101. A uniform rod of length \(2L\) has its left half made of material with mass \(m\) and its right half made of material with mass \(3m\). The centre of mass from the left end is
ⓐ. \(\frac{5L}{4}\)
ⓑ. \(L\)
ⓒ. \(\frac{3L}{2}\)
ⓓ. \(\frac{L}{2}\)
Correct Answer: \(\frac{5L}{4}\)
Explanation: \( \textbf{Divide the rod:} \) Treat the left half and right half as two uniform parts.
\( \textbf{Masses:} \) Left half has mass \(m\), and right half has mass \(3m\).
\( \textbf{Centres of the parts:} \) The left half extends from \(0\) to \(L\), so \(x_1=\frac{L}{2}\).
\( \textbf{Centre of right half:} \) The right half extends from \(L\) to \(2L\), so \(x_2=\frac{3L}{2}\).
\( \textbf{Composite relation:} \)
\[
X_{\text{CM}}=\frac{m_1x_1+m_2x_2}{m_1+m_2}
\]
\( \textbf{Substitution:} \)
\[
X_{\text{CM}}=\frac{m\left(\frac{L}{2}\right)+3m\left(\frac{3L}{2}\right)}{m+3m}
\]
\( \textbf{Numerator:} \)
\[
\frac{mL}{2}+\frac{9mL}{2}=5mL
\]
\( \textbf{Denominator:} \)
\[
m+3m=4m
\]
\( \textbf{Calculation:} \)
\[
X_{\text{CM}}=\frac{5mL}{4m}=\frac{5L}{4}
\]
\( \textbf{Final answer:} \) The centre of mass is \(\frac{5L}{4}\) from the left end, shifted toward the heavier right half.
102. In the negative-mass method for a lamina with a hole, the removed part is treated as
ⓐ. zero mass placed only at the chosen origin
ⓑ. a force acting at the edge of the hole
ⓒ. a positive mass at the centre of mass of the removed part
ⓓ. a negative mass at the removed part's centre of mass
Correct Answer: a negative mass at the removed part's centre of mass
Explanation: A body with a hole can be treated as a complete body plus a removed part having negative mass. The negative mass is placed at the centre of mass of the removed portion. This method keeps the weighted-average formula usable without separately describing the complicated remaining shape. The total mass used in the denominator is the mass of the complete body minus the removed mass. The sign is only a calculation device for removed material, not a claim that real negative mass exists in the lamina.
103. A uniform square lamina of side \(2a\) has a small square of side \(a\) removed from one corner. To set up the centre-of-mass calculation using the subtraction method, the correct idea is to use
ⓐ. the geometrical centre of the original square without any correction
ⓑ. only the outer boundary of the remaining shape
ⓒ. the full square as positive mass and the removed square as negative mass
ⓓ. the removed square as positive mass and the full square as negative mass
Correct Answer: the full square as positive mass and the removed square as negative mass
Explanation: In the subtraction method, the original complete lamina is first treated as a positive mass located at its own centre of mass. The removed portion is then treated as a negative mass located at the centre of mass of that removed portion. The centre of mass of the remaining body is found by applying the usual weighted-average formula with these signed masses. This works because removing material reduces the mass contribution from that region. The original geometrical centre alone is no longer enough because the missing corner breaks the symmetry.
104. A composite body is made by joining two point-replaceable parts. Part P has mass \(2M\) and its own centre at \(x=0\). Part Q has mass \(M\) and its own centre at \(x=6a\). The centre of mass of the composite body is
ⓐ. \(2a\)
ⓑ. \(3a\)
ⓒ. \(4a\)
ⓓ. \(6a\)
Correct Answer: \(2a\)
Explanation: \( \textbf{Composite-body model:} \) Treat each part as a point mass at its own centre of mass.
\( \textbf{Given data:} \) Part P has mass \(2M\) at \(x=0\), and part Q has mass \(M\) at \(x=6a\).
\( \textbf{Required quantity:} \) \(X_{\text{CM}}\) of the two-part composite body.
