JEE Physics Centre of Mass MCQs Set 01

Answer: (b) can lie inside or outside the body

Explanation:
1. A ring has its centre of mass at the empty centre, outside the material
2. A uniform solid sphere has it inside
3. Hence it may lie either inside or outside

Teacher's Note:
1. Think of a ring or a boomerang as the counter-example to 'always inside'.
2. The centre of mass depends on the mass distribution, not on the shape alone.

Answer: (a) \( \frac12(3\hat i+\hat j-\hat k) \)

Explanation:
1. \( x=\frac{1+2+6}6=\frac32 \)
2. \( y=\frac{4+2-3}6=\frac12 \)
3. \( z=\frac{1+2-6}6=-\frac12 \), so \( \vec r_{cm}=\frac12(3\hat i+\hat j-\hat k) \)

Teacher's Note:
1. Treat each component separately.
2. The total mass in the denominator is 6 kg.

Answer: (c) (0.5 m, 0.3 m)

Explanation:
1. \( x=\frac{1(0)+2(1)+3(1)+4(0)}{10}=0.5 \)
2. \( y=\frac{1(1)+2(1)+3(0)+4(0)}{10}=0.3 \)

Teacher's Note:
1. Read the coordinates of each mass carefully before substituting.
2. Total mass is 10 kg.

Answer: (c) \( \frac{m_1}{m_2}d \)

Explanation:
1. The centre of mass does not move: \( m_1d=m_2d' \)
2. \( d'=\frac{m_1}{m_2}d \)

Teacher's Note:
1. With no external force the product mass x displacement is balanced.
2. The lighter particle moves farther.

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

Explanation:
1. \( x_{cm}=\frac{\int_0^L x^3dx}{\int_0^L x^2dx} \)
2. \( =\frac{L^4/4}{L^3/3} \)
3. \( =\frac{3L}4 \)

Teacher's Note:
1. Increase the power of x by one in numerator and denominator when integrating.

Answer: (b) \( a/6 \) from the centre, on the side away from the hole

Explanation:
1. Remaining area \( \pi a^2-\frac{\pi a^2}4=\frac{3\pi a^2}4 \)
2. Hole centre at a/2 from the origin
3. \( x=\frac{0-\frac{\pi a^2}4\cdot\frac a2}{\frac{3\pi a^2}4}=-\frac a6 \)

Teacher's Note:
1. Treat the hole as a negative mass.
2. The centre moves away from the removed part.

Answer: (d) \( \frac R{20} \) from A, on the side opposite to the hole

Explanation:
1. \( A_1=\pi R^2,\ A_2=\frac{\pi R^2}{16} \)
2. \( x=\frac{0-\frac{\pi R^2}{16}\cdot\frac{3R}4}{\pi R^2-\frac{\pi R^2}{16}} \)
3. \( =-\frac{3R/64}{15/16}=-\frac R{20} \)

Teacher's Note:
1. Subtract the moment of the hole.
2. The negative sign shows the shift is away from the hole.

Answer: (c) \( \frac{2R}{\pi} \)

Explanation:
1. Using \( dm=\lambda R\,d\theta \) and \( y=R\sin\theta \)
2. \( y_{cm}=\frac{\int_0^\pi R\sin\theta\,R\,d\theta}{\pi R}=\frac{2R}\pi \)

Teacher's Note:
1. A wire gives 2R/pi, a flat semicircular disc gives 4R/3pi.

Answer: (a) \( \frac{4R}{3\pi} \)

Explanation:
1. Take thin strips of arc length and integrate over the area
2. The result is \( \frac{4R}{3\pi}\approx0.42R \)

Teacher's Note:
1. Remember the pair: wire 2R/pi, disc 4R/3pi.

Answer: (a) \( \frac{3R}8 \)

Explanation:
1. Integrate discs of radius \( \sqrt{R^2-y^2} \)
2. \( y_{cm}=\frac{3R}8 \)

Teacher's Note:
1. Solid hemisphere: 3R/8; hollow hemispherical shell: R/2.

