Concept of Physics | H VC Verma | Solution | Centre of Mass | Linear Momentum | Collision

1. Three particles of masses \(1.0 \, \text{kg}\), \(2.0 \, \text{kg}\) and \(3.0 \, \text{kg}\) are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge \(1 \, \text{m}\). Locate the centre of mass of the system.

Solution: To locate the center of mass, we first establish a coordinate system. We take corner A as the origin \((0, 0)\) and the side AB along the X-axis.

1. Identify the Coordinates of the Masses:

Since the triangle is equilateral with a side length of \(1 \, \text{m}\):

• Mass \(m_1 = 1.0 \, \text{kg}\) at A: \((x_1, y_1) = (0, 0)\)
• Mass \(m_2 = 2.0 \, \text{kg}\) at B: \((x_2, y_2) = (1, 0)\)
• Mass \(m_3 = 3.0 \, \text{kg}\) at C:
  Using properties of an equilateral triangle:
  \(x_3 = 0.5\),
  \(y_3 = 1 \cdot \sin 60^\circ = \frac{\sqrt{3}}{2}\)Therefore, \((x_3, y_3) = \left(0.5, \frac{\sqrt{3}}{2}\right)\)

2. Calculate the Total Mass (\(M\)):

\[
M = m_1 + m_2 + m_3 = 1.0 + 2.0 + 3.0 = 6.0 \, \text{kg}
\]

3. Apply the Center of Mass Formula:

\[
R = \frac{1}{M} \sum m_i r_i
\]

X-coordinate (\(X_{cm}\)):

\[
X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}
\]

\[
X_{cm} = \frac{(1.0 \times 0) + (2.0 \times 1) + (3.0 \times 0.5)}{6.0}
\]

\[
X_{cm} = \frac{0 + 2.0 + 1.5}{6.0} = \frac{3.5}{6.0} = \frac{7}{12} \, \text{m}
\]

Y-coordinate (\(Y_{cm}\)):

\[
Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}
\]

\[
Y_{cm} = \frac{(1.0 \times 0) + (2.0 \times 0) + \left(3.0 \times \frac{\sqrt{3}}{2}\right)}{6.0}
\]

\[
Y_{cm} = \frac{\frac{3\sqrt{3}}{2}}{6.0} = \frac{3\sqrt{3}}{12} = \frac{\sqrt{3}}{4} \, \text{m}
\]

Final Answer:

The center of mass of the system is located at
\[
\left( \frac{7}{12} \, \text{m}, \frac{\sqrt{3}}{4} \, \text{m} \right).
\]

Solution: 

We use the standard geometry of a water molecule \((H_2O)\).

1. Given Data:

• Bond length \((O\!-\!H)\): \(d = 0.96 \times 10^{-10}\,\text{m}\)
• Bond angle: \(\theta = 104^\circ\)
• Mass of oxygen: \(m_O = 16m\)
• Mass of hydrogen: \(m_H = m\)
• Total mass:
  \[
  M = 16m + m + m = 18m
  \]

2. Choice of Coordinate System:

• Place oxygen at origin: \((0,0)\)
• Take Y-axis along the bisector of the \(H\!-\!O\!-\!H\) angle
• Each bond makes an angle:
  \[
  \frac{104^\circ}{2} = 52^\circ
  \]
  with the Y-axis

3. Coordinates of Atoms:

• Oxygen: \((0,0)\)
• Hydrogen atoms:
  \[
  (-d\sin 52^\circ,\; d\cos 52^\circ), \quad (d\sin 52^\circ,\; d\cos 52^\circ)
  \]

4. Centre of Mass:

\[
Y_{cm} = \frac{\sum m_i y_i}{M}
\]

\[
Y_{cm} = \frac{m\cdot d\cos 52^\circ + m\cdot d\cos 52^\circ}{18m}
\]

\[
Y_{cm} = \frac{2d\cos 52^\circ}{18} = \frac{d\cos 52^\circ}{9}
\]

5. Numerical Calculation:

\[
Y_{cm} = \frac{0.96 \times 10^{-10} \times 0.6157}{9}
\]

\[
Y_{cm} \approx 6.6 \times 10^{-12}\,\text{m}
\]

Final Answer:

Distance of centre of mass from oxygen atom:
\[
\boxed{6.6 \times 10^{-12}\,\text{m}}
\]

3. Seven homogeneous bricks, each of length \(L\), are arranged as shown in figure (9-E2). Each brick is displaced with respect to the one in contact by \( \frac{L}{10} \). Find the x-coordinate of the centre of mass relative to the origin shown.

Solution: 

We treat each brick as a point mass located at its geometric centre.

