec r}_{Block}}}}{{{m_{Wedge}} + {m_{Block}}}} \\ {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{M\left( {a\widehat i + b\widehat j} \right) + m\left( {0\hat i + h\hat j} \right)}}{{M + m}} \\ {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{M\left( {a\widehat i + b\widehat j} \right) + mh\widehat j}}{{M + m}}\quad ...(i) \\\end{aligned} \end{equation} $$

Now, when the block has reached the bottom, let us assume that the wedge is displaced by a distance $x$.

Final position of centre of mass of the system is, $$\begin{equation} \begin{aligned} {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{{m_{Wedge}}{{\vec r}_{Wedge}} + {m_{Block}}{{\vec r}_{Block}}}}{{{m_{Wedge}} + {m_{Block}}}} \\ {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{M\left[ {\left( {a + x} \right)\widehat i + b\widehat j} \right] + m\left[ {\left( {h\cot \theta + x} \right)\widehat i + 0\widehat j} \right]}}{{M + m}} \\ {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{M\left[ {\left( {a + x} \right)\widehat i + b\widehat j} \right] + m\left( {h\cot \theta + x} \right)\widehat i}}{{M + m}}\quad ...(ii) \\\end{aligned} \end{equation} $$

In $y$ direction the centre of mass is not at rest as normal force due to ground acts as an external force.

Since no external force acts on the system in the $x$ direction, so the centre of mass will be at rest in $x$ direction. Therefore the position of centre of mass (COM) will remain same in $x$ direction.

$${x_{CO{M_{Syeste{m_i}}}}} = {x_{CO{M_{Syeste{m_f}}}}}$$ or $$\frac{{Ma\widehat i}}{{M + m}} = \frac{{M\left( {a + x} \right) + m\left( {h\cot \theta + x} \right)}}{{M + m}}\widehat i$$ On solving, $$x = - \frac{{mh\cot \theta }}{{M + m}}\widehat i\quad ...(iii)$$

The wedge will be displaced by a distance $x = - \frac{{mh\cot \theta }}{{M + m}}\widehat i$ towards negative $x$ axis as shown in figure.

Note: $-ve$ sign in equation $(iii)$ suggests that the displacement of wedge is in direction opposite to what we have considered.

Question 11. A man of mass $20kg$ is standing on a light boat of mass $2kg$. The boat is $10 m$ away from the shore. Find the length of the boat so that the boat reaches the shore when the man walks from one end to the other end.

As the centre of mass of the boat and mass is at point $P$ i.e. $\left( {10 + \frac{l}{2},a} \right)$ and point $Q$ i.e. $(10,b)$.

Initial position of centre of mass of the systems is, $$\begin{equation} \begin{aligned} {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{{m_{boat}}{{\vec r}_{boat}} + {m_{man}}{{\vec r}_{man}}}}{{{m_{boat}} + {m_{man}}}} \\ {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{2\left[ {\left( {10 + \frac{l}{2}} \right)\widehat i + a\widehat j} \right] + 20\left( {10\hat i + b\hat j} \right)}}{{2 + 20}} \\ {{\vec r}_{CO{M_{syste{m_i}}}}} = \frac{{\left( {220 + l} \right)\widehat i + \left( {2a + 20b} \right)\widehat j}}{{22}}\quad ...(i) \\\end{aligned} \end{equation} $$

Now, when the man has reached the end, the boat has also reached the shore.

Final position of centre of mass of the system is, $$\begin{equation} \begin{aligned} {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{{m_{boat}}{{\vec r}_{boat}} + {m_{man}}{{\vec r}_{man}}}}{{{m_{boat}} + {m_{man}}}} \\ {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{2\left( {\frac{l}{2}\widehat i + a\widehat j} \right) + 20\left( {l\widehat i + b\widehat j} \right)}}{{2 + 20}} \\ {{\vec r}_{CO{M_{syste{m_f}}}}} = \frac{{\left( {l + 20l} \right)\widehat i + \left( {2a + 20b} \right)\widehat j}}{{22}}\quad ...(ii) \\\end{aligned} \end{equation} $$

