JEE Physics Centre of Mass MCQs Set 05

Answer: (c) \( R/5 \)

Explanation:
1. Masses are in the ratio of areas: large \( 4\pi R^2 \), small \( \pi R^2 \), i.e. 4 : 1.
2. The centre of the small disc is at distance R from the centre of the large disc.
3. \( x_{cm}=\frac{4\cdot0+1\cdot R}{5}=\frac{R}{5} \) towards the small disc.

Teacher's Note:
1. When the two discs overlap, the stacked system is simply a heavy disc plus a light disc at an offset.
2. The thickness and density cancel because both discs are identical in material.

Answer: (d) \( R/3 \)

Explanation:
1. Mass is proportional to area: whole disc \( 4\pi R^2 \), hole \( \pi R^2 \), remaining \( 3\pi R^2 \).
2. The hole centre is at distance R from the original centre. Treat the hole as negative mass and take the original centre as origin: \( x=\frac{4\pi R^2(0)-\pi R^2(R)}{3\pi R^2} \).
3. \( =-\frac R3 \), so the centre of mass is R/3 from the original centre, on the side away from the hole.

Teacher's Note:
1. A cavity is handled by subtracting its mass at its own centre.
2. The centre of mass of the remaining plate shifts away from the missing part.

Answer: (a) \( \frac{3AL+2BL^2}{3(2A+BL)} \)

Explanation:
1. Total mass: \( m=\int_0^L(A+Bx)dx=AL+\tfrac{BL^2}{2} \).
2. First moment: \( \int_0^L x(A+Bx)dx=\tfrac{AL^2}{2}+\tfrac{BL^3}{3} \).
3. \( x_{cm}=\frac{AL^2/2+BL^3/3}{AL+BL^2/2}=\frac{3AL+2BL^2}{3(2A+BL)} \).

Teacher's Note:
1. For a continuous body, replace the sum by an integral of x dm with dm = rho dx.
2. If B = 0 the formula reduces to L/2, as it should for a uniform rod.

Answer: (b) \( \frac{4(R_2^3-R_1^3)}{3\pi(R_2^2-R_1^2)} \)

Explanation:
1. A solid half-disc of radius R has its centre of mass at \( 4R/3\pi \) and mass proportional to \( R^2 \).
2. The annulus is the large half-disc minus the small one: \( y=\frac{R_2^2\cdot\frac{4R_2}{3\pi}-R_1^2\cdot\frac{4R_1}{3\pi}}{R_2^2-R_1^2} \).
3. \( y=\frac{4(R_2^3-R_1^3)}{3\pi(R_2^2-R_1^2)} \).

Teacher's Note:
1. Subtraction of a smaller half-disc from a larger one avoids any integration.
2. Letting R1 tend to zero recovers the solid half-disc value 4R/(3 pi).

Answer: (c) 1 cm downward

Explanation:
1. The centre of mass rises by \( \Delta y_{cm}=\frac{10(7)+30\,d}{40}=1 \).
2. \( 70+30d=40\Rightarrow d=-1\ \text{cm} \).
3. The negative sign means the 30 kg block is moved 1 cm downward.

Teacher's Note:
1. Write the displacement of the centre of mass as a mass-weighted average of the displacements.
2. Moving the heavy block down partly cancels the rise caused by the light block.

Answer: (a) 0.4 m

Explanation:
1. Relative to the platform, A moves 3 m forward and B moves 3 m backward. Let the platform move by X.
2. No external horizontal force, so \( 40(3+X)+60(-3+X)+50X=0 \).
3. \( 150X=60\Rightarrow X=0.4 \) m, in the direction in which A (the lighter man) walks relative to the platform.

Teacher's Note:
1. The centre of mass of the whole system cannot move on a smooth floor.
2. Use displacements relative to the platform plus the unknown platform shift.

Answer: (c) 10 m/s and an increase of 1/4

Explanation:
1. Momentum: \( 2m(20)=m(30)+m v\Rightarrow v=10 \) m/s.
2. Initial KE \( =\tfrac12(2m)(400)=400m \); final KE \( =\tfrac12m(900+100)=500m \).
3. Fractional change \( =\frac{500-400}{400}=+\frac14 \) (an increase, supplied by the explosion).

