Q.In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = $10^{-10}$ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Showing the 12 most recent of 31 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A body A is projected with a velocity of $40\ ms^{-1}$ at an angle of $60^0$ with the horizontal. At some other time, another body B is thrown vertically upwards with a velocity of $40\sqrt{3}\ ms^{-1}$ such that it collides body A at a height of 60 m. If the velocity of centre of mass of the system of the two bodies after collision is $20\ ms^{-1}$, then the ratio of the masses of the bodies A and B is (Acceleration due to gravity $=10\ ms^{-2}$) (A) 1:3 (B) 1:2 (C) 4:1 (D) 3:2

   ›Reveal solutionSolution

   This tests projectile kinematics combined with conservation of momentum through a collision — find each body's velocity components at the meeting point, then equate the vector sum of momenta to $(m_A+m_B)v_{cm}$. Answer: (C) 4:1.

   Concept and Intuition

   Gravity is a finite (non-impulsive) force, so during the brief instant of collision it contributes negligible impulse — total momentum is conserved right through the collision, exactly as in an ordinary collision problem, even though both bodies are simultaneously in projectile motion. The center-of-mass velocity right after collision must equal $\dfrac{m_A\vec v_A+m_B\vec v_B}{m_A+m_B}$, where $\vec v_A,\vec v_B$ are the individual velocities the instant before collision. So the real work is purely kinematic: find where and how fast each body is moving when it reaches height 60 m.

   Step-by-Step Solution

   1. Body A's vertical velocity component: $v_{y0}=40\sin60^\circ=40\cdot\frac{\sqrt3}{2}=20\sqrt3\ m/s$. Its maximum height is $H_A=\dfrac{v_{y0}^2}{2g}=\dfrac{1200}{20}=60\ m$ — exactly the collision height! So A is momentarily at the top of its trajectory: $v_{y,A}=0$, and its horizontal component is unchanged throughout flight: $v_{x,A}=40\cos60^\circ=20\ m/s$. So $\vec v_A=(20,\,0)$ m/s.
   2. Body B's velocity at $h=60$: using $v^2=u^2-2gh$ with $u=40\sqrt3$: $v_B^2=4800-2(10)(60)=4800-1200=3600$, so $v_B=60\ m/s$ (magnitude; direction purely vertical, up or down — but squared, the sign won't matter below). So $\vec v_B=(0,\,\pm60)$ m/s.
   3. Momentum conservation through the collision (vector form):

   $\vec v_{cm}=\frac{m_A\vec v_A+m_B\vec v_B}{m_A+m_B}=\left(\frac{20m_A}{m_A+m_B},\ \frac{\pm60m_B}{m_A+m_B}\right).$

   4. Apply $|\vec v_{cm}|=20$:

   $\left(\frac{20m_A}{m_A+m_B}\right)^2+\left(\frac{60m_B}{m_A+m_B}\right)^2=20^2.$

   Let $M=m_A+m_B$. Then $400m_A^2+3600m_B^2=400M^2=400(m_A+m_B)^2=400m_A^2+800m_Am_B+400m_B^2$.

   Simplify: $3200m_B^2=800m_Am_B \Rightarrow m_A=4m_B$.

   5. Ratio: $m_A:m_B=4:1$.

   Common Mistakes

   • Missing that A's height of 60 m exactly equals its maximum height — this is the crucial simplification that makes $v_{y,A}=0$, without which the problem would need a time variable.
   • Forgetting momentum is a vector quantity here — A's velocity is purely horizontal and B's is purely vertical, so they must be combined by components (Pythagorean-style), not simply added as scalars.
   • Sign of $v_B$ (up vs down) doesn't matter since it only enters squared — no need to determine which direction B was moving at the collision instant.

   ✓Final answer

   The correct option is (C) — 4:1.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If three particles of masses $2m$, $m$ and $4m$ are moving in three mutually perpendicular directions with velocities $3\ ms^{-1}$, $4\ ms^{-1}$ and $3\ ms^{-1}$ respectively, then the magnitude of the velocity of the center of mass of the system of three particles is (A) $3.5\ ms^{-1}$ (B) $2\ ms^{-1}$ (C) $2.5\ ms^{-1}$ (D) $3\ ms^{-1}$

   ›Reveal solutionSolution

   Adding the three mutually perpendicular momenta as vector components and dividing by the total mass gives $v_{cm}=2$ m/s.

