WBCHSE Class 11 Physics Statics Short Answer Questions

Statics Short Answer Type Questions

Question 1. Two particles of mass 2 g and 3 g are situated at the locations (3 cm, 5 cm) and (4 cm, 6 cm) respectively. Find the position vector of the centre of mass of the two particles.
Answer:

x = \(\frac{2 \times 3+3 \times 4}{2+3}=3.6 \mathrm{~cm}\)

and y = \(\frac{2 \times 5+3 \times 6}{2+3}=5.6 \mathrm{~cm}\)

Hence, position vector, \(\vec{r}=3.6 \hat{i}+5.6 \hat{j} \mathrm{~cm}\)

Question 2. Explain why the concept of the centre of mass is much more fundamental than the centre of gravity.
Answer:

As the mass of a body remains constant anywhere in the universe, a body always has a definite centre of mass. But at places where there is no gravity (for example, in outer space, inside an artificial satellite rotating around Earth or in case of a free-falling object), the weight of the body becomes zero i.e., there is no centre of gravity.

Hence the concept of the centre of mass is much more fundamental than the centre of gravity.

Question 3.

1. Find the velocity of the centre of mass of two identical particles moving with velocities v₁ and v₂.
2. Show that in the absence of any external force the centre of mass of two moving particles moves with uniform velocity.

Answer:

1. If the mass of each particle is m, the resultant momentum = \(\left(m \vec{v}_1+m \vec{v}_2\right)=m\left(\vec{v}_1+\vec{v}_2\right)\)

If the velocity of the centre of mass of the two particles is \(\vec{v}\), then the momentum of the centre of mass =(m+ m)\(\vec{v}\) = 2m\(\vec{v}\).

∴ 2m\(\vec{v}\) = m(\(\vec{v_1}\)+\(\vec{v_2}\))

Hence, \(\vec{v}=\frac{\vec{v}_1+\vec{v}_2}{2}\)

2. If no external force is applied to the particles, the resultant momentum of the particles remains conserved. That means, on the above example, \(m\left(\vec{v}_1+\vec{v}_2\right)\) = constant. Clearly, both the velocities \(\vec{v}_1\) and \(\vec{v}_2\) can be changed in such a way that the above condition is not violated.

On the other hand, for the centre of mass of the two particles, 2 m\(\vec{v}\) = m(\(\vec{v}_1\)+ \(\vec{v}_2\)) = constant.

As mass m is constant, so \(\vec{v}\) = constant

Hence, in the absence of any external force the centre of mass of two moving particles moves with uniform velocity.

Question 4. Three forces F₁, F₂, F₃ —of which F₂ and F₃ are mutually perpendicular act on a particle of mass m so that the particle is stationary. Find the acceleration of the particle when F₁ is withdrawn.
Answer:

As the particle is stationary, the resultant force \(\vec{F}_1+\vec{F}_2+\vec{F}_3=\overrightarrow{0}\)

Hence, \(\vec{F}_2+\vec{F}_3=-\vec{F}_1\)

∴ \(\vec{F}_2+\vec{F}_3\) is equal and opposite to \(\vec{F}_1\).

∴ \(\vec{F}_2\) and \(\vec{F}_3\) are mutually perpendicular; so the magnitude of \(\left(\vec{F}_2+\vec{F}_3\right)\) is \(\sqrt{F_2^2+F_3^2}\).

When \(\vec{F}_1\) is withdrawn, the resultant force on the particle \(=\vec{F}_2+\vec{F}_3\); in that case the acceleration of the particle will be in the opposite direction of \(\vec{F}_1\).

∴ Magnitude of the acceleration \(=\frac{\sqrt{F_2^2+F_3^2}}{m}\)

Question 5. Write down its mathematical form. Itoo particles of masses 2g and 3g are situated at the positions (2 cm, 3 cm) and (4 cm, 5 cm) respectively. Find the position vector of the centre of mass of the two particles.
Answer:

x \(=\frac{2 \times 2+3 \times 4}{2+3}=\frac{4+12}{5}=\frac{16}{5}=3.2 \mathrm{~cm}\)

y = \(\frac{2 \times 3+3 \times 5}{2+3}=\frac{6+15}{5}=4.2 \mathrm{~cm}\)

Position vector of the centre of mass, \(\vec{r}=x \hat{i}+y \hat{j}=(3.2 \hat{i}+4.2 \hat{j}) \mathrm{cm}\)

Question 6. A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the centre of mass of the system from the origin is

1. =R
2. ≤R
3. >R
4. ≥F

Answer:

• If the particles are uniformly placed at a distance R from the origin, then the centre of mass will be situated at the origin.
• Again, when all the particles are placed at a distance R from the origin at one side of it, then the centre of mass will be situated at a distance of R from the origin.
• For any other kind of mass distribution, the centre of mass will be situated between the origin and R.

The option 2 is correct.

