Concept Card: Motion of the Centre of Mass
1. Kinematics & Dynamics of Centre of Mass:
The velocity and acceleration of the centre of mass of a system of particles of total mass $M = \Sigma m_i$ are given by the first and second time derivatives of its position vector:
$$\mathbf{\vec{v}_{\text{cm}} = \frac{d\vec{r}_{\text{cm}}}{dt} = \frac{\Sigma m_i \vec{v}_i}{M} = \frac{\vec{P}_{\text{total}}}{M}}$$
$$\mathbf{\vec{a}_{\text{cm}} = \frac{d\vec{v}_{\text{cm}}}{dt} = \frac{\Sigma m_i \vec{a}_i}{M} = \frac{\Sigma \vec{F}_i}{M}}$$
- Total Momentum Equivalence: The total linear momentum of any system of particles is identically equal to the product of total mass and the velocity of its centre of mass:
$$\mathbf{\vec{P}_{\text{total}} = M \vec{v}_{\text{cm}}}$$ - Newton’s Second Law for a System: All internal forces between particles cancel out pairwise (Newton’s Third Law, $\Sigma \vec{F}_{\text{int}} = \vec{0}$):
$$\Sigma \vec{F}_{\text{net}} = \Sigma \vec{F}_{\text{ext}} + \Sigma \vec{F}_{\text{int}} = \mathbf{\Sigma \vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}}$$
The centre of mass accelerates IF AND ONLY IF a non-zero resultant external force acts on the system.
2. Conservation of Linear Momentum & CM State
- Zero External Force ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$):
$$\vec{a}_{\text{cm}} = \vec{0} \implies \vec{v}_{\text{cm}} = \text{constant} \iff \vec{P}_{\text{total}} = \text{constant}$$ - Initially Stationary System ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$ and $\vec{v}_{\text{cm}}(0) = \vec{0}$):
$$\vec{v}_{\text{cm}} = \vec{0} \implies \Delta \vec{r}_{\text{cm}} = \vec{0}$$
The centre of mass remains permanently at rest in space regardless of internal collisions, relative motion, or explosions. - Directional Decoupling: If the net external force along a given axis is zero (e.g., $\Sigma F_{\text{ext}, x} = 0$), then $a_{\text{cm}, x} = 0$, $v_{\text{cm}, x} = \text{constant}$, and $P_x = \text{constant}$, even if gravity or other external forces act along the $y$-axis.
3. High-Yield Classical JEE Paradigms
- Exploding Projectile in Mid-Air:
When a shell explodes in flight under gravity, explosive forces are internal ($\Sigma \vec{F}_{\text{int}} = \vec{0}$). The only external force acting on the fragments is gravity ($M\vec{g}$):
$$\vec{a}_{\text{cm}} = \vec{g}$$
The centre of mass of all fragments continues to move along the original uninterrupted parabolic trajectory until the first fragment strikes the ground. - Atwood Machine CM Acceleration:
For two masses $m_1 > m_2$ connected over an ideal pulley, individual acceleration is $a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g$. The centre of mass acceleration is downward with magnitude:
$$\mathbf{a_{\text{cm}} = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g}$$ - Kinetic Energy Decomposition (König’s Theorem):
$$K_{\text{lab}} = K_{\text{cm}} + K_{\text{rel}} = \frac{1}{2}M v_{\text{cm}}^2 + \Sigma \frac{1}{2}m_i v_i’^2$$
For a two-body system: $K_{\text{rel}} = \frac{1}{2}\mu v_{\text{rel}}^2$, where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.
4. Common JEE Pitfalls & Traps
- Trap 1 (Internal Forces Do Work): While internal forces cannot change $\vec{P}_{\text{total}}$ or $\vec{v}_{\text{cm}}$, they DO perform work ($\Sigma W_{\text{int}} = \Delta K_{\text{rel}} \ne 0$). In an explosion, chemical energy transforms into relative kinetic energy.
- Trap 2 (Ground Impact Disruption): The CM follows the original parabola ONLY until a fragment hits the ground. Once a fragment touches the earth, external normal reaction and friction act from the ground, invalidating $\Sigma \vec{F}_{\text{ext}} = M\vec{g}$.
- Trap 3 (Atwood Sign Error): Because masses move in opposite directions, $a_{\text{cm}}$ involves the difference of accelerations, not their sum.