\( \textbf{Relation:} \)
\[
X_{\text{CM}}=\frac{m_Px_P+m_Qx_Q}{m_P+m_Q}
\]
\( \textbf{Substitution:} \)
\[
X_{\text{CM}}=\frac{(2M)(0)+(M)(6a)}{2M+M}
\]
\( \textbf{Simplification:} \)
\[
X_{\text{CM}}=\frac{6Ma}{3M}
\]
\( \textbf{Cancel the common mass factor:} \)
\[
X_{\text{CM}}=2a
\]
\( \textbf{Physical check:} \) The result is closer to the heavier part P at \(x=0\), so \(2a\) is reasonable.
\( \textbf{Final answer:} \) The centre of mass is at \(x=2a\).
105. A uniform square lamina of side \(2a\) has a square of side \(a\) removed from its upper-right corner. Take the centre of the original square as origin, with axes parallel to its sides. The centre of the removed square is at \(\left(\frac{a}{2},\frac{a}{2}\right)\). If the full square has mass \(4m\), the centre of mass of the remaining lamina is
ⓐ. \(\left(-\frac{a}{6},-\frac{a}{6}\right)\)
ⓑ. \(\left(\frac{a}{6},\frac{a}{6}\right)\)
ⓒ. \(\left(-\frac{a}{3},-\frac{a}{3}\right)\)
ⓓ. \(\left(\frac{a}{3},\frac{a}{3}\right)\)
Correct Answer: \(\left(-\frac{a}{6},-\frac{a}{6}\right)\)
Explanation: \( \textbf{Complete lamina:} \) The full square has mass \(4m\) and centre of mass at \((0,0)\).
\( \textbf{Removed part:} \) The removed square has side \(a\), so its mass is \(m\), and it is treated as negative mass at \(\left(\frac{a}{2},\frac{a}{2}\right)\).
\( \textbf{Remaining mass:} \)
\[
M_{\text{rem}}=4m-m=3m
\]
\( \textbf{x-coordinate:} \)
\[
X_{\text{CM}}=\frac{(4m)(0)-m\left(\frac{a}{2}\right)}{3m}
\]
\[
X_{\text{CM}}=-\frac{a}{6}
\]
\( \textbf{y-coordinate:} \)
\[
Y_{\text{CM}}=\frac{(4m)(0)-m\left(\frac{a}{2}\right)}{3m}
\]
\[
Y_{\text{CM}}=-\frac{a}{6}
\]
\( \textbf{Direction check:} \) Mass is removed from the upper-right corner, so the remaining centre of mass shifts toward the lower-left side.
\( \textbf{Final answer:} \) The centre of mass is \(\left(-\frac{a}{6},-\frac{a}{6}\right)\).
106. A uniform disc has a small circular hole cut out away from its centre. For locating the centre of mass of the remaining lamina, the most suitable modelling step is to
ⓐ. ignore the removed part because it contains no material now
ⓑ. keep the centre of mass at the original disc centre in every off-centre case
ⓒ. treat the hole as a negative mass placed at the hole's centre
ⓓ. place the whole remaining mass at the rim of the large disc
Correct Answer: treat the hole as a negative mass placed at the hole's centre
Explanation: A hole is handled by treating the removed portion as negative mass in the centre-of-mass formula. For a circular hole, the removed part's mass may be placed at the centre of that small circle because the removed circular part is itself uniform and symmetric. The full disc is treated as positive mass at its original centre. The remaining lamina's centre of mass comes from the weighted average of the positive full disc and the negative removed disc. If the hole is off-centre, the original symmetry is broken, so the centre of mass generally shifts away from the hole.
107. A composite body is formed by joining two uniform rods at right angles. Rod P of mass \(m\) and length \(2a\) lies along the \(x\)-axis from \(x=0\) to \(x=2a\). Rod Q of mass \(2m\) and length \(2a\) lies along the \(y\)-axis from \(y=0\) to \(y=2a\). The centre of mass of the composite body is
ⓐ. \((a,a)\)
ⓑ. \(\left(\frac{2a}{3},\frac{a}{3}\right)\)
ⓒ. \(\left(\frac{a}{3},\frac{2a}{3}\right)\)
ⓓ. \(\left(\frac{a}{2},\frac{a}{2}\right)\)
Correct Answer: \(\left(\frac{a}{3},\frac{2a}{3}\right)\)
Explanation: \( \textbf{Rod P:} \) It lies from \(x=0\) to \(x=2a\), so its centre is at \((a,0)\), and its mass is \(m\).
