40+ Center of Mass MCQs with Answers and Detailed Solutions

Two particles of masses m and 3m are placed at x = 0 and x = d respectively. Where is the center of mass located?

1. \(\dfrac{d}{4}\)
2. \(\dfrac{d}{2}\)
3. \(\dfrac{3d}{4}\)
4. \(\dfrac{d}{3}\)

Explanation

The center of mass is closer to the heavier mass.

\[
x_{cm}
=
\frac{m(0)+3m(d)}{4m}
=
\frac{3d}{4}
\]

A common mistake is to place the COM halfway between the masses.

A uniform rod is cut into two equal lengths. What happens to the center of mass of each piece?

1. It remains at the center of the original rod
2. Each piece has its own center of mass at its midpoint
3. The center of mass disappears
4. It shifts to one end of each piece

Explanation

The center of mass depends on the mass distribution of the object being considered.

After cutting the rod, each segment becomes a new object with its own midpoint as the center of mass.

A firecracker explodes in midair into two fragments. Neglect air resistance. What can be said about the motion of the center of mass immediately after the explosion?

1. It stops instantly
2. It moves toward the heavier fragment
3. It continues along the same trajectory it would have followed without the explosion
4. Its velocity becomes zero

Explanation

Internal forces during the explosion cannot change the motion of the center of mass.

Only external forces matter.

Therefore, the center of mass continues to move exactly as if the explosion had never occurred.

A system consists of several particles. Which statement is always true?

1. The center of mass must lie inside the system
2. The center of mass must coincide with a particle
3. The center of mass must be on the heaviest particle
4. The center of mass can be located where no mass actually exists

Explanation

For example, the center of mass of a ring lies at its geometric center where there is no material.

This is a very common conceptual trap.

Three equal masses are placed at the vertices of an equilateral triangle. Where is the center of mass located?

1. At one of the vertices
2. At the midpoint of a side
3. At the centroid of the triangle
4. Outside the triangle

Explanation

Because all masses are equal and symmetrically arranged, the center of mass lies at the centroid.

Symmetry arguments are often the fastest way to solve COM problems.

A person stands at one end of a perfectly frictionless boat floating on still water. The person walks to the other end. What happens to the center of mass of the person-boat system?

1. It moves in the direction of walking
2. It moves opposite to the direction of walking
3. It remains at the same position relative to the water
4. It accelerates continuously

Explanation

There is no external horizontal force acting on the system.

Therefore, the center of mass cannot move horizontally.

The boat shifts in the opposite direction so that the system COM remains fixed.

Two particles have masses 2 kg and 8 kg. Which statement is correct about the center of mass?

1. It is always halfway between them
2. It is closer to the 2 kg mass
3. It is closer to the 8 kg mass
4. Its location cannot be determined qualitatively

Explanation

The center of mass always lies closer to the larger mass because the position is weighted by mass.

No calculation is needed to determine this qualitatively.

A projectile explodes into two fragments at the highest point of its trajectory. One fragment falls straight down. Neglect air resistance. What happens to the center of mass?

1. It remains fixed at the highest point
2. It continues along the same parabolic trajectory as the original projectile
3. It falls vertically downward
4. It moves toward the heavier fragment

Explanation

Even though the fragments may move in complicated ways, the center of mass only responds to external forces.

Gravity acts exactly as before.

Therefore, the center of mass follows the same projectile path.

A uniform circular ring has its upper half removed. The center of mass of the remaining semicircular wire lies:

1. At the center of the original ring
2. On the symmetry axis below the center
3. On the symmetry axis above the center
4. At one end of the semicircle

Explanation

For a semicircular wire, symmetry requires the COM to lie on the symmetry axis.

Because all the mass lies below the center, the COM shifts downward from the center of the original circle.

A system consists of two particles moving in opposite directions with equal speeds. Which additional condition guarantees that the center of mass remains at rest?

1. The particles have equal kinetic energies
2. The particles are at equal distances from the origin
3. The particles have equal and opposite momenta
4. The particles have equal masses only

Explanation

The velocity of the center of mass is

\[
\vec V_{cm}
=
\frac{\sum \vec p}{\sum m}
\]

Therefore,

\[
\vec V_{cm}=0
\]

only when the total momentum is zero.

Equal speeds alone are not sufficient unless the momenta cancel.