   Concept and Intuition

   When velocities are along mutually perpendicular directions, the total momentum vector's magnitude follows directly from the Pythagorean sum of the individual (mass times velocity) components — no angle resolution is needed since the axes are already orthogonal.

   Step-by-Step Solution

   1. Momentum components (each along its own perpendicular axis): $p_x=2m(3)=6m$, $p_y=m(4)=4m$, $p_z=4m(3)=12m$.
   2. Magnitude of total momentum: $|\vec p|=m\sqrt{6^2+4^2+12^2}=m\sqrt{36+16+144}=m\sqrt{196}=14m$.
   3. Total mass: $M=2m+m+4m=7m$.
   4. Velocity of center of mass: $v_{cm}=\dfrac{|\vec p|}{M}=\dfrac{14m}{7m}=2\ ms^{-1}$.

   Common Mistakes

   • Simply averaging the three speeds instead of vector-adding the momenta.
   • Forgetting to divide by the total mass ($7m$), not one of the individual masses.

   ✓Final answer

   The correct option is (B) — $2\ ms^{-1}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Two particles of masses 4 g and 2 g are separated by a distance of 60 cm. The center of mass of the system of these two particles is (A) Lies at a distance of 30 cm from 4 g particle (B) Lies at a distance of 40 cm from 4 g particle (C) Lies at a distance of 40 cm from 2 g particle (D) Lies at a distance of 20 cm from 2 g particle

   ›Reveal solutionSolution

   The center of mass lies closer to the heavier particle; computing the distances gives 20 cm from the 4 g mass and 40 cm from the 2 g mass, matching option (C).

   Concept and Intuition

   The center of mass of a two-particle system divides the line joining them in the inverse ratio of their masses — it sits closer to the heavier mass. This is a direct application of the weighted-average definition of center of mass.

   Step-by-Step Solution

   1. Let the 4 g particle be at $x=0$ and the 2 g particle be at $x=60\ cm$.
   2. Center of mass: $x_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1+m_2} = \dfrac{4(0) + 2(60)}{4+2} = \dfrac{120}{6} = 20\ cm$.
   3. So the COM is at 20 cm from the 4 g particle (measuring from $x=0$).
   4. Distance from the 2 g particle (at $x=60$): $60 - 20 = 40\ cm$.
   5. Checking against the options: distance from 4 g particle is 20 cm (not listed exactly as such, but option (C) states 'distance of 40 cm from 2 g particle', which is exactly what we found).

   Common Mistakes

   • Mixing up which particle the COM is closer to — it must be closer to the HEAVIER (4 g) mass, i.e., only 20 cm away from it, not 40 cm.
   • Misreading option (D) '20 cm from 2 g particle' as correct — that's the distance from the 4 g particle, not the 2 g particle.

   ✓Final answer

   The correct option is (C) — Lies at a distance of 40 cm from 2 g particle.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Two particles A and B initially at rest move towards each other under a mutual force of attraction. At the instant, the speed of A is V and the speed of B is 2V, find the speed of centre of mass of the system, if the masses of A and B are in the ratio of 2 : 1 (A) Zero (B) $\dfrac{4V}{3}$ (C) $\dfrac{3V}{4}$ (D) $\dfrac{3V}{2}$

   ›Reveal solutionSolution

   With no external force and zero initial momentum, the centre of mass of the two-particle system stays at rest forever — its speed is Zero, option (A).

   Concept and Intuition

   The velocity of the centre of mass of an isolated system changes only under an external force. Here the only force is the mutual (internal) attraction between A and B, which by Newton's third law cancels out when you sum forces over the whole system. Since both particles start at rest, the total initial momentum is zero — and it must stay zero forever, meaning the centre of mass never moves at all, at any later instant.