\( \textbf{Rod Q:} \) It lies from \(y=0\) to \(y=2a\), so its centre is at \((0,a)\), and its mass is \(2m\).
\( \textbf{Total mass:} \)
\[
M=m+2m=3m
\]
\( \textbf{x-coordinate:} \)
\[
X_{\text{CM}}=\frac{m(a)+2m(0)}{3m}
\]
\[
X_{\text{CM}}=\frac{a}{3}
\]
\( \textbf{y-coordinate:} \)
\[
Y_{\text{CM}}=\frac{m(0)+2m(a)}{3m}
\]
\[
Y_{\text{CM}}=\frac{2a}{3}
\]
\( \textbf{Mass-distribution check:} \) The heavier vertical rod pulls the centre of mass closer to the \(y\)-axis and upward.
\( \textbf{Final answer:} \) The centre of mass is \(\left(\frac{a}{3},\frac{2a}{3}\right)\).
108. A uniform lamina is made by joining a square plate of mass \(4m\) with a smaller square plate of mass \(m\) along one side. Their individual centres lie on the same horizontal line and are separated by \(3a\). Measured from the centre of the larger square toward the smaller square, the centre of mass is
ⓐ. \(\frac{4a}{5}\)
ⓑ. \(\frac{3a}{4}\)
ⓒ. \(\frac{3a}{5}\)
ⓓ. \(\frac{5a}{3}\)
Correct Answer: \(\frac{3a}{5}\)
Explanation: \( \textbf{Composite-body idea:} \) Replace each square plate by a point mass at its own centre of mass.
\( \textbf{Choose origin:} \) Take the centre of the larger square as \(x=0\).
\( \textbf{Mass-position data:} \) Larger square: mass \(4m\), position \(0\). Smaller square: mass \(m\), position \(3a\).
\( \textbf{Required coordinate:} \) \(X_{\text{CM}}\) measured from the larger square's centre.
\( \textbf{Formula:} \)
\[
X_{\text{CM}}=\frac{m_1x_1+m_2x_2}{m_1+m_2}
\]
\( \textbf{Substitution:} \)
\[
X_{\text{CM}}=\frac{(4m)(0)+m(3a)}{4m+m}
\]
\( \textbf{Simplification:} \)
\[
X_{\text{CM}}=\frac{3ma}{5m}
\]
\[
X_{\text{CM}}=\frac{3a}{5}
\]
\( \textbf{Final answer:} \) The centre of mass lies \(\frac{3a}{5}\) from the larger square's centre toward the smaller square.
109. Study the table and choose the row that gives a suitable composite-body replacement.
| Row | Body part | Replacement for centre-of-mass calculation |
|---|
| P | Uniform rod segment | Point mass at its midpoint |
| Q | Uniform circular disc | Point mass at any point on the rim |
| R | Removed uniform part | Positive mass at the removed part's centre |
| S | Uniform rectangular plate | Point mass at one corner |
ⓐ. Row S
ⓑ. Row P
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row P
Explanation: A uniform rod segment has its centre of mass at its midpoint, so it can be replaced by a point mass at that point for a composite-body calculation. A uniform disc is replaced by a point mass at its centre, not at an arbitrary point on the rim. A removed part is treated as negative mass, not positive mass, when using the subtraction method. A uniform rectangular plate is replaced by a point mass at the intersection of its diagonals, not at a corner. The replacement point must be the centre of mass of that individual part.
110. A body is made by removing a small mass \(m\) at \(x=4a\) from a larger uniform body of mass \(5m\) whose centre is at \(x=0\). The centre of mass of the remaining body is
ⓐ. \(a\)
ⓑ. \(-\frac{4a}{5}\)
ⓒ. \(\frac{4a}{5}\)
ⓓ. \(-a\)
Correct Answer: \(-a\)
Explanation: \( \textbf{Full body:} \) Treat the larger body as mass \(5m\) at \(x=0\).
\( \textbf{Removed part:} \) Treat the removed mass as negative mass \(-m\) at \(x=4a\).