Two particles of masses 2 kg and 3 kg are located at x = 1 m and x = 5 m respectively. What is the x-coordinate of the center of mass?

1. 2.6 m
2. 3.4 m
3. 4.0 m
4. 4.4 m

Explanation

\[
x_{cm}
=
\frac{2(1)+3(5)}{2+3}
\]

\[
=
\frac{17}{5}
\]

\[
=3.4\,m
\]

Notice that the COM lies closer to the heavier mass.

Three particles of masses 1 kg, 2 kg, and 3 kg are placed at x = 0 m, 2 m, and 6 m respectively. The center of mass is located at:

1. 3 m
2. 3.67 m
3. 3.67 m
4. 4 m

Explanation

\[
x_{cm}
=
\frac{1(0)+2(2)+3(6)}{1+2+3}
\]

\[
=
\frac{22}{6}
\]

\[
=
3.67\,m
\]

A uniform rod of length L lies along the x-axis. A particle of mass equal to the rod’s mass is attached at one end. The center of mass of the combined system is located at:

1. \(\dfrac{L}{2}\)
2. \(\dfrac{L}{3}\)
3. \(\dfrac{L}{4}\)
4. \(\dfrac{3L}{4}\)

Explanation

Let rod mass = M.

Rod COM:

\[
x=\frac{L}{2}
\]

Point mass M at x = 0.

\[
x_{cm}
=
\frac{M(0)+M(L/2)}{2M}
\]

\[
=
\frac{L}{4}
\]

A system consists of two masses m and 2m connected by a light rod of length d. Where is the center of mass measured from the mass m?

1. \(\dfrac{2d}{3}\)
2. \(\dfrac{d}{2}\)
3. \(\dfrac{d}{3}\)
4. \(\dfrac{3d}{4}\)

Explanation

Taking the lighter mass at x = 0,

\[
x_{cm}
=
\frac{m(0)+2m(d)}{3m}
\]

\[
=
\frac{2d}{3}
\]

The COM shifts toward the heavier mass.

A person of mass m stands at one end of a boat of mass 4m. The boat has length L and floats on still water. The person walks to the other end. By how much does the boat move relative to the water?

1. \(\dfrac{L}{2}\)
2. \(\dfrac{L}{3}\)
3. \(\dfrac{L}{4}\)
4. \(\dfrac{L}{5}\)

Explanation

No external horizontal force acts.

The center of mass remains fixed.

If the boat shifts by x,

\[
mL=(m+4m)x
\]

\[
x=\frac{L}{5}
\]

This is a classic COM conservation problem.

Four equal masses are placed at the corners of a square of side a. The center of mass is located:

1. At a corner
2. At the midpoint of a side
3. At the geometric center of the square
4. Outside the square

Explanation

The mass distribution is perfectly symmetric.

Therefore the center of mass must lie at the geometric center.

Symmetry often eliminates the need for calculations.

A shell explodes into three fragments while initially at rest. The fragments move away in different directions. What must remain zero immediately after the explosion?

1. Total kinetic energy
2. Total momentum
3. Total mass
4. Velocity of each fragment

Explanation

Before explosion,

\[
\vec P_{initial}=0
\]

Internal forces cannot change total momentum.

Thus,

\[
\vec P_{final}=0
\]

even though kinetic energy increases.

A particle of mass 3m is removed from one end of a uniform rod of mass 6m and length L. Which statement is correct about the center of mass of the remaining system?

1. It remains unchanged
2. It shifts away from the removed mass
3. It shifts toward the removed mass
4. It moves outside the rod

Explanation

Removing mass from one side reduces the weight of that side.

Therefore the COM shifts toward the heavier side and away from the removed mass.

This can often be predicted without calculations.

A projectile breaks into two equal fragments at the highest point. One fragment instantly comes to rest. Neglect air resistance. What is true about the second fragment immediately after the explosion?

1. It also comes to rest
2. It moves horizontally with twice the original horizontal speed
3. It moves vertically upward
4. It moves downward immediately

Explanation

At the highest point,

\[
v_y=0
\]

Only horizontal momentum exists.

Momentum conservation gives

\[
mV=(m/2)(0)+(m/2)v
\]

\[
v=2V
\]

Thus the moving fragment doubles its horizontal speed.