   Step-by-Step Solution

   1. Total momentum of an isolated two-body system is conserved because internal (mutual) forces contribute zero net external force.
   2. Initially, both particles are at rest, so total initial momentum $p_i=0$.
   3. By conservation of momentum, total momentum at any later time must also be zero: $m_Av_A - m_Bv_B=0$ (opposite directions, since they move towards each other).
   4. Velocity of centre of mass: $v_{cm}=\dfrac{m_Av_A+m_Bv_B(\text{with sign})}{m_A+m_B}=\dfrac{p_{total}}{m_A+m_B}=\dfrac{0}{m_A+m_B}=0$.
   5. (Consistency check: with $m_A:m_B=2:1$, i.e. $m_A=2m,\ m_B=m$, momentum balance gives $2m\cdot V = m\cdot 2V$, which holds exactly — confirming the given speeds are consistent with zero net momentum.)

   Common Mistakes

   • Trying to compute $v_{cm}$ by plugging in the given speeds $V$ and $2V$ as if they were simultaneous velocities to average — the correct approach is recognizing momentum conservation makes $v_{cm}$ zero at every instant, not just initially.
   • Getting distracted by the mass ratio and speed values, which are provided only for internal consistency, not to be substituted into a centre-of-mass velocity formula.

   ✓Final answer

   The correct option is (A) — Zero.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Two identical particles move towards each other with velocity 2V and V respectively. The velocity of center of mass of this system is (A) $V$ (B) $\dfrac{V}{3}$ (C) $\dfrac{V}{2}$ (D) $\dfrac{V}{4}$

   ›Reveal solutionSolution

   This tests the definition of centre-of-mass velocity as the momentum-weighted average velocity, with opposite-direction velocities for a head-on approach. The answer is (C), V/2.

   Concept and Intuition

   The velocity of the centre of mass of a system is the total momentum divided by total mass: $v_{cm}=\dfrac{\sum m_iv_i}{\sum m_i}$. Since the two particles move "towards each other," their velocities must be taken with opposite signs along the line joining them.

   Step-by-Step Solution

   1. Let both particles have mass $m$ (identical particles).
   2. Take the direction of the particle moving with speed $2V$ as positive; since the other moves towards it (opposite direction), its velocity is $-V$.
   3. Total momentum: $p=m(2V)+m(-V)=mV$.
   4. Total mass: $2m$.
   5. Velocity of centre of mass: $v_{cm}=\dfrac{p}{2m}=\dfrac{mV}{2m}=\dfrac{V}{2}$.

   Common Mistakes

   • Adding the speeds without accounting for opposite directions (giving $v_{cm}=\dfrac{3V}{2}$), which ignores that "towards each other" means opposite signs.
   • Forgetting to divide by the total mass $2m$ (using just $m$), which would double the actual answer.

   ✓Final answer

   The correct option is (C) — $\dfrac{V}{2}$.

   ANSWER: C

6. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A thin square shaped copper plate of uniform mass distribution of side 4 m has its centre of mass at (2, 2). If at its top right corner, a square of 2 m side is cut from the plate, the centre of mass of the remaining plate is (A) $\dfrac{5}{6}, \dfrac{5}{6}$ (B) $\dfrac{5}{3}, \dfrac{5}{3}$ (C) $\dfrac{5}{6}, \dfrac{5}{3}$ (D) $\dfrac{5}{3}, \dfrac{5}{6}$

   ›Reveal solutionSolution

   Removing the top-right quarter-square from the plate and using the

   "negative mass" centre-of-mass trick gives the new COM at

   $\left(\tfrac53,\tfrac53\right)$. Answer: (B).

   Concept and Intuition

   When a piece is removed from a uniform lamina, treat the remaining shape as

   the original full shape minus the removed piece, and use "negative

   mass" for the removed part in the centre-of-mass formula:

   $x_{cm} = \dfrac{M_{full}x_{full} - m_{cut}x_{cut}}{M_{full} - m_{cut}}$.

   This avoids having to integrate over the oddly-shaped remaining region

   directly.