Solved Examples
Example 1 (Direct Conceptual Application – Multi-Particle Acceleration & Velocity):
Three particles of masses $m_1 = 1.0\text{ kg}$, $m_2 = 2.0\text{ kg}$, and $m_3 = 3.0\text{ kg}$ have instantaneous velocities:
$$\vec{v}_1 = (3.0\hat{i} – 2.0\hat{j})\text{ m/s}, \qquad \vec{v}_2 = (4.0\hat{i} + \hat{j})\text{ m/s}, \qquad \vec{v}_3 = (-2.0\hat{i} + 3.0\hat{j})\text{ m/s}$$
External forces acting on them at this instant are:
$$\vec{F}_1 = (2.0\hat{i} + 4.0\hat{j})\text{ N}, \qquad \vec{F}_2 = (-\hat{i} – 2.0\hat{j})\text{ N}, \qquad \vec{F}_3 = (5.0\hat{i} + 4.0\hat{j})\text{ N}$$
(a) Find the total linear momentum $\vec{P}_{\text{total}}$ of the system.
(b) Find the velocity vector $\vec{v}_{\text{cm}}$ of the centre of mass.
(c) Find the acceleration vector $\vec{a}_{\text{cm}}$ of the centre of mass.
Solution:
(a) Total mass: $M = 1.0 + 2.0 + 3.0 = 6.0\text{ kg}$.
Total linear momentum:
$$\vec{P}_{\text{total}} = 1.0(3\hat{i} – 2\hat{j}) + 2.0(4\hat{i} + \hat{j}) + 3.0(-2\hat{i} + 3\hat{j}) = (3 + 8 – 6)\hat{i} + (-2 + 2 + 9)\hat{j} = (5.0\hat{i} + 9.0\hat{j})\text{ kg}\cdot\text{m/s}$$
(b) Velocity of the centre of mass:
$$\vec{v}_{\text{cm}} = \frac{\vec{P}_{\text{total}}}{M} = \frac{5.0\hat{i} + 9.0\hat{j}}{6.0} = \left(\frac{5}{6}\hat{i} + \frac{3}{2}\hat{j}\right)\text{ m/s} = (0.833\hat{i} + 1.50\hat{j})\text{ m/s}$$
(c) Resultant external force:
$$\Sigma \vec{F}_{\text{ext}} = (2 – 1 + 5)\hat{i} + (4 – 2 + 4)\hat{j} = (6.0\hat{i} + 6.0\hat{j})\text{ N}$$
Acceleration of the centre of mass:
$$\vec{a}_{\text{cm}} = \frac{\Sigma \vec{F}_{\text{ext}}}{M} = \frac{6.0\hat{i} + 6.0\hat{j}}{6.0} = (\hat{i} + \hat{j})\text{ m/s}^2$$
Example 2 (Mathematical Derivation – Centre of Mass Acceleration in an Atwood Machine):
Two masses $m_1 = 3.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$ are connected by a light inextensible cord passing over a fixed frictionless pulley. The system is released from rest. Taking $g = 10.0\text{ m/s}^2$:
(a) Determine the acceleration $a$ of the individual blocks.
(b) Derive and compute the acceleration vector $\vec{a}_{\text{cm}}$ of the centre of mass.
(c) Find the downward velocity of the centre of mass at $t = 2.0\text{ s}$ and find the net external force on the system.
Solution:
(a) Block acceleration:
$$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g = \left(\frac{3.0 – 1.0}{3.0 + 1.0}\right)(10.0) = \frac{2}{4}(10.0) = 5.0\text{ m/s}^2$$
(b) Choosing upward as $+\hat{j}$: $\vec{a}_1 = -5.0\hat{j}\text{ m/s}^2$ and $\vec{a}_2 = +5.0\hat{j}\text{ m/s}^2$.
$$\vec{a}_{\text{cm}} = \frac{3.0(-5.0\hat{j}) + 1.0(+5.0\hat{j})}{4.0} = -\frac{10.0}{4.0}\hat{j} = -2.5\hat{j}\text{ m/s}^2$$
Magnitude: $a_{\text{cm}} = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g = (0.50)^2(10.0) = 2.50\text{ m/s}^2$ downward.
(c) Downward velocity at $t = 2.0\text{ s}$:
$$\vec{v}_{\text{cm}} = \vec{a}_{\text{cm}} t = (-2.5\hat{j})(2.0) = -5.0\hat{j}\text{ m/s}$$
Net external force:
$$\Sigma \vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}} = (4.0\text{ kg})(-2.5\hat{j}\text{ m/s}^2) = -10.0\hat{j}\text{ N}$$
*(Gravity is $40\text{ N}$ downward; upward support reaction from pulley is $2T = 2(15) = 30\text{ N}$. Net external force is $30 – 40 = -10\text{ N}$).*
Example 3 (Standard JEE Advanced Scenario – Mid-Air Projectile Explosion):
A shell of mass $M = 3.0\text{ kg}$ is fired from the origin on flat ground with speed $u = 50.0\text{ m/s}$ at $\theta = 37^\circ$ ($\cos 37^\circ = 0.80, \sin 37^\circ = 0.60$). At the highest point of its trajectory, an explosion splits it into piece 1 ($m_1 = 1.0\text{ kg}$) and piece 2 ($m_2 = 2.0\text{ kg}$). Piece 1 falls vertically downward with zero horizontal speed. Take $g = 10.0\text{ m/s}^2$.