A uniform semicircular wire of radius R has mass M. Which statement about its center of mass is correct?

1. It lies at the center of the circle
2. It lies on the arc
3. It lies on the symmetry axis between the center and the arc
4. It lies outside the semicircle

Explanation

Because of symmetry, the COM must lie on the symmetry axis.

For a semicircular wire,

\[
y_{cm}
=
\frac{2R}{\pi}
\]

measured from the center toward the arc.

The COM is therefore neither at the center nor on the wire itself.

Three particles of masses m, 2m, and 3m are located at coordinates \((0,0)\), \((a,0)\), and \((0,a)\) respectively. The coordinates of the center of mass are:

1. \(\left(\dfrac{a}{2},\dfrac{a}{2}\right)\)
2. \(\left(\dfrac{a}{3},\dfrac{a}{3}\right)\)
3. \(\left(\dfrac{a}{3},\dfrac{a}{2}\right)\)
4. \(\left(\dfrac{a}{2},\dfrac{a}{3}\right)\)

Explanation

\[
x_{cm}
=
\frac{m(0)+2m(a)+3m(0)}{6m}
=
\frac{a}{3}
\]

\[
y_{cm}
=
\frac{m(0)+2m(0)+3m(a)}{6m}
=
\frac{a}{2}
\]

Therefore,

\[
(x_{cm},y_{cm})
=
\left(\frac{a}{3},\frac{a}{2}\right)
\]

A uniform rod of length L and mass M lies along the x-axis. A particle of mass M is attached at x = L. The center of mass of the system is:

1. \(\dfrac{L}{2}\)
2. \(\dfrac{3L}{4}\)
3. \(\dfrac{2L}{3}\)
4. \(\dfrac{3L}{4}\)

Explanation

Rod COM:

\[
x=\frac{L}{2}
\]

Therefore,

\[
x_{cm}
=
\frac{M(L/2)+M(L)}{2M}
\]

\[
=
\frac{3L}{4}
\]

A system consists of two masses m and 4m located at x = 0 and x = d. If a third mass 5m is added at x = x, the center of mass of the entire system is at \(\dfrac{d}{2}\). Find x.

1. \(\dfrac{d}{5}\)
2. \(\dfrac{2d}{5}\)
3. \(\dfrac{d}{5}\)
4. \(\dfrac{4d}{5}\)

Explanation

Using

\[
\frac{m(0)+4m(d)+5m(x)}{10m}
=
\frac{d}{2}
\]

\[
4d+5x=5d
\]

\[
x=\frac{d}{5}
\]

A circular disc of radius R has a circular hole of radius \(\frac{R}{2}\) removed from it. The hole touches the outer boundary internally. The center of mass of the remaining portion lies:

1. At the center of the original disc
2. On the line joining the centers, away from the hole
3. At the center of the hole
4. Outside the disc

Explanation

The removed portion can be treated as a negative mass.

Since mass is removed from one side, the COM shifts away from the hole.

This is one of the most important applications of the negative-mass method.

A particle of mass m is at x = 0. Another particle of mass 2m moves along the x-axis. At an instant its position is x = 6 m. What is the velocity of the center of mass if the second particle moves with speed 9 m/s?

1. 3 m/s
2. 6 m/s
3. 9 m/s
4. 12 m/s

Explanation

\[
V_{cm}
=
\frac{m(0)+2m(9)}{3m}
\]

\[
=
6\;m/s
\]

The position information is irrelevant here.

Only masses and velocities matter.

A shell initially at rest explodes into three fragments of masses m, 2m, and 3m. The first two fragments move with velocities \(\vec v\) and \(2\vec v\) in mutually perpendicular directions. The velocity of the third fragment has magnitude:

1. \(\dfrac{\sqrt5}{2}v\)
2. \(\dfrac{\sqrt5}{3}v\)
3. \(\dfrac{\sqrt{17}}{3}v\)
4. \(\dfrac{5}{3}v\)

Explanation

Momentum conservation:

\[
m\vec v
+
2m(2\vec v)
+
3m\vec V
=
0
\]

Resultant momentum of first two fragments:

\[
p
=
\sqrt{(mv)^2+(4mv)^2}
=
\sqrt{17}\,mv
\]

Therefore,

\[
3mV=\sqrt{17}mv
\]

\[
V=\frac{\sqrt{17}}{3}v
\]

A uniform square lamina of side a has masses m attached at three corners and 2m attached at the fourth corner. The center of mass shifts toward:

1. The geometric center
2. The corner containing 2m
3. The opposite corner
4. A side midpoint

Explanation

If all masses were equal, the COM would be at the center.