1. Coordinate System and Mass:

• Origin is at the left edge of the bottom brick
• Each brick has length \(L\), so its centre is at a distance \( \frac{L}{2} \) from its left edge
• Let mass of each brick be \(m\)
• Total mass:
  \[
  M = 7m
  \]

2. Positions of Centres of Bricks:

Left edges shift by \( \frac{L}{10} \) successively and then reverse:

• \(x_1 = \frac{L}{2}\)
• \(x_2 = \frac{L}{10} + \frac{L}{2} = 0.6L\)
• \(x_3 = \frac{2L}{10} + \frac{L}{2} = 0.7L\)
• \(x_4 = \frac{3L}{10} + \frac{L}{2} = 0.8L\)
• \(x_5 = \frac{2L}{10} + \frac{L}{2} = 0.7L\)
• \(x_6 = \frac{L}{10} + \frac{L}{2} = 0.6L\)
• \(x_7 = \frac{L}{2}\)

3. Centre of Mass Formula:

\[
X_{cm} = \frac{1}{M} \sum m_i x_i
\]

Since all masses are equal:

\[
X_{cm} = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7}{7}
\]

\[
X_{cm} = \frac{0.5L + 0.6L + 0.7L + 0.8L + 0.7L + 0.6L + 0.5L}{7}
\]

\[
X_{cm} = \frac{4.4L}{7}
\]

4. Simplification:

\[
X_{cm} = \frac{44L}{70} = \frac{22L}{35}
\]

Final Answer:

\[
\boxed{X_{cm} = \frac{22L}{35}}
\]

4. A uniform disc of radius \(R\) is put over another uniform disc of radius \(2R\) of the same thickness and density. The peripheries of the two discs touch each other. Locate the centre of mass of the system.

Solution: 

We treat each disc as a point mass located at its geometric centre.

1. Choice of Coordinate System:

• Take the centre of the larger disc as origin: \((0,0)\)
• Both discs have the same thickness and density, so mass is proportional to area \((\pi r^2)\)

2. Mass of the Discs:

\[
m_l \propto \pi (2R)^2 = 4\pi R^2, \quad m_s \propto \pi R^2
\]

Let mass of smaller disc be \(m\), then:

\[
m_s = m, \quad m_l = 4m
\]

3. Position of Centres:

• Centre of larger disc: \((0,0)\)
• Since the peripheries touch externally, distance between centres:
  \[
  2R – R = R
  \]
• Hence, centre of smaller disc is at:
  \[
  (R,0)
  \]

4. Centre of Mass Formula:

\[
X_{cm} = \frac{\sum m_i x_i}{\sum m_i}
\]

\[
X_{cm} = \frac{(4m \cdot 0) + (m \cdot R)}{4m + m}
\]

\[
X_{cm} = \frac{mR}{5m} = \frac{R}{5}
\]

Final Answer:

The centre of mass lies at a distance
\[
\boxed{\frac{R}{5}}
\]
from the centre of the larger disc, towards the smaller disc.

5. A disc of radius \(R\) is cut out from a larger disc of radius \(2R\) in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.

Solution: 

We use the negative mass method, treating the removed portion (hole) as a disc of negative mass.

1. Choice of Coordinate System:

• Take the centre of the original disc as origin: \((0,0)\)
• Assume the system lies along the X-axis for simplicity

2. Masses and Positions:

Let \(\sigma\) be the mass per unit area.

• Original disc:
  • Radius = \(2R\)
  • Mass:
    \[
    m_1 = \sigma \pi (2R)^2 = 4\sigma \pi R^2
    \]
  • Position: \((0,0)\)
• Removed disc (hole):
  • Radius = \(R\)
  • Mass:
    \[
    m_2 = \sigma \pi R^2
    \]
  • Since edges touch externally, distance between centres:
    \[
    2R – R = R
    \]
  • Position: \((R,0)\)

3. Centre of Mass of Residual Disc:

\[
X_{cm} = \frac{m_1 x_1 – m_2 x_2}{m_1 – m_2}
\]

\[
X_{cm} = \frac{(4\sigma \pi R^2 \cdot 0) – (\sigma \pi R^2 \cdot R)}{4\sigma \pi R^2 – \sigma \pi R^2}
\]

\[
X_{cm} = \frac{-\sigma \pi R^3}{3\sigma \pi R^2} = -\frac{R}{3}
\]

4. Interpretation:

The negative sign indicates that the centre of mass lies on the side opposite to the hole.

Final Answer:

The centre of mass is located at a distance
\[
\boxed{\frac{R}{3}}
\]
from the centre of the original disc, away from the hole.

