JEE Physics Centre of Mass MCQs Set 09

Answer: (c) \( \left(\frac a3, \frac a3\right) \)

Explanation:
1. Each body acts as a point mass M at its own geometric centre.
2. \( x_{cm} = \frac{M(0)+M(0)+M(a)}{3M} = \frac a3 \)
3. \( y_{cm} = \frac{M(a)+M(0)+M(0)}{3M} = \frac a3 \)

Teacher's Note:
1. A symmetric body can be replaced by a point mass at its centre when finding the centre of mass of a system.
2. Equal masses reduce the weighted average to a plain average of the coordinates.

Answer: (b) \( \frac52 \)

Explanation:
1. The first three particles have their centre of mass at the origin, so they contribute zero to the weighted sum.
2. \( \vec r_{cm} = \frac{4\,\vec r_4}{1+2+3+4} = \frac{4\alpha}{10}(\hat i+2\hat j+3\hat k) \)
3. Setting this equal to \( \hat i+2\hat j+3\hat k \): \( \frac{4\alpha}{10}=1 \Rightarrow \alpha = \frac52 \)

Teacher's Note:
1. A group of particles can be treated as one particle of total mass at its own centre of mass.
2. Here the group (6 g) sits at the origin, so only the fourth particle shifts the result.

Answer: (c) \( \left(\frac{13}{3}, \frac{10}{3}\right) \)

Explanation:
1. \( x_{cm} = \frac{5(5)+10(4)}{15} = \frac{65}{15} = \frac{13}{3} \)
2. \( y_{cm} = \frac{5(4)+10(3)}{15} = \frac{50}{15} = \frac{10}{3} \)

Teacher's Note:
1. Weight each coordinate by the mass and divide by the total mass.
2. The centre of mass lies on the rod, closer to the heavier mass.

Answer: (a) \( 2a\hat i + \frac{4a}{3}\hat j \)

Explanation:
1. Masses are proportional to the densities: \( \sigma, 2\sigma, 3\sigma \) (equal areas). Centres: top plate \( (a, 3a) \), plate of density 2σ \( (a, a) \), plate of density 3σ \( (3a, a) \).
2. \( x_{cm} = \frac{1(a)+2(a)+3(3a)}{6} = 2a \)
3. \( y_{cm} = \frac{1(3a)+2(a)+3(a)}{6} = \frac{4a}{3} \)

Teacher's Note:
1. Each uniform square plate acts as a point mass at its centre.
2. Because the areas are equal, the weights are simply the density ratios 1 : 2 : 3.

Answer: (d) 5 cm

Explanation:
1. For a solid cone the centre of mass is at \( h/4 \) from the base along the axis.
2. \( \frac{20}{4} = 5 \) cm

Teacher's Note:
1. Solid cone: h/4 from the base. Hollow cone (curved surface only): h/3 from the base.
2. The centre of mass of a solid cone always lies inside it.

Answer: (a) \( \frac{3L}{4} \)

Explanation:
1. For constant cross-section, \( dm \propto \rho\,dx \propto x^2dx \).
2. \( x_{cm} = \frac{\int_0^L x^3dx}{\int_0^L x^2dx} = \frac{L^4/4}{L^3/3} = \frac{3L}{4} \)

Teacher's Note:
1. With uniform cross-section the mass element is proportional to density times dx.
2. Integrals of powers of x are the whole job here.

Answer: (d) Zero

Explanation:
1. Only internal forces act, so the total momentum stays at its initial value, zero.
2. \( M v_{cm} = 0 \Rightarrow v_{cm} = 0 \)

Teacher's Note:
1. Internal forces cannot change the velocity of the centre of mass.
2. The particles have unequal masses, with the faster one lighter; momenta are equal and opposite.

Answer: (b) \( -\frac52\ \mathrm{m/s^2} \)

Explanation:
1. Acceleration of each mass: \( a = \frac{m_1-m_2}{m_1+m_2}g = \frac{10}{20}(10) = 5\ \mathrm{m/s^2} \) (g = 10 m/s²).
2. 15 kg goes down, 5 kg goes up: \( a_{cm} = \frac{15(-5)+5(+5)}{20} = -\frac52\ \mathrm{m/s^2} \)
3. Check: \( a_{cm} = \left(\frac{m_1-m_2}{m_1+m_2}\right)^2 g = \frac14(10) = 2.5 \) downward.

Teacher's Note:
1. The centre of mass accelerates downward because the heavier mass moves faster in that direction.
2. The general result is a_cm = ((m1 - m2)/(m1 + m2))^2 g.