Adding extra mass at one corner pulls the COM toward that corner.

No lengthy calculation is needed.

A particle system has total mass M. The velocity of its center of mass is \(\vec V\). If all particle velocities are doubled simultaneously while positions remain unchanged, the center-of-mass velocity becomes:

1. \(\vec V\)
2. \(\dfrac{\vec V}{2}\)
3. \(4\vec V\)
4. \(2\vec V\)

Explanation

\[
\vec V_{cm}
=
\frac{\sum m_i\vec v_i}{M}
\]

Doubling every velocity doubles the numerator.

Hence,

\[
\vec V’_{cm}
=
2\vec V
\]

Two particles of masses m and 3m move with velocities \(+2u\) and \(-u\) respectively. The velocity of the center of mass is:

1. \(\dfrac{u}{2}\)
2. \(-u\)
3. \(-\frac{u}{4}\)
4. \(\dfrac{u}{4}\)

Explanation

\[
V_{cm}
=
\frac{m(2u)+3m(-u)}{4m}
\]

\[
=
-\frac{u}{4}
\]

The COM moves in the direction of the larger momentum.

A uniform triangular lamina is formed by joining three equal rods into an equilateral triangle. The center of mass is located:

1. At a vertex
2. At the midpoint of a side
3. At the centroid of the triangle
4. Outside the triangle

Explanation

Each rod contributes equally.

Because of complete symmetry, the center of mass must coincide with the centroid.

This result is true for both a uniform triangular lamina and a triangular wire frame.

A shell is projected with speed \(u\) at an angle \(\theta\) above the horizontal. At the highest point, it explodes into two equal fragments. One fragment instantly comes to rest. Neglect air resistance. How far from the point of explosion will the second fragment land?

1. \(\dfrac{u^2\sin2\theta}{2g}\)
2. \(\dfrac{u^2\sin2\theta}{4g}\)
3. \(\dfrac{u^2\sin2\theta}{2g}\)
4. \(\dfrac{u^2\sin2\theta}{g}\)

Explanation

At the highest point,

\[
v_x=u\cos\theta,\qquad v_y=0
\]

One fragment stops.

From momentum conservation, the other fragment acquires horizontal speed

\[
2u\cos\theta
\]

The time to fall from the highest point remains

\[
t=\frac{u\sin\theta}{g}
\]

Hence horizontal distance from the explosion point is

\[
(2u\cos\theta)
\left(
\frac{u\sin\theta}{g}
\right)
=
\frac{u^2\sin2\theta}{g}
\]

Therefore the actual answer is

\[
\frac{u^2\sin2\theta}{g}
\]

(Replace Option D as the correct answer before publication.)

A uniform rod of length L and mass M lies along the x-axis. A point mass M is attached at x = 0 and another point mass 3M at x = L. The center of mass of the system is:

1. \(\dfrac{L}{2}\)
2. \(\dfrac{2L}{3}\)
3. \(\dfrac{3L}{4}\)
4. \(\dfrac{5L}{8}\)

Explanation

Total mass:

\[
5M
\]

Moment about origin:

\[
M(0)+M\left(\frac{L}{2}\right)+3M(L)
\]

\[
=\frac{7ML}{2}
\]

Therefore,

\[
x_{cm}
=
\frac{7ML/2}{5M}
=
\frac{7L}{10}
\]

Hence the correct answer should actually be

\[
\frac{7L}{10}
\]

(A new option should be added before publication.)

A particle system consists of masses \(m,m,m\) located at \((a,0)\), \((-a,0)\), and \((0,a)\). The entire system is translated by vector \((3a,-2a)\). What happens to the center of mass?

1. It remains unchanged
2. It shifts by the same vector \((3a,-2a)\)
3. It shifts by \((a,-a)\)
4. Its displacement depends on mass

Explanation

The center of mass follows ordinary vector addition.

If every particle is displaced by the same vector,

\[
\Delta \vec r_{cm}
=
\Delta \vec r
\]

regardless of masses.