Question 7. Two bodies of masses m₁ and m₂ are separated by a distance R. The distance of the centre of mass of the bodies from the mass m₁ is

1. \(\frac{m_2 R}{m_1+m_2}\)
2. \(\frac{m_1 R}{m_1+m_2}\)
3. \(\frac{m_1 m_2}{m_1+m_2} R\)
4. \(\frac{m_1+m_2}{m_1} R\)

Answer:

The coordinates of masses m₁ and m₂ are (0,0) and (F, 0) respectively.

∴ The position of the centre of mass, \(x_{\mathrm{cm}}=\frac{m_1 \times 0+m_2 \times R}{m_1+m_2}=\frac{m_2 R}{m_1+m_2}\)

The option 1 is correct.

Question 8. Two particles A and B (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2 v, the speed of the centre of mass is

1. Zero
2. v
3. \(\frac{3 v}{2}\)
4. –\(\frac{3 v}{2}\)

Answer:

As no external force is applied on the two particles, the speed of the centre of mass of these two particles does not change. Therefore, the centre of mass is at rest.

The option 1 is correct.

Question 9. The distance of the centre of mass of a solid uniform cone from its vertex is z₀. If the radius of its base is R and its height is h, then z₀ is equal to

1. \(\frac{h^2}{4 R}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{5 h}{8}\)
4. \(\frac{3 h^2}{8 R}\)

Answer:

The centre of mass of a uniform cone lies at a height of \(\frac{h}{4}\) from the base on the line joining the vertex and the centre of the base.

Hence, it is situated at a depth of \(\frac{3 h}{4}\) from the vertex.

The option 2 is correct.

Question 10. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

1. \(\frac{W x}{d}\)
2. \(\frac{W d}{x}\)
3. \(\frac{W(d-x)}{x}\)
4. \(\frac{W(d-x)}{d}\)

Answer:

Let the normal reactions on A and B be N_A and N_B.

At equilibrium, N_A+N_B=W

Again, when the resultant torques with respect to the centre of mass is zero, then \(N_A x+W .0-N_B(d-x)=0 \quad \text { or, } N_A \frac{x}{d-x}-N_B=0\)

So, \(N_A+N_A \frac{x}{d-x}=W\)

or, \(N_A\left(1+\frac{x}{d-x}\right)=W\)

or, \(N_A \frac{d}{d-x}=W\)

∴ \(N_A=\frac{W(d-x)}{d}\)

The option 4 is correct.

Question 11. Two spherical bodies of mass M and 5 M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is

1. 2.5R
2. 4.5R
3. 7.5R
4. 1.5R

Answer:

Their centre of mass will remain stationary due to the absence of any external force.

Hence, if the displacements of the smaller and the larger body are x₁ and x₂ respectively, \(M x_1=5 M x_2 \quad \text { or, } x_2=\frac{x_1}{5}\)

Since, at the time of the collision, the distance between their centres =R + 2R = 3R, so, the net path traversed by the spheres, x₁ + x₂ = 12R- 3R = 9 R

∴ \(x_1+\frac{x_1}{5}=9 R \quad \text { or, } \frac{6}{5} x_1=9 R \quad \text { or, } x_1=7.5 R\)

The option 3 is correct.

Question 12. Four bodies have been arranged at the comers of a rectangle shown. Find the centre of mass of the system.

Answer:

The masses at the corners O, A, B and C of the rectangle are respectively, m₁ = m, m₂ = 2m, m₃ = 3m and m₄ = 2m. the x-axis is chosen along OC and the y-axis along OA.

So the coordinates of the masses are respectively (0,0); (0, b), (a,h) and (a,0).

If (x, y) are the coordinated of the centre of mass, then

x = \(\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4}\)

= \(\frac{m \cdot 0+2 m \cdot 0+3 m \cdot a+2 m \cdot a}{m+2 m+3 m+2 m}\)

= \(\frac{5 m a}{8 m}=\frac{5}{8} a\)

Similarly, y = \(\frac{m \cdot 0+2 m \cdot b+3 m \cdot b+2 m \cdot 0}{m+2 m+3 m+2 m}\)

= \(\frac{5 m b}{8 m}=\frac{5}{8} b\)

So the centre of mass is at \(\left(\frac{5}{8} a, \frac{5}{8} b\right)\).

Question 13. Why are we not able to rotate a wheel by pulling or pushing along its radius?
Answer:

The centre of mass is at the centre of a wheel and every radial direction passes through this centre. As per definition, any force acting through the centre of mass can produce translation, but no rotation.

Question 14. Why do we use a wrench of a long arm to unscrew a nut tightly fitted to a bolt?
Answer:

Unscrewing a nut means a rotation about the axis of the nut. The application of a higher moment of force will facilitate this rotation. If the force applied remains the same, the moment of force is higher when the arm is longer. So we use a wrench of a long arm to unscrew a nut tightly fitted to a bolt.