(a) Determine the horizontal velocity of piece 2 immediately after the explosion.
(b) Calculate the horizontal distance from the origin where piece 2 lands on the ground.
(c) Calculate the energy released during the explosion.
Solution:
(a) Velocity at apex: $v_{\text{apex}} = u\cos 37^\circ = 50.0 \times 0.80 = 40.0\text{ m/s}$.
Total horizontal momentum before explosion: $P_x = M v_{\text{apex}} = 3.0(40.0) = 120.0\text{ kg}\cdot\text{m/s}$.
Conservation of horizontal momentum:
$$120.0 = m_1(0) + m_2 v_{2x} \implies 2.0 v_{2x} = 120.0 \implies v_{2x} = +60.0\text{ m/s}$$
(b) Horizontal range of uninterrupted projectile:
$$R = \frac{2 (u\cos 37^\circ)(u\sin 37^\circ)}{g} = \frac{2(40.0)(30.0)}{10.0} = 240.0\text{ m}$$
Apex location: $x_{\text{apex}} = R/2 = 120.0\text{ m}$. Time to fall from apex: $t_{\text{fall}} = \frac{u\sin 37^\circ}{g} = 3.0\text{ s}$.
During this time, piece 1 lands at $x_1 = 120.0\text{ m}$. Piece 2 lands at:
$$x_2 = x_{\text{apex}} + v_{2x} t_{\text{fall}} = 120.0 + 60.0(3.0) = 120.0 + 180.0 = 300.0\text{ meters}$$
*(CM check: $x_{\text{cm}} = \frac{1.0(120) + 2.0(300)}{3.0} = \frac{720}{3} = 240.0\text{ m} = R$. Verified!).*
(c) Kinetic energy just before explosion: $K_i = \frac{1}{2}(3.0)(40.0)^2 = 2400.0\text{ J}$.
Kinetic energy immediately after explosion: $K_f = 0 + \frac{1}{2}(2.0)(60.0)^2 = 3600.0\text{ J}$.
Energy released in explosion:
$$\Delta E = K_f – K_i = 3600.0 – 2400.0 = 1200.0\text{ Joules}$$
Example 4 (Edge Case – Simultaneous vs. Sequential Jumping & Rocket Analogy):
A flatcar of mass $M = 100.0\text{ kg}$ is initially at rest on a frictionless straight track. Two persons $A$ and $B$, each of mass $m = 50.0\text{ kg}$, stand on the flatcar.
(a) Both persons jump off simultaneously toward the right with horizontal speed $u = 6.0\text{ m/s}$ relative to the flatcar. Find the final recoil speed $v_1$ of the flatcar.
(b) The persons jump off sequentially one after the other, each with horizontal speed $u = 6.0\text{ m/s}$ relative to the flatcar at the moment of jumping. Find the final recoil speed $v_2$ of the flatcar.
(c) Compare the two recoil speeds and explain the underlying physical principle.
Solution:
(a) Simultaneous jump:
$$(2m)(u – v_1) – M v_1 = 0 \implies v_1 = \left(\frac{2m}{M + 2m}\right)u = \left(\frac{100.0}{200.0}\right)(6.0) = 3.0\text{ m/s}$$
(b) Sequential jumps:
– First jump (person $A$ leaps):
$$m(u – v_a) – (M + m)v_a = 0 \implies v_a = \left(\frac{m}{M + 2m}\right)u = \left(\frac{50.0}{200.0}\right)(6.0) = 1.50\text{ m/s}$$
– Second jump (person $B$ leaps from moving flatcar of remaining mass $M + m = 150\text{ kg}$):
$$-(M + m)v_a = m(u – v_2) – M v_2 = m u – (M + m)v_2$$
$$(M + m)v_2 = m u + (M + m)v_a \implies 150.0 v_2 = 50(6.0) + 150(1.50) = 300.0 + 225.0 = 525.0$$
$$v_2 = \frac{525.0}{150.0} = 3.50\text{ m/s}$$
(c) Comparison:
$$v_2 = 3.50\text{ m/s} > v_1 = 3.0\text{ m/s}$$
When jumping sequentially, the first departure reduces the system mass to $150\text{ kg}$. The second jumper pushes against a lighter cart, imparting a larger second velocity increment ($2.0\text{ m/s}$ vs $1.5\text{ m/s}$). This demonstrates the staging advantage in multi-stage rocketry.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
If the resultant external force acting on a system of particles is identically zero ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$), which of the following must be true?