Since no external force acts on the system in the $x$ direction, so the position of centre of mass will remain unchanged in $x$ direction. In the $y$ direction, buoyant force acts as an external force.

$${x_{CO{M_{Syste{m_i}}}}} = {x_{CO{M_{Syste{m_f}}}}}$$ or $$\frac{{\left( {220 + l} \right)\widehat i + \left( {2a + 20b} \right)\widehat j}}{{22}} = \frac{{\left( {l + 20l} \right)\widehat i + \left( {2a + 20b} \right)\widehat j}}{{22}}$$ On solving, $$l=11m$$

Illustration 2: Trajectory of centre of mass remains unchanged during explosion

When the internal energy is released instantaneously, explosion occurs.

During explosion no external force acts on the system. So, the trajectory of the centre of mass remains unchanged.

Let us understand this concept with the help of following two cases.

Consider a particle of mass $5m$ is projected with an initial velocity $u$ at an angle $\theta $ with the horizontal.

Since this system is a one particle system, so the motion of centre of mass $(COM)$ is same as the motion of the particle.

The horizontal distance covered by the particle during its flight is known as the range $(R)$ of the projectile motion.

Therefore, the range of the centre of mass $(COM)$ is same as the particle.

So, $$Rang{e_{COM}} = Rang{e_{Particle}}$$ Also, $${x_{CO{M_{Syste{m_i}}}}} = \frac{{{m_{particle}}{x_{particle}}}}{{{m_{particle}}{x_{particle}}}} = R\quad ...(i)$$

When the same particle during its projectile motion explodes into 3 parts $P$,$Q$ & $R$ having masses $m$,$2m$ & $3m$ respectively.

During explosion only internal energy is released. So, no external force acts on the system. Therefore, the motion of the centre of mass will remain unchanged.

Note: In the vertical $(y)$ direction gravitational force acts as an external force. So, the motion of centre of mass will be conserved or remain unchanged only in the horizontal $(x)$ direction.

So we can write, $$\begin{equation} \begin{aligned} {x_{CO{M_{Syste{m_f}}}}} = \frac{{{m_P}{x_1} + {m_Q}{x_2} + {m_R}{x_3}}}{{{m_P} + {m_Q} + {m_R}}} \\ {x_{CO{M_{Syste{m_f}}}}} = \frac{{m{x_1} + 2m{x_2} + 3m{x_3}}}{{5m}}\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(ii)$, we get, $$R = \frac{{m{x_1} + 2m{x_2} + 3m{x_3}}}{{5m}}$$

Question 12. A missile of mass $m$ is fired from the base station and is expected to hit the target at a distance $R$. During its flight it explodes into 3 wreckage of masses $\left( {\frac{m}{2}} \right)$, $\left( {\frac{m}{3}} \right)$ & $\left( {\frac{m}{6}} \right)$. The wreckages of masses $\left( {\frac{m}{2}} \right)$ & $\left( {\frac{m}{3}} \right)$ hits the ground at a distance $\left( {\frac{R}{2}} \right)$ & $\left( {\frac{3R}{2}} \right)$ respectively from the base station. Find the distance where the third wreckage hits the ground.

Solution: Since during explosion no external force acts on the system, therefore the path of the centre of mass will remain unchanged.

Note: In the vertical $(y)$ direction gravitational force acts as an external force. So, the motion of centre of mass will be conserved or remain unchanged only in the horizontal $(x)$ direction.

Initial position of the $COM$ of the system,

$${x_{CO{M_{Syste{m_i}}}}} = R\quad ...(i)$$

After explosion, $${x_{CO{M_{Syste{m_f}}}}} = \frac{{\frac{m}{2}\frac{R}{2} + \frac{m}{6}x + \frac{m}{3}\frac{{3R}}{2}}}{m}\quad ...(ii)$$

From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {x_{CO{M_{Syste{m_i}}}}} = {x_{CO{M_{Syste{m_f}}}}} \\ R = \frac{{\frac{m}{2}\frac{R}{2} + \frac{m}{6}x + \frac{m}{3}\frac{{3R}}{2}}}{m} \\\end{aligned} \end{equation} $$ or $$x = \frac{{3R}}{2}$$