6. A square plate of edge \(d\) and a circular disc of diameter \(d\) are placed touching each other at the midpoint of an edge of the plate as shown in figure (9-Q2). Locate the centre of mass of the combination, assuming same mass per unit area for the two plates.

Solution: 

We treat both the square plate and the circular disc as point masses located at their geometric centres.

1. Masses of the Plates:

Let \(\sigma\) be the mass per unit area.

• Circular disc:
  • Radius \(r = \frac{d}{2}\)
  • Area:
    \[
    A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}
    \]
  • Mass:
    \[
    m_1 = \sigma A_1 = \frac{\sigma \pi d^2}{4}
    \]
• Square plate:
  • Area:
    \[
    A_2 = d^2
    \]
  • Mass:
    \[
    m_2 = \sigma d^2
    \]

2. Coordinate System:

• Take the centre of the circular disc as origin: \((0,0)\)
• The disc and square touch at the midpoint of the square’s side
• Distance from disc centre to contact point = \(\frac{d}{2}\)
• Distance from contact point to square centre = \(\frac{d}{2}\)
• Hence, centre of square is at:
  \[
  (d, 0)
  \]

3. Centre of Mass Formula:

\[
X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}
\]

\[
M = m_1 + m_2 = \frac{\sigma \pi d^2}{4} + \sigma d^2 = \sigma d^2 \left(1 + \frac{\pi}{4}\right)
\]

\[
X_{cm} = \frac{\left(\frac{\sigma \pi d^2}{4} \cdot 0\right) + (\sigma d^2 \cdot d)}{\sigma d^2 \left(1 + \frac{\pi}{4}\right)}
\]

\[
X_{cm} = \frac{\sigma d^3}{\sigma d^2 \left(\frac{4 + \pi}{4}\right)} = \frac{d}{\frac{4 + \pi}{4}}
\]

\[
X_{cm} = \frac{4d}{4 + \pi}
\]

Final Answer:

The centre of mass is located at a distance
\[
\boxed{\frac{4d}{4 + \pi}}
\]
to the right of the centre of the circular disc.

7. Calculate the velocity of the centre of mass of the system of particles shown in figure (9-E3).

Solution: 

To find the velocity of the centre of mass \(\vec{v}_{cm}\), we use:

\[
\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i}
\]

Use approximations: \(\cos 37^\circ \approx 0.8\), \(\sin 37^\circ \approx 0.6\).

1. Velocity Components of Each Particle

2. Centre of Mass Velocity

Total Mass:

\[
M = 1.0 + 1.2 + 1.5 + 0.5 + 1.0 = 5.2 \, \text{kg}
\]

X-component:

\[
v_{cm,x} = \frac{1.0(-1.2) + 1.2(0) + 1.5(-0.8) + 0.5(3.0) + 1.0(1.6)}{5.2}
\]

\[
v_{cm,x} = \frac{-1.2 + 0 – 1.2 + 1.5 + 1.6}{5.2}
= \frac{0.7}{5.2}
\approx 0.135 \, \text{m/s}
\]

Y-component:

\[
v_{cm,y} = \frac{1.0(-0.9) + 1.2(0.4) + 1.5(0.6) + 0.5(0) + 1.0(-1.2)}{5.2}
\]

\[
v_{cm,y} = \frac{-0.9 + 0.48 + 0.9 + 0 – 1.2}{5.2}
= \frac{-0.72}{5.2}
\approx -0.138 \, \text{m/s}
\]

Final Result

Velocity (vector form):

\[
\vec{v}_{cm} \approx (0.135\,\hat{i} – 0.138\,\hat{j}) \, \text{m/s}
\]

Magnitude:

\[
v_{cm} = \sqrt{(0.135)^2 + (-0.138)^2} \approx 0.193 \, \text{m/s}
\]

Direction:

Approximately \(45.6^\circ\) below the positive X-axis.

8. Two blocks of masses \(10\,\text{kg}\) and \(20\,\text{kg}\) are placed on the X-axis. The first mass is moved on the axis by a distance of \(2\,\text{cm}\). By what distance should the second mass be moved to keep the position of the centre of mass unchanged?

Solution: 

To keep the centre of mass unchanged, the net shift must be zero.

1. Given Data:

• \(m_1 = 10\,\text{kg}\)
• \(m_2 = 20\,\text{kg}\)
• \(\Delta x_1 = 2\,\text{cm}\)
• \(\Delta x_2 = ?\)

2. Condition for No Shift in Centre of Mass:

\[
m_1 \Delta x_1 + m_2 \Delta x_2 = 0
\]

3. Substitute Values:

\[
10 \times 2 + 20 \times \Delta x_2 = 0
\]

\[
20 + 20\Delta x_2 = 0
\]

\[
20\Delta x_2 = -20
\]

\[
\Delta x_2 = -1\,\text{cm}
\]

4. Interpretation:

The negative sign indicates motion in the opposite direction.