Answer: (d) \( \frac g4 \)

Explanation:
1. Each block has acceleration \( a = \frac{3m-m}{3m+m}g = \frac g2 \).
2. \( a_{cm} = \frac{3m(-\frac g2)+m(\frac g2)}{4m} = -\frac g4 \)
3. Magnitude \( \frac g4 \), directed downwards.

Teacher's Note:
1. The pulley is held by the support, so the system is not isolated and the centre of mass accelerates.
2. Net external force = 4mg - T_support; the formula ((m1 - m2)/(m1 + m2))^2 g gives the same result.

Answer: (a) 7.5R

Explanation:
1. They touch when the centre separation is \( R + 2R = 3R \), so the centres approach by \( 12R - 3R = 9R \).
2. Only internal forces act and the system starts from rest, so the centre of mass stays fixed. Displacements are in inverse ratio of masses: \( s_1 : s_2 = 5 : 1 \).
3. \( s_1 = \frac56 \times 9R = 7.5R \)

Teacher's Note:
1. Use the fixed centre of mass instead of solving the variable-force motion.
2. Distance covered by each body is inversely proportional to its mass.

Answer: (a) \( \frac{-a^3b}{R^3-a^3} \)

Explanation:
1. Treat the cavity as a negative mass: masses proportional to volumes, \( m_{full}\propto R^3 \), \( m_{hole} \propto -a^3 \).
2. \( x_{cm} = \frac{R^3(0) - a^3 (b)}{R^3 - a^3} = \frac{-a^3 b}{R^3-a^3} \)

Teacher's Note:
1. Removing a part is the same as adding a negative mass at the centre of the removed part.
2. The remaining body's centre of mass shifts away from the cavity, hence the negative sign.

Answer: (d) \( -\frac a6 \)

Explanation:
1. Mass is proportional to area: full plate \( 4a^2 \), removed part \( a^2 \) with centre at \( y = +\frac a2 \).
2. \( y_{cm} = \frac{4a^2(0) - a^2\left(\frac a2\right)}{4a^2 - a^2} = -\frac{a}{6} \)

Teacher's Note:
1. Removed region above the centre pulls the centre of mass downward.
2. Always subtract area-weighted moments of the removed part.

Answer: (c) \( \frac R6 \)

Explanation:
1. Hole mass is \( \frac{m}{4} \) (area ratio \( \left(\frac12\right)^2 \)), its centre is at distance \( \frac R2 \) from O.
2. \( d = \frac{\frac m4 \cdot \frac R2}{m - \frac m4} = \frac{R/8}{3/4} = \frac R6 \)

Teacher's Note:
1. The centre of mass of the remaining disc lies on the line from the hole to O, on the opposite side from the hole.
2. Area scales as the square of the radius.

Answer: (b) \( 15\sqrt2 \) m/s

Explanation:
1. Masses are 0.25 kg, 0.25 kg and 0.5 kg.
2. Momentum of the two lighter parts: \( 0.25\times30 = 7.5 \) kg m/s each, at right angles, so the resultant is \( 7.5\sqrt2 \).
3. By conservation of momentum, the heavier part has the same magnitude: \( 0.5\,v = 7.5\sqrt2 \Rightarrow v = 15\sqrt2 \) m/s

Teacher's Note:
1. The total momentum of the fragments must be zero because the bomb was at rest.
2. The heavy fragment moves opposite to the resultant of the other two.

Answer: (d) \( \frac{Lm}{M+m} \)

Explanation:
1. No horizontal external force acts, so the centre of mass stays fixed.
2. Let the plank move x opposite to the person. The person moves \( L - x \) relative to the ground: \( m(L-x) = Mx \)
3. \( x = \frac{mL}{M+m} \)

Teacher's Note:
1. Relative displacement is L, shared in inverse ratio of the masses.
2. Remember that the person's ground displacement is less than L.

Answer: (d) \( \frac53 \) m, to the right

Explanation:
1. Take the left end as origin in the plank's frame. Before: \( x_{cm}' = \frac{50(0)+100(10)+150(5)}{300} = \frac{1750}{300} \).
2. After: \( x_{cm}' = \frac{50(10)+100(0)+150(5)}{300} = \frac{1250}{300} \). Centre of mass moves \( \frac{500}{300} = \frac53 \) m to the left relative to the plank.
3. The centre of mass is fixed on the ground, so the plank moves \( \frac53 \) m to the right.

Teacher's Note:
1. Find how the centre of mass shifts relative to the plank, then move the plank the opposite way.
2. Net external horizontal force on the system is zero.