(A) The acceleration of the centre of mass is zero
(B) The velocity of the centre of mass must be zero
(C) The linear momentum of each individual particle must remain constant
(D) The total kinetic energy of the system must remain constant
Problem 2 (JEE Main – Single Correct):
Two masses $m_1 = 2.0\text{ kg}$ and $m_2 = 3.0\text{ kg}$ are connected across an ideal pulley in an Atwood machine. Taking acceleration due to gravity as $g$, the magnitude of the acceleration of their centre of mass is:
(A) $\frac{g}{5}$
(B) $\frac{g}{10}$
(C) $\frac{g}{25}$
(D) $\frac{g}{50}$
Problem 3 (JEE Main – Single Correct):
A bomb moving in a parabolic trajectory under gravity explodes in mid-air into several fragments. Before any fragment touches the ground, the centre of mass of all the fragments:
(A) Plunges vertically downward with acceleration $g$
(B) Continues to move along the original uninterrupted parabolic trajectory
(C) Moves horizontally at a constant speed
(D) Comes to rest at the position of the explosion
Problem 4 (JEE Main – Single Correct):
A person of mass $m$ stands on a cart of mass $M$ initially at rest on a smooth horizontal floor. If the person jumps off with horizontal velocity $u$ relative to the cart, the recoil speed of the cart relative to the floor is:
(A) $\frac{m u}{M}$
(B) $\frac{m u}{m + M}$
(C) $\frac{M u}{m + M}$
(D) $\frac{(m + M)u}{m}$
Problem 5 (JEE Main – Single Correct):
Two particles of masses $2.0\text{ kg}$ and $4.0\text{ kg}$ move toward each other along a straight line with speeds $4.0\text{ m/s}$ and $2.0\text{ m/s}$ respectively. The velocity of the centre of mass of the two-particle system is:
(A) $2.0\text{ m/s}$
(B) $1.0\text{ m/s}$
(C) $0\text{ m/s}$
(D) $3.0\text{ m/s}$
Problem 6 (JEE Advanced – One or More Correct):
An artillery shell fired from a cannon explodes in mid-air into three unequal fragments:
(A) The total linear momentum of the fragments immediately after explosion equals the momentum of the shell immediately before.
(B) The total mechanical kinetic energy of the fragments immediately after explosion is strictly greater than the kinetic energy immediately before.
(C) The acceleration of the centre of mass remains equal to $\vec{g}$ downward until a fragment strikes the ground.
(D) The internal explosive forces perform net positive work during the fragmentation.
Problem 7 (JEE Advanced – One or More Correct):
Two blocks of masses $m_1$ and $m_2$ on a frictionless horizontal floor are connected by an ideal spring. The spring is compressed and the blocks are released from rest:
(A) The centre of mass remains stationary at its initial position at all times.
(B) The total linear momentum of the two-block system is zero at all times.
(C) The kinetic energy of the centre of mass is zero at all times.
(D) The total mechanical energy (kinetic + elastic potential) of the system is conserved.
Problem 8 (JEE Advanced – One or More Correct):
A person of mass $m$ stands at the back of a flatcar of mass $M$ moving with constant speed $v_0$ along a frictionless straight track. The person runs to the front with speed $u$ relative to the flatcar:
(A) The velocity of the centre of mass of the (person + flatcar) system remains strictly $v_0$ throughout.
(B) The speed of the flatcar relative to the track decreases while the person is running forward.
(C) When the person stops upon reaching the front, the speed of the flatcar returns to $v_0$.
(D) The net displacement of the flatcar relative to the track while the person traverses length $L$ of the car is $-\frac{m L}{m + M}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Two blocks of masses $m_1 = 3.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$ are connected by a light cord over a frictionless fixed pulley in an Atwood machine. Taking $g = 10.0\text{ m/s}^2$, find the magnitude of the downward acceleration of the centre of mass of the system in $\text{m/s}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A projectile of mass $2.0\text{ kg}$ launched from the origin has a horizontal range of $R = 80.0\text{ m}$. At its highest point, an internal explosion splits it into two equal fragments of $1.0\text{ kg}$ each. One fragment falls vertically downward from rest relative to the ground. Calculate the distance (in meters) from the origin at which the second fragment strikes the ground.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (C) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 2.5 | 10. 120
Solution 1:
By Newton’s second law for a system, $\Sigma \vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}$. If $\Sigma \vec{F}_{\text{ext}} = \vec{0}$, then $\vec{a}_{\text{cm}} = \vec{0}$. The velocity $\vec{v}_{\text{cm}}$ is constant (not necessarily zero). Individual particle momenta and kinetic energy can change due to internal forces.