   Step-by-Step Solution

   1. Set up coordinates so the original $4\,\text{m}\times4\,\text{m}$ square spans $(0,0)$ to $(4,4)$; its COM is at its geometric centre $(2,2)$ — consistent with the given data. Let its mass be $M$ (area 16, so mass per unit area $\sigma = M/16$).
   2. The cut-out is a $2\text{m}\times2\text{m}$ square at the "top right corner" — spanning $(2,2)$ to $(4,4)$ — with its own centre at $(3,3)$ and area 4, so its mass is $m_{cut} = 4\sigma = M/4$.
   3. Remaining mass: $M_{rem} = M - M/4 = \dfrac{3M}{4}$.
   4. Apply the "negative mass" COM formula for the x-coordinate: $x_{cm} = \dfrac{M(2) - (M/4)(3)}{3M/4} = \dfrac{2M - 0.75M}{0.75M} = \dfrac{1.25}{0.75} = \dfrac{5}{3}$.
   5. By the symmetry of the square and the corner cut (both x and y play identical roles), $y_{cm} = \dfrac{5}{3}$ as well.
   6. New centre of mass: $\left(\dfrac53, \dfrac53\right)$ — option (B).

   Common Mistakes

   • Trying to average the corner coordinates directly instead of properly weighting by mass via the "subtract a negative mass" method.
   • Misplacing the cut square's centre (it's at $(3,3)$, not at the corner point $(4,4)$ itself).

   ✓Final answer

   The correct option is (B) — $\left(\dfrac53, \dfrac53\right)$.

   ANSWER: B

7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Three bodies A, B, and C of masses 2 kg, 3 kg, and 5 kg respectively are projected simultaneously with the same speed from the roof of a tower. The body A is thrown vertically upwards, body B is thrown vertically downwards and body C is projected horizontally. The acceleration of the centre of mass of the system of three bodies is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) $6\ \mathrm{m\,s^{-2}}$ (B) $10\ \mathrm{m\,s^{-2}}$ (C) $8\ \mathrm{m\,s^{-2}}$ (D) $12\ \mathrm{m\,s^{-2}}$

   ›Reveal solutionSolution

   This tests the idea that the acceleration of the centre of mass of a system depends only on the net external force per unit total mass — here every body is in free fall, so the COM accelerates at exactly $g$, independent of masses or launch directions.

   Concept and Intuition

   The acceleration of the centre of mass of a system is given by $a_{cm} = F_{ext}/M_{total}$, where $F_{ext}$ is the total external force. After being launched, each of the three bodies is in free fall — the only force on each is gravity, $m_i g$ downward, irrespective of the direction it was thrown (up, down, or horizontal) since projectile motion under gravity alone always has acceleration $g$ downward for every body. So the total external force on the system is $(m_A+m_B+m_C)g$ downward, and dividing by the total mass just gives $g$ again. The individual masses and throw directions are irrelevant distractors — as long as gravity is the only force, every body (and hence their COM) accelerates at $g$.

   Step-by-Step Solution

   1. Each body, once released, experiences only gravity: $a_A = a_B = a_C = g$ (downward), regardless of initial velocity direction.
   2. Total external force on the 3-body system: $F_{ext} = (m_A+m_B+m_C)\,g$, downward.
   3. Acceleration of centre of mass: $a_{cm} = F_{ext}/(m_A+m_B+m_C) = g = 10\ \mathrm{ms^{-2}}$, downward.

   Common Mistakes

   • Trying to average the individual accelerations weighted by mass and getting confused about directions — since all three individual accelerations are identically $g$ downward, the mass-weighted average is trivially also $g$, and there's no need for vector decomposition of the different launch velocities.
   • Assuming the horizontally-thrown body C somehow contributes a different (lower) vertical acceleration — its vertical acceleration is also exactly $g$, its horizontal velocity component is irrelevant to the vertical/net acceleration of the COM (the COM does drift horizontally too, but the acceleration magnitude asked for is just $g$, vertically).

   ✓Final answer

   The correct option is (B) — $10\ \mathrm{ms^{-2}}$.