Final Answer:

The second block should be moved by
\[
\boxed{1\,\text{cm}}
\]
in the opposite direction.

9. Two blocks of masses \(10\,\text{kg}\) and \(30\,\text{kg}\) are placed along a vertical line. The first block is raised through a height of \(7\,\text{cm}\). By what distance should the second mass be moved to raise the centre of mass by \(1\,\text{cm}\)?

Solution: 

We use the relation between the displacement of individual masses and the displacement of the centre of mass.

1. Given Data:

• \(m_1 = 10\,\text{kg}\)
• \(m_2 = 30\,\text{kg}\)
• \(\Delta z_1 = +7\,\text{cm}\)
• \(\Delta Z_{cm} = +1\,\text{cm}\)

2. Formula:

\[
\Delta Z_{cm} = \frac{m_1 \Delta z_1 + m_2 \Delta z_2}{m_1 + m_2}
\]

3. Substitute Values:

\[
1 = \frac{10 \times 7 + 30 \times \Delta z_2}{40}
\]

4. Solve:

\[
40 = 70 + 30\Delta z_2
\]

\[
-30 = 30\Delta z_2
\]

\[
\Delta z_2 = -1\,\text{cm}
\]

5. Interpretation:

The negative sign shows that the second mass must move downward.

Final Answer:

The second block should be moved
\[
\boxed{1\,\text{cm downward}}
\]

10. Consider a gravity-free hall in which a tray of mass \(M\), carrying a cubical block of ice of mass \(m\) and edge \(L\), is at rest in the middle (figure 9-E4). If the ice melts, by what distance does the centre of mass of “the tray plus the ice” system descend?

Solution: 

1. External Forces Analysis:

• The system is in a gravity-free environment, so no gravitational force acts on it.
• Melting of ice is an internal process within the system.
• No external force acts on the system.

2. Principle of Centre of Mass Motion:

\[
\vec{a}_{CM} = \frac{\vec{F}_{\text{ext}}}{M_{\text{total}}}
\]

Since \(\vec{F}_{\text{ext}} = 0\):

\[
\vec{a}_{CM} = 0
\]

This means:

• The centre of mass has no acceleration
• Since it was initially at rest, its velocity remains zero
• Therefore, its position remains unchanged

3. Conclusion:

Even though the ice changes shape during melting, this is an internal rearrangement of mass. Without external forces, the centre of mass of the system does not move.

Final Answer:

The centre of mass descends by
\[
\boxed{0}
\]

11. Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii \(R_1\) and \(R_2\) (figure 9-E5).

Solution: We treat the given plate as a large semicircle of radius \(R_2\) with a smaller semicircle of radius \(R_1\) removed from it.

1. Masses and Individual Centres:

Let \(\sigma\) be the mass per unit area.

• Large semicircle:
  • Area:
    \[
    A_2 = \frac{1}{2}\pi R_2^2
    \]
  • Mass:
    \[
    m_2 = \sigma \frac{\pi R_2^2}{2}
    \]
  • Centre of mass:
    \[
    y_2 = \frac{4R_2}{3\pi}
    \]
• Removed semicircle:
  • Area:
    \[
    A_1 = \frac{1}{2}\pi R_1^2
    \]
  • Mass:
    \[
    m_1 = \sigma \frac{\pi R_1^2}{2}
    \]
  • Centre of mass:
    \[
    y_1 = \frac{4R_1}{3\pi}
    \]

2. Centre of Mass of Residual Plate:

\[
Y_{cm} = \frac{m_2 y_2 – m_1 y_1}{m_2 – m_1}
\]

\[
Y_{cm} = \frac{\left(\sigma \frac{\pi R_2^2}{2} \cdot \frac{4R_2}{3\pi}\right) – \left(\sigma \frac{\pi R_1^2}{2} \cdot \frac{4R_1}{3\pi}\right)}{\sigma \frac{\pi}{2}(R_2^2 – R_1^2)}
\]

3. Simplification:

\[
Y_{cm} = \frac{\frac{4}{3\pi}(R_2^3 – R_1^3)}{R_2^2 – R_1^2}
\]

Using identities:
\[
R_2^3 – R_1^3 = (R_2 – R_1)(R_2^2 + R_1R_2 + R_1^2)
\]
\[
R_2^2 – R_1^2 = (R_2 – R_1)(R_2 + R_1)
\]

\[
Y_{cm} = \frac{4}{3\pi} \cdot \frac{R_2^2 + R_1R_2 + R_1^2}{R_2 + R_1}
\]

Final Answer:

The centre of mass is located at
\[
\boxed{\frac{4(R_1^2 + R_1R_2 + R_2^2)}{3\pi (R_1 + R_2)}}
\]
above the centre along the axis of symmetry.

12. Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?

Solution: 

1. Physical Principle:

• No external horizontal force acts on the system (friction neglected).
• Therefore, the centre of mass remains at the same position.

2. Initial Positions (take left end as \(x=0\)):

• Mr. Verma (50 kg): \(x = 0\)
• Mr. Mathur (60 kg): \(x = 4\,\text{m}\)
• Boat (40 kg): centre at \(x = 2\,\text{m}\)

3. Let the boat move by \(x\) to the right.

• Final position of both men (middle of boat): \(2 + x\)
• Final position of boat’s centre: \(2 + x\)

4. Displacements:

• \(\Delta x_V = 2 + x\)
• \(\Delta x_M = (2 + x) – 4 = x – 2\)
• \(\Delta x_b = x\)

5. Apply Centre of Mass Condition:

\[
50(2 + x) + 60(x – 2) + 40x = 0
\]

\[
100 + 50x + 60x – 120 + 40x = 0
\]

\[
150x – 20 = 0
\]

\[
x = \frac{20}{150} = \frac{2}{15}\,\text{m}
\]

6. Conversion:

\[
x = \frac{2}{15} \times 100 \approx 13.3\,\text{cm}
\]

Final Answer:

The boat moves by
\[
\boxed{\frac{2}{15}\,\text{m} \approx 13.3\,\text{cm}}
\]

13. A cart of mass \(M\) is at rest on a frictionless horizontal surface and a pendulum bob of mass \(m\) hangs from the roof of the cart (figure 9-E6). The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is \(L\). Find the displacement of the cart during this process.

Solution: 

1. External Forces:

• The surface is frictionless, so no external horizontal force acts.
• Hence, the centre of mass has no horizontal acceleration.

2. Principle Used:

Since the system is initially at rest and no external horizontal force acts,
the horizontal position of the centre of mass remains unchanged.

3. Displacements:

• Let displacement of the cart be \(x\).
• The bob moves a distance \(L\) relative to the cart.
• So, displacement of the bob relative to ground = \(x + L\).

4. Apply Centre of Mass Condition:

\[
Mx + m(x + L) = 0
\]

\[
Mx + mx + mL = 0
\]

\[
(M + m)x = -mL
\]

\[
x = -\frac{mL}{M + m}
\]

5. Interpretation:

The negative sign indicates that the cart moves in the opposite direction to the motion of the bob.

Final Answer:

The displacement of the cart is
\[
\boxed{\frac{mL}{M + m}}
\]
in the direction opposite to the bob’s motion.

14. The balloon, the light rope and the monkey shown in figure (9-E7) are at rest in the air. If the monkey reaches the top of the rope, by what distance does the balloon descend? Mass of the balloon = \(M\), mass of the monkey = \(m\) and the length of the rope ascended by the monkey = \(L\).

Solution: 

1. Physical Principle:

• The system (balloon + monkey) is initially at rest.
• The buoyant force balances the total weight \((M + m)g\), so the net external force is zero.
• Hence, the centre of mass remains at the same position.

2. Displacements:

• Let the balloon descend by \(y\) (downward).
• The monkey climbs a distance \(L\) relative to the balloon.
• So, displacement of monkey relative to ground = \(L – y\) (upward).

3. Apply Centre of Mass Condition:

(Taking upward as positive)

\[
M(-y) + m(L – y) = 0
\]

\[
-My + mL – my = 0
\]

\[
mL = (M + m)y
\]

\[
y = \frac{mL}{M + m}
\]

Final Answer:

The balloon descends by
\[
\boxed{\frac{mL}{M + m}}
\]

15. Find the ratio of the linear momenta of two particles of masses \(1.0\,\text{kg}\) and \(4.0\,\text{kg}\) if their kinetic energies are equal.

Solution: 

1. Relation Between Momentum and Kinetic Energy:

\[
p = \sqrt{2mK}
\]

2. Given:

• \(m_1 = 1.0\,\text{kg}\)
• \(m_2 = 4.0\,\text{kg}\)
• \(K_1 = K_2 = K\)

3. Ratio of Momenta:

\[
\frac{p_1}{p_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}}
\]

\[
\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}
\]

4. Substitution:

\[
\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
\]

Final Answer:

\[
\boxed{p_1 : p_2 = 1 : 2}
\]