Answer: (a) 10 cm

Explanation:
1. For a solid cone \( y_{cm}=\frac h4 \) from the base
2. \( =\frac{40}4=10 \) cm

Teacher's Note:
1. Solid cone: h/4 from the base; hollow cone: h/3.

Answer: (d) \( \frac h3 \)

Explanation:
1. The centroid is the intersection of the medians
2. It lies one third of the height above the base

Teacher's Note:
1. The centroid divides each median in the ratio 2 : 1 from the vertex.

Answer: (b) \( \frac{a}{\sqrt3} \)

Explanation:
1. Equal masses give the centroid
2. Altitude \( =\frac{\sqrt3}{2}a \); the centroid is \( \frac23 \) of it from the vertex
3. \( \frac23\cdot\frac{\sqrt3}2a=\frac a{\sqrt3} \)

Teacher's Note:
1. The centroid is two thirds of the way along the median from the vertex.

Answer: (a) may be positive for some particles and negative for others

Explanation:
1. \( \sum m_ix_i=0 \)
2. If all were positive the sum could not vanish
3. So there must be both signs (or all zero)

Teacher's Note:
1. Weighted sum zero requires positive and negative contributions.

Answer: (c) 1.33 cm along the 4 cm side and 1.5 cm along the 3 cm side from the right-angle corner

Explanation:
1. Take the right-angle corner as origin: A(0,0), B(4,0), C(0,3)
2. \( x=\frac{0.5(0)+1.0(4)+1.5(0)}{3}=\frac43=1.33 \) cm
3. \( y=\frac{0+0+1.5(3)}3=1.5 \) cm

Teacher's Note:
1. Choose axes along the perpendicular sides to simplify.
2. Total mass is 3 kg.

Answer: (c) \( \le R \)

Explanation:
1. \(
2. \vec r_{cm}
3. =\left
4. \frac{\sum m_i\vec r_i}{\sum m_i}\right
5. \le\frac{\sum m_i
6. \vec r_i
7. }{\sum m_i}=R \)
8. Equality only if all particles lie at one point

Teacher's Note:
1. The centre of mass is a weighted average of points on a sphere, so it lies inside the sphere or on it.

Answer: (d) continue on the parabolic path the unexploded bomb would have followed

Explanation:
1. The explosion forces are internal
2. Gravity is the only external force
3. So the centre of mass keeps its original parabolic trajectory

Teacher's Note:
1. Only external forces change the motion of the centre of mass.

Answer: (b) the ball and the Earth both move away from each other

Explanation:
1. Without external forces the centre of mass of the ball-Earth system stays at rest
2. The ball goes up, so the Earth must go down by a tiny amount
3. They move apart (the ball and Earth separate)

Teacher's Note:
1. The Earth's displacement is negligible because of its huge mass.

Answer: (b) the kinetic energy of the system but not its linear momentum

Explanation:
1. Internal forces occur in equal and opposite pairs, so net momentum stays the same
2. They can do net work (e.g. an explosion), changing kinetic energy

Teacher's Note:
1. Think of an explosion: momentum same, kinetic energy increases.

Answer: (c) equals g

Explanation:
1. \( a_{cm}=\frac{m_1g+m_2g}{m_1+m_2}=g \)

Teacher's Note:
1. Only gravity acts externally; the result is independent of the masses.

Answer: (a) \( \frac{10}3 \) m/s

Explanation:
1. \( v_{cm}=\frac{1(2)+0.5(6)}{1.5} \)
2. \( =\frac5{1.5}=\frac{10}3 \) m/s

Teacher's Note:
1. Mass-weighted average of velocities.

Answer: (a) zero

Explanation:
1. Take east as positive
2. \( v_{cm}=\frac{2(3)+3(-2)}{5}=0 \)

Teacher's Note:
1. Total momentum is zero.

Answer: (b) 2 m

Explanation:
1. \( a_{cm}=\frac{30}{10+20}=1 \) m/s\(^2\)
2. \( s=\frac12(1)(2)^2=2 \) m

Teacher's Note:
1. The internal string tension does not affect the motion of the centre of mass.