   ANSWER: B

8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Two particles of masses '$m_1$' and '$m_2$' ($m_1 > m_2$) are separated by a distance '$d$'. When the positions of the two particles are interchanged, the shift in the centre of mass is (A) $\left(\dfrac{m_1+m_2}{m_1-m_2}\right)d$ (B) $\left(\dfrac{m_1-m_2}{m_1+m_2}\right)d$ (C) zero (D) $\left(\dfrac{m_1}{m_1-m_2}\right)d$

   ›Reveal solutionSolution

   Compute the centre of mass before and after swapping positions; the shift is $\left(\dfrac{m_1-m_2}{m_1+m_2}\right)d$.

   Concept and Intuition

   The centre of mass of a two-particle system is a weighted average of position by mass. Interchanging the particles' positions changes which mass sits at which coordinate, shifting the COM toward whichever particle is now on the heavier side.

   Step-by-Step Solution

   1. Let $m_1$ initially at $x=0$ and $m_2$ at $x=d$.
   2. Initial COM: $x_{cm,i}=\dfrac{m_1(0)+m_2(d)}{m_1+m_2}=\dfrac{m_2d}{m_1+m_2}$.
   3. After interchange, $m_1$ is at $x=d$ and $m_2$ at $x=0$: $x_{cm,f}=\dfrac{m_1(d)+m_2(0)}{m_1+m_2}=\dfrac{m_1d}{m_1+m_2}$.
   4. Shift $=x_{cm,f}-x_{cm,i}=\dfrac{(m_1-m_2)d}{m_1+m_2}$. Since $m_1>m_2$, this is positive, and its magnitude is $\left(\dfrac{m_1-m_2}{m_1+m_2}\right)d$.

   Common Mistakes

   • Getting the sign/ratio backwards (writing $m_2-m_1$ or $m_1+m_2$ over $m_1-m_2$).
   • Assuming the shift is always $d$ regardless of the masses.

   ✓Final answer

   The correct option is (B) — $\left(\dfrac{m_1-m_2}{m_1+m_2}\right)d$.

   ANSWER: B

9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If two bodies of masses 2 kg and 3 kg are moving at right angles with velocities $20\ \mathrm{m\,s^{-1}}$ and $10\ \mathrm{m\,s^{-1}}$ respectively, then the velocity of the centre of mass of the system of the two bodies is (A) $5\ \mathrm{m\,s^{-1}}$ (B) $30\ \mathrm{m\,s^{-1}}$ (C) $10\ \mathrm{m\,s^{-1}}$ (D) $14\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   The centre-of-mass velocity is the vector sum of individual momenta divided by total mass. With perpendicular momenta of 40 and 30 kg·m/s, the resultant is 50 kg·m/s, giving (C) $10\ \mathrm{m\,s^{-1}}$.

   Concept and Intuition

   The velocity of the centre of mass of a system is $\vec v_{cm}=\dfrac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}$, which is the same as saying $\vec v_{cm}=\dfrac{\vec p_{total}}{M_{total}}$, where $\vec p_{total}$ is the vector sum of the individual momenta. When the two velocities are perpendicular, their momenta are also perpendicular, so we combine them using the Pythagorean theorem.

   Step-by-Step Solution

   1. Momentum of body 1: $p_1=m_1v_1=2\times20=40$ kg·m/s.
   2. Momentum of body 2: $p_2=m_2v_2=3\times10=30$ kg·m/s.
   3. Since the velocities (and hence momenta) are at right angles, the resultant momentum magnitude is $p_{total}=\sqrt{p_1^2+p_2^2}=\sqrt{40^2+30^2}=\sqrt{1600+900}=\sqrt{2500}=50$ kg·m/s.
   4. Total mass: $M=m_1+m_2=2+3=5$ kg.
   5. Velocity of centre of mass: $v_{cm}=\dfrac{p_{total}}{M}=\dfrac{50}{5}=10\ \mathrm{m\,s^{-1}}$.

   Common Mistakes

   • Simply adding the two velocities (20+10=30) instead of combining momenta vectorially.
   • Forgetting to use the Pythagorean rule for perpendicular vectors and instead adding momenta algebraically.
   • Dividing by the wrong mass (e.g. using only one body's mass instead of the total).