\( \textbf{Remaining mass:} \)
\[
M_{\text{rem}}=5m-m=4m
\]
\( \textbf{Use signed-mass average:} \)
\[
X_{\text{CM}}=\frac{(5m)(0)+(-m)(4a)}{4m}
\]
\( \textbf{Numerator:} \)
\[
(5m)(0)-4ma=-4ma
\]
\( \textbf{Calculation:} \)
\[
X_{\text{CM}}=\frac{-4ma}{4m}=-a
\]
\( \textbf{Direction check:} \) Removing mass from the positive side shifts the remaining centre of mass toward the negative side.
\( \textbf{Final answer:} \) The centre of mass of the remaining body is at \(x=-a\).
111. Consider the following statements about composite-body centre-of-mass calculations.
I. Each simple part may be replaced by its mass placed at its own centre of mass.
II. The coordinate origin can be chosen for convenience, but all part coordinates must then use the same origin.
III. Removed material is treated as positive mass at the centre of the removed part.
ⓐ. II and III only
ⓑ. I, II, and III
ⓒ. I and III only
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is true because the detailed shape of a simple part can be replaced by its total mass at its own centre of mass. Statement II is true because the coordinate origin is a calculation choice, but consistency is necessary once it is chosen. Statement III is false in the subtraction method because removed material is treated as negative mass. Using positive mass for the removed part would shift the centre of mass toward the missing region instead of away from it. The sign of the removed mass is the key difference between adding parts and cutting parts out.
112. A uniform rod of length \(4a\) has a segment of length \(a\) cut off from its right end. If the original rod's left end is taken as \(x=0\), the centre of mass of the remaining rod is
ⓐ. \(a\)
ⓑ. \(\frac{5a}{2}\)
ⓒ. \(2a\)
ⓓ. \(\frac{3a}{2}\)
Correct Answer: \(\frac{3a}{2}\)
Explanation: \( \textbf{Remaining part:} \) Cutting off the rightmost length \(a\) leaves a uniform rod from \(x=0\) to \(x=3a\).
\( \textbf{Direct symmetry method:} \) The remaining rod is uniform and straight.
\( \textbf{Centre of a uniform rod:} \) Its centre of mass lies at the midpoint of its remaining length.
\( \textbf{Remaining length:} \)
\[
L_{\text{rem}}=3a
\]
\( \textbf{Midpoint from the left end:} \)
\[
X_{\text{CM}}=\frac{3a}{2}
\]
\( \textbf{Subtraction interpretation:} \) The removed part was on the right side, so the centre of mass moves left from the original midpoint \(2a\).
\( \textbf{Final answer:} \) The centre of mass of the remaining rod is at \(\frac{3a}{2}\).
113. The velocity of the centre of mass of a system of particles is given by
ⓐ. \(\vec{V}_{\text{CM}}=\frac{\sum m_i\vec{r}_i}{\sum m_i}\)
ⓑ. \(\vec{V}_{\text{CM}}=\frac{\sum m_i\vec{v}_i}{\sum m_i}\)
ⓒ. \(\vec{V}_{\text{CM}}=\sum \frac{\vec{v}_i}{m_i}\)
ⓓ. \(\vec{V}_{\text{CM}}=\left(\sum m_i\right)\left(\sum \vec{v}_i\right)\)
Correct Answer: \(\vec{V}_{\text{CM}}=\frac{\sum m_i\vec{v}_i}{\sum m_i}\)
Explanation: The velocity of the centre of mass is the mass-weighted average of the velocities of all particles in the system. The term \(m_i\vec{v}_i\) is the momentum contribution of the \(i^{\text{th}}\) particle. Dividing the sum of these terms by the total mass \(M\) gives a velocity. The formula is vectorial, so directions of the individual velocities matter. It is not a simple average unless all particles have equal masses.