Correct Option: (A)
Solution 2:
For an Atwood machine, the acceleration of the individual blocks is $a = \left(\frac{m_2 – m_1}{m_1 + m_2}\right)g = \frac{3 – 2}{3 + 2}g = \frac{g}{5}$.
The centre of mass acceleration is:
$$a_{\text{cm}} = \left(\frac{m_2 – m_1}{m_1 + m_2}\right)a = \left(\frac{1}{5}\right)\left(\frac{g}{5}\right) = \frac{g}{25}$$
Correct Option: (C)
Solution 3:
The explosion is caused by internal forces which cancel in pairs. The only external force acting on the fragments is gravity ($M\vec{g}$). Thus $\vec{a}_{\text{cm}} = \vec{g}$, and the centre of mass continues moving along the original parabolic path until the first fragment hits the ground.
Correct Option: (B)
Solution 4:
Let the cart recoil with speed $v_c$ to the left. The ground speed of the man is $u – v_c$ to the right.
Linear momentum conservation:
$$m(u – v_c) – M v_c = 0 \implies m u – (m + M)v_c = 0 \implies v_c = \frac{m u}{m + M}$$
Correct Option: (B)
Solution 5:
Choose the direction of the $2.0\text{ kg}$ mass as positive $+x$:
$$v_{\text{cm}} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{2.0(+4.0) + 4.0(-2.0)}{2.0 + 4.0} = \frac{8.0 – 8.0}{6.0} = 0\text{ m/s}$$
Correct Option: (C)
Solution 6:
– (A) True: Explosion duration is negligible, so external impulse is zero; momentum is conserved.
– (B) True: Chemical potential energy is converted into kinetic energy, so $K_f > K_i$.
– (C) True: External force is purely gravity, so $\vec{a}_{\text{cm}} = \vec{g}$.
– (D) True: Chemical work converts to kinetic energy: $W_{\text{int}} = \Delta K > 0$.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $\Sigma \vec{F}_{\text{ext}} = \vec{0}$ and system starts from rest, so $\vec{a}_{\text{cm}} = \vec{0}, \vec{v}_{\text{cm}} = \vec{0}$.
– (B) True: $\vec{P}_{\text{total}} = M \vec{v}_{\text{cm}} = \vec{0}$.
– (C) True: $K_{\text{cm}} = \frac{1}{2}M v_{\text{cm}}^2 = 0$.
– (D) True: Spring force is conservative and floor is smooth, so mechanical energy is conserved.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: No net horizontal external force acts on the flatcar-person system; $v_{\text{cm}} = v_0$ is constant.
– (B) True: While the person runs forward at relative speed $u$, the flatcar recoils backward relative to the CM, so its ground speed drops to $v_0 – \frac{m u}{m + M}$.
– (C) True: When the person stops, relative velocity is zero, so both move at $v_0$.
– (D) True: In the CM frame, the displacement of the flatcar is $-\frac{m L}{m + M}$.
Correct Options: (A, B, C, D)
Solution 9:
$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)g = \left(\frac{3.0 – 1.0}{3.0 + 1.0}\right)(10.0) = \frac{2}{4}(10.0) = 5.0\text{ m/s}^2$.
The magnitude of the acceleration of the centre of mass is:
$$a_{\text{cm}} = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g = \left(\frac{2.0}{4.0}\right)^2 (10.0) = (0.50)^2(10.0) = 0.25 \times 10.0 = 2.5\text{ m/s}^2$$
Correct Answer: 2.5
Solution 10:
The explosion occurs at the apex, which is at horizontal position $x_{\text{apex}} = R/2 = 40.0\text{ m}$.
Fragment 1 drops vertically from the apex, so it lands at $x_1 = 40.0\text{ m}$.
Because the external vertical acceleration is $g$ for both fragments, they take the identical time $t = \sqrt{2H/g}$ to hit the ground.
Thus, at the instant of impact, the centre of mass lands at the original range $x_{\text{cm}} = R = 80.0\text{ m}$.
$$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \implies 80.0 = \frac{1.0(40.0) + 1.0(x_2)}{2.0}$$
$$160.0 = 40.0 + x_2 \implies x_2 = 120.0\text{ meters}$$
Correct Answer: 120