   ✓Final answer

   The correct option is (C) — $10\ \mathrm{m\,s^{-1}}$.

   ANSWER: C

10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.Three blocks A, B and C are arranged as shown in the figure such that the distance between two successive blocks is 10 m. Block A is displaced towards block B by 2 m and block C is displaced towards block B by 3 m. The distance through which the block B should be moved so that the centre of mass of the system does not change is [FIGURE] (three blocks on a horizontal line, in order A (10 kg), B (25 kg), C (15 kg), each successive pair 10 m apart) (A) 1.4 m, towards block C (B) 1.5 m, towards block A (C) 2 m, towards block A (D) 1 m, towards block C

    ›Reveal solutionSolution

    A centre-of-mass-invariance problem: with A and C's new positions fixed, solving for B's displacement that keeps the weighted average position unchanged gives 1 m towards C.

    Concept and Intuition

    The centre of mass of a system of point masses is the mass-weighted average position. If we're told the COM must stay fixed while two of the three masses move, we can treat the third mass's position as the unknown and solve the COM equation for it — exactly like solving for an unknown coordinate in a weighted average.

    Step-by-Step Solution

    1. Set up initial coordinates along the line: A at $0$, B at $10$, C at $20$ (each successive pair 10 m apart), with masses $m_A=10$, $m_B=25$, $m_C=15$ kg (total $50$ kg).
    2. Initial COM: $x_{cm}=\dfrac{10(0)+25(10)+15(20)}{50}=\dfrac{0+250+300}{50}=11$ m.
    3. A moves 2 m towards B (i.e. towards increasing $x$): new position $=0+2=2$.
    4. C moves 3 m towards B (i.e. towards decreasing $x$): new position $=20-3=17$.
    5. Let B's new position be $10+d$ (positive $d$ = towards C). For the COM to stay at 11 m: $\dfrac{10(2)+25(10+d)+15(17)}{50}=11$.
    6. $20+250+25d+255 = 550 \Rightarrow 525+25d=550 \Rightarrow d = 1$.
    7. $d=+1$ (positive, i.e. in the direction from A towards C), so B moves 1 m towards C.

    Common Mistakes

    • Getting the sign of A's and C's displacement directions backwards (both move towards B, i.e. towards each other).
    • Assuming B must move towards A just because A and C's masses/effects seem to "pull" it that way, without solving the equation.

    ✓Final answer

    The correct option is (D) — 1 m, towards block C.

    ANSWER: D

11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.The coordinates of the centre of mass of a uniform L shaped plate of mass 3 kg shown in the figure is [FIGURE] (an L-shaped plate on the x-y plane (metres): the outline runs from the origin (0,0) up to (0,2), across to (1,2), down to (1,1), across to (2,1), down to (2,0), and back to (0,0) — i.e. a 2x2 square with the 1x1 square at the bottom-right corner removed) (A) $\left(\dfrac{5}{6}\ \mathrm{m}, \dfrac{5}{6}\ \mathrm{m}\right)$ (B) $\left(\dfrac{3}{2}\ \mathrm{m}, \dfrac{3}{2}\ \mathrm{m}\right)$ (C) $\left(\dfrac{1}{2}\ \mathrm{m}, \dfrac{1}{2}\ \mathrm{m}\right)$ (D) $\left(\dfrac{6}{5}\ \mathrm{m}, \dfrac{6}{5}\ \mathrm{m}\right)$

    ›Reveal solutionSolution

    Tests centre of mass of a composite uniform lamina by decomposing it into simple rectangles. Answer: $(5/6\ \mathrm{m}, 5/6\ \mathrm{m})$.

    Concept and Intuition

    For a uniform (constant surface density) lamina made of an irregular shape, the standard trick is to break it into simple pieces whose individual centroids and areas are easy to write down, then combine them as a weighted average — weighting by area (since uniform density means mass is proportional to area).