114. Two particles of masses \(2\,\text{kg}\) and \(3\,\text{kg}\) move along the \(x\)-axis with velocities \(4\,\text{m s}^{-1}\) and \(-1\,\text{m s}^{-1}\), respectively. The velocity of the centre of mass is
ⓐ. \(5\,\text{m s}^{-1}\)
ⓑ. \(1\,\text{m s}^{-1}\)
ⓒ. \(2\,\text{m s}^{-1}\)
ⓓ. \(3\,\text{m s}^{-1}\)
Correct Answer: \(1\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(m_1=2\,\text{kg}\), \(v_1=4\,\text{m s}^{-1}\), \(m_2=3\,\text{kg}\), and \(v_2=-1\,\text{m s}^{-1}\).
\( \textbf{Required quantity:} \) \(V_{\text{CM}}\) along the \(x\)-axis.
\( \textbf{Useful relation:} \)
\[
V_{\text{CM}}=\frac{m_1v_1+m_2v_2}{m_1+m_2}
\]
\( \textbf{Why signs matter:} \) The velocities are along the same axis, and the negative sign shows motion opposite to the positive direction.
\( \textbf{Substitution:} \)
\[
V_{\text{CM}}=\frac{(2)(4)+(3)(-1)}{2+3}\,\text{m s}^{-1}
\]
\( \textbf{Numerator:} \)
\[
8-3=5
\]
\( \textbf{Denominator:} \)
\[
2+3=5
\]
\( \textbf{Calculation:} \)
\[
V_{\text{CM}}=\frac{5}{5}\,\text{m s}^{-1}=1\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The centre of mass moves with velocity \(1\,\text{m s}^{-1}\) in the positive \(x\)-direction.
115. A two-particle system has equal masses. One particle moves east with speed \(v\), and the other moves west with the same speed \(v\). The velocity of the centre of mass is
ⓐ. \(v\) east
ⓑ. \(v\) west
ⓒ. zero
ⓓ. \(2v\) east
Correct Answer: zero
Explanation: Equal masses moving with equal speeds in opposite directions have equal and opposite momentum contributions. The velocity of the centre of mass depends on the vector sum \(m\vec{v}+m(-\vec{v})\). This sum is zero, so \(\vec{V}_{\text{CM}}=\vec{0}\). The particles themselves are moving, but the representative point of the whole system remains at rest if no other motion is involved. This shows that internal relative motion and centre-of-mass motion are different descriptions.
116. The acceleration of the centre of mass is expressed as
ⓐ. \(\vec{A}_{\text{CM}}=\frac{\sum m_i\vec{a}_i}{\sum m_i}\)
ⓑ. \(\vec{A}_{\text{CM}}=\sum \frac{\vec{a}_i}{m_i}\)
ⓒ. \(\vec{A}_{\text{CM}}=\frac{\sum m_i\vec{v}_i}{\sum m_i}\)
ⓓ. \(\vec{A}_{\text{CM}}=\left(\sum m_i\right)\left(\sum \vec{a}_i\right)\)
Correct Answer: \(\vec{A}_{\text{CM}}=\frac{\sum m_i\vec{a}_i}{\sum m_i}\)
Explanation: The acceleration of the centre of mass follows the same weighted-average structure as position and velocity. Each particle acceleration is weighted by the corresponding mass. The denominator is the total mass \(M\), so the result has the unit of acceleration, \(\text{m s}^{-2}\). This relation is especially useful because it connects naturally with the net external force on a system. The acceleration of the centre of mass is controlled by the system-level balance, not by simply choosing the largest individual acceleration.
117. A system consists of two particles of masses \(m\) and \(3m\). Their accelerations along a line are \(4a\) and \(-2a\), respectively. The acceleration of the centre of mass is
ⓐ. \(-\frac{a}{2}\)
ⓑ. \(+\frac{a}{2}\)
ⓒ. \(-\frac{3a}{2}\)
ⓓ. \(+\frac{3a}{2}\)
Correct Answer: \(-\frac{a}{2}\)
Explanation: \( \textbf{Given data:} \) The masses are \(m\) and \(3m\).
\( \textbf{Accelerations:} \) \(a_1=4a\) and \(a_2=-2a\).
\( \textbf{Required quantity:} \) \(A_{\text{CM}}\) along the given line.