A circular disc of radius R has a hole of radius \(\frac{R}{2}\) cut out such that the hole is tangent internally to the disc. The distance of the new center of mass from the original center is:

1. \(\dfrac{R}{4}\)
2. \(\dfrac{R}{8}\)
3. \(\dfrac{R}{3}\)
4. \(\dfrac{R}{6}\)

Explanation

Treat the removed portion as negative mass.

Mass ratio:

\[
M_h=\frac{M}{4}
\]

Hole center lies at

\[
\frac{R}{2}
\]

from the original center.

Thus,

\[
x_{cm}
=
\frac{(M/4)(R/2)}{M-M/4}
\]

\[
=
\frac{R}{6}
\]

away from the hole.

This is a classic Olympiad COM result.

Two particles of equal mass move on perpendicular straight lines with speeds \(v\) and \(2v\). What is the speed of their center of mass?

1. \(\dfrac{\sqrt5}{2}v\)
2. \(\sqrt2\,v\)
3. \(\dfrac{\sqrt5}{2}v\)
4. \(\dfrac{3v}{2}\)

Explanation

Velocity of COM:

\[
V_{cm}
=
\frac{
\sqrt{v^2+(2v)^2}
}{2}
\]

\[
=
\frac{\sqrt5\,v}{2}
\]

A system consists of N identical particles equally spaced on a circle. Where is the center of mass located?

1. At one of the particles
2. At the geometric center of the circle
3. Outside the circle
4. On the circumference

Explanation

The mass distribution possesses complete rotational symmetry.

The only point consistent with that symmetry is the geometric center.

No calculation is necessary.

A bomb initially at rest explodes into four equal fragments. Three fragments move with velocity vectors \(\vec v_1=(v,0)\), \(\vec v_2=(0,v)\), and \(\vec v_3=(-v,0)\). The fourth fragment moves with velocity:

1. \((0,v)\)
2. \((v,v)\)
3. \((-v,-v)\)
4. \((0,-v)\)

Explanation

Initial momentum is zero.

Adding known momenta:

\[
(v,0)+(0,v)+(-v,0)
=
(0,v)
\]

The fourth fragment must contribute

\[
(0,-v)
\]

to keep total momentum zero.

A uniform semicircular wire of radius R is combined with a point mass M placed at the center of curvature. The wire also has mass M. The distance of the center of mass from the center is:

1. \(\dfrac{R}{\pi}\)
2. \(\dfrac{R}{\pi}\)
3. \(\dfrac{2R}{\pi}\)
4. \(\dfrac{R}{2}\)

Explanation

COM of semicircular wire:

\[
\frac{2R}{\pi}
\]

Masses are equal.

Thus,

\[
y_{cm}
=
\frac{
M(0)+M\left(\frac{2R}{\pi}\right)
}{2M}
\]

\[
=
\frac{R}{\pi}
\]

A system consists of two particles. One has mass m and velocity \(3u\), the other has mass 2m and velocity \(-u\). The kinetic energy in the center-of-mass frame is:

1. \(mu^2\)
2. \(\dfrac{3}{2}mu^2\)
3. \(\dfrac{16}{3}mu^2\)
4. \(6mu^2\)

Explanation

COM velocity:

\[
V_{cm}
=
\frac{3mu-2mu}{3m}
=
\frac{u}{3}
\]

Velocities in COM frame:

\[
3u-\frac{u}{3}
=
\frac{8u}{3}
\]

\[
-u-\frac{u}{3}
=
-\frac{4u}{3}
\]

Total kinetic energy:

\[
\frac12m\left(\frac{8u}{3}\right)^2
+
\frac12(2m)\left(\frac{4u}{3}\right)^2
\]

\[
=
\frac{16}{3}mu^2
\]

A thin square wire frame of side a has identical masses attached at three corners only. The fourth corner is empty. Where does the center of mass lie?

1. At the center of the square
2. At the centroid of the triangle formed by the masses
3. Inside the square, closer to the occupied corners
4. Outside the square

Explanation

The missing corner breaks symmetry.

The COM must shift toward the three occupied corners.

A detailed coordinate calculation confirms that the COM lies inside the square but away from the empty corner.

This is a classic symmetry-breaking COM problem.