    Step-by-Step Solution

    1. From the figure, the L-shape is bounded by the path $(0,0)\to(0,2)\to(1,2)\to(1,1)\to(2,1)\to(2,0)\to(0,0)$. This decomposes into:
       • Rectangle 1: $x\in[0,1]$, $y\in[0,2]$ — a $1\ \mathrm{m}\times2\ \mathrm{m}$ block, area $A_1 = 2\ \mathrm{m^2}$, centroid at $(0.5, 1)$.
       • Rectangle 2: $x\in[1,2]$, $y\in[0,1]$ — a $1\ \mathrm{m}\times1\ \mathrm{m}$ block, area $A_2 = 1\ \mathrm{m^2}$, centroid at $(1.5, 0.5)$.
    2. Total area $= A_1+A_2 = 3\ \mathrm{m^2}$, matching the total mass of 3 kg (so surface density $= 1\ \mathrm{kg/m^2}$, and each rectangle's mass equals its area in kg): $m_1 = 2$ kg, $m_2 = 1$ kg.
    3. Centre of mass:

    $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2} = \frac{2(0.5)+1(1.5)}{3} = \frac{1+1.5}{3} = \frac{2.5}{3} = \frac{5}{6}\ \mathrm{m}$

    $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1+m_2} = \frac{2(1)+1(0.5)}{3} = \frac{2+0.5}{3} = \frac{2.5}{3} = \frac{5}{6}\ \mathrm{m}$

    Common Mistakes

    • Mis-decomposing the L-shape (mixing up which corner is missing) — always trace the boundary path carefully to identify the two rectangles.
    • Forgetting that uniform density lets you weight by area directly instead of needing an explicit density value.

    ✓Final answer

    The correct option is (A) — $\left(\dfrac{5}{6}\ \mathrm{m}, \dfrac{5}{6}\ \mathrm{m}\right)$.

    ANSWER: A

12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.A body of mass 2 kg is moving towards north with a velocity of $20\ \mathrm{m\,s^{-1}}$ and another body of mass 3 kg is moving towards east with a velocity of $10\ \mathrm{m\,s^{-1}}$. The magnitude of the velocity of the centre of mass of the system of the two bodies is (A) $20\ \mathrm{m\,s^{-1}}$ (B) $10\ \mathrm{m\,s^{-1}}$ (C) $15\ \mathrm{m\,s^{-1}}$ (D) $2\sqrt{5}\ \mathrm{m\,s^{-1}}$

    ›Reveal solutionSolution

    The centre of mass moves with velocity equal to total momentum over total mass; since the two momenta are perpendicular they add as a 3-4-5 vector triangle, giving $10\ \mathrm{m\,s^{-1}}$ — option (B).

    Concept and Intuition

    The velocity of the centre of mass of a system is defined by $\vec v_{cm}=\dfrac{\sum m_i\vec v_i}{\sum m_i}$ — it is exactly the total (vector) momentum of the system divided by the total mass. Because the two bodies move along perpendicular directions (north and east), their momenta must be combined as vectors, not scalars — this is the crux of the problem.

    Step-by-Step Solution

    1. Momentum of body 1 (north): $p_1=m_1v_1=2\times20=40\ \mathrm{kg\,m\,s^{-1}}$ (north).
    2. Momentum of body 2 (east): $p_2=m_2v_2=3\times10=30\ \mathrm{kg\,m\,s^{-1}}$ (east).
    3. Since north and east are perpendicular, the resultant momentum magnitude is $p=\sqrt{p_1^2+p_2^2}=\sqrt{40^2+30^2}=\sqrt{1600+900}=\sqrt{2500}=50\ \mathrm{kg\,m\,s^{-1}}$.
    4. Total mass $M=m_1+m_2=2+3=5\ \mathrm{kg}$.
    5. Centre-of-mass speed: $v_{cm}=p/M=50/5=10\ \mathrm{m\,s^{-1}}$.

    Common Mistakes

    • Adding the speeds algebraically ($20+10=30$, then dividing by 5 to get 6) instead of combining momenta as perpendicular vectors.
    • Forgetting to weight each velocity by its own mass before combining.

    ✓Final answer

    The correct option is (B) — $10\ \mathrm{m\,s^{-1}}$.

    ANSWER: B