\( \textbf{Formula:} \)
\[
A_{\text{CM}}=\frac{m_1a_1+m_2a_2}{m_1+m_2}
\]
\( \textbf{Substitution:} \)
\[
A_{\text{CM}}=\frac{m(4a)+3m(-2a)}{m+3m}
\]
\( \textbf{Numerator:} \)
\[
4ma-6ma=-2ma
\]
\( \textbf{Denominator:} \)
\[
m+3m=4m
\]
\( \textbf{Calculation:} \)
\[
A_{\text{CM}}=\frac{-2ma}{4m}=-\frac{a}{2}
\]
\( \textbf{Sign meaning:} \) The heavier particle has acceleration in the negative direction, so it dominates the weighted average.
\( \textbf{Final answer:} \) The acceleration of the centre of mass is \(-\frac{a}{2}\).
118. A firecracker initially at rest explodes into two fragments on a smooth horizontal surface. During the short explosion, horizontal external force is negligible. The centre of mass
ⓐ. starts moving in the direction of the heavier fragment
ⓑ. starts moving in the direction of the lighter fragment
ⓒ. remains at rest
ⓓ. moves upward with increasing speed
Correct Answer: remains at rest
Explanation: The explosion produces large internal forces between the fragments. These internal forces can change the velocities of the individual fragments, sending them in different directions. However, if the net external horizontal force is negligible, the motion of the centre of mass is not changed by those internal forces. Since the firecracker was initially at rest, the centre of mass remains at rest during and after the explosion in the horizontal description. The fragments may separate rapidly even though the centre-of-mass motion is unchanged.
119. A system has total mass \(M\). If the net external force on it is \(\vec{F}_{\text{ext}}\), the centre-of-mass acceleration is connected by
ⓐ. \(\vec{F}_{\text{ext}}=\sum m_i\vec{r}_i\)
ⓑ. \(\vec{F}_{\text{ext}}=M\vec{V}_{\text{CM}}\)
ⓒ. \(\vec{F}_{\text{ext}}=M\vec{A}_{\text{CM}}\)
ⓓ. \(\vec{F}_{\text{ext}}=\frac{\vec{A}_{\text{CM}}}{M}\)
Correct Answer: \(\vec{F}_{\text{ext}}=M\vec{A}_{\text{CM}}\)
Explanation: The centre of mass of a system moves as if the total external force acts on the total mass of the system. The relation is \(\vec{F}_{\text{ext}}=M\vec{A}_{\text{CM}}\). Internal forces may redistribute motion among the particles, but they do not determine the acceleration of the centre of mass when they cancel in action-reaction pairs. The equation is a system-level form of Newton's second law. It uses external force, not the sum of internal pushes and pulls.
120. A \(10\,\text{kg}\) system has a net external force of \(30\,\text{N}\) acting toward the positive \(x\)-direction. The acceleration of its centre of mass is
ⓐ. \(3\,\text{m s}^{-2}\)
ⓑ. \(20\,\text{m s}^{-2}\)
ⓒ. \(0.3\,\text{m s}^{-2}\)
ⓓ. \(300\,\text{m s}^{-2}\)
Correct Answer: \(3\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given data:} \) Total mass \(M=10\,\text{kg}\), and net external force \(F_{\text{ext}}=30\,\text{N}\).
\( \textbf{Required quantity:} \) Acceleration of the centre of mass \(A_{\text{CM}}\).
\( \textbf{System-level Newton's second law:} \)
\[
F_{\text{ext}}=MA_{\text{CM}}
\]
\( \textbf{Rearrange for acceleration:} \)
\[
A_{\text{CM}}=\frac{F_{\text{ext}}}{M}
\]
\( \textbf{Substitution:} \)
\[
A_{\text{CM}}=\frac{30\,\text{N}}{10\,\text{kg}}
\]
\( \textbf{Unit conversion check:} \) Since \(1\,\text{N}=1\,\text{kg m s}^{-2}\), \(\frac{\text{N}}{\text{kg}}=\text{m s}^{-2}\).
\( \textbf{Calculation:} \)
\[
A_{\text{CM}}=3\,\text{m s}^{-2}
\]
\( \textbf{Direction:} \) The acceleration is in the positive \(x\)-direction because the net external force is in that direction.
\( \textbf{Final answer:} \) The centre of mass accelerates at \(3\,\text{m s}^{-2}\) toward positive \(x\).