Teacher's Note:
1. Internal explosive forces conserve momentum but add kinetic energy.
2. The kinetic energy can only increase in an explosion.

Answer: (d) \( 3.0\times10^{-23} \) m/s

Explanation:
1. Momentum conservation: \( 5\times36=6\times10^{24}\,V \).
2. \( V=\frac{180}{6\times10^{24}}=3.0\times10^{-23} \) m/s.

Teacher's Note:
1. The enormous mass of the Earth makes its recoil utterly negligible.
2. The same law applies when a gun fires a bullet or a rocket expels gas.

Answer: (a) \( 2.5\hat i+15\hat j+5\hat k \)

Explanation:
1. Initial momentum (g cm/s): \( 400\hat i+600\hat j+800\hat k \).
2. Final: first = 0, second \( =30(10\hat i+20\hat k)=300\hat i+600\hat k \).
3. Third: \( 40\vec v_3=(400-300)\hat i+600\hat j+(800-600)\hat k=100\hat i+600\hat j+200\hat k \), so \( \vec v_3=2.5\hat i+15\hat j+5\hat k \).

Teacher's Note:
1. No external force acts, so the vector sum of momenta is unchanged.
2. Work component by component to avoid mistakes.

Answer: (c) 10 cm

Explanation:
1. Horizontal momentum: \( 1(10\cos45^\circ)=100\,V\Rightarrow V=0.0707 \) m/s.
2. Time of flight of the stone: \( T=\frac{2(10)\sin45^\circ}{10}=\sqrt2 \) s.
3. Displacement \( =0.0707\times1.414=0.10 \) m = 10 cm, opposite to the stone's horizontal motion.

Teacher's Note:
1. Only the horizontal component of momentum is conserved because the ground exerts a vertical force.
2. Platform and boy move together, so their combined mass 100 kg is used.

Answer: (d) \( \frac{m_2v_0}{m_1+m_2},\ \ v_0\sqrt{\frac{m_1m_2}{(m_1+m_2)k}} \)

Explanation:
1. Total momentum \( m_2v_0 \) gives \( v_{cm}=\frac{m_2v_0}{m_1+m_2} \).
2. At maximum elongation both blocks move with \( v_{cm} \). Energy: \( \tfrac12m_2v_0^2=\tfrac12(m_1+m_2)v_{cm}^2+\tfrac12kx^2 \).
3. \( kx^2=m_2v_0^2\left(1-\frac{m_2}{m_1+m_2}\right)=\frac{m_1m_2v_0^2}{m_1+m_2}\Rightarrow x=v_0\sqrt{\frac{m_1m_2}{(m_1+m_2)k}} \).

Teacher's Note:
1. At maximum extension the relative velocity of the blocks is zero.
2. The reduced mass appears naturally: x = v0 sqrt(mu/k).

Answer: (c) \( K/2 \)

Explanation:
1. Maximum deformation occurs when both particles move with common velocity \( v/2 \).
2. KE then \( =\tfrac12(2m)(v/2)^2=\tfrac14mv^2=\tfrac K2 \).
3. The rest of the energy is stored as potential energy: \( U_{max}=K-\tfrac K2=\tfrac K2 \).

Teacher's Note:
1. At maximum compression the relative velocity is zero.
2. Half of the kinetic energy is temporarily stored; it is returned as the bodies separate.

Answer: (d) \( \frac{1}{\sqrt2} \)

Explanation:
1. Velocities after impact: \( v_1=\frac{(1-e)v}{2},\ v_2=\frac{(1+e)v}{2} \).
2. KE ratio \( =\frac{v_1^2+v_2^2}{v^2}=\frac{1+e^2}{2}=\frac34 \).
3. \( e^2=\frac12\Rightarrow e=\frac1{\sqrt2} \).

Teacher's Note:
1. For equal masses the final kinetic energy fraction is (1+e squared)/2.
2. The loss of energy is the difference between 1 and that fraction.

Answer: (c) \( -2\ \text{m/s},\ +\tfrac34\ \text{m/s},\ \tfrac{99}{4}\ \text{J} \)

Explanation:
1. Take A's direction as positive: momentum \( 3(4)+8(-1.5)=0 \), so the centre of mass is at rest.
2. Relative speed of separation \( =e\times5.5=2.75 \): \( v_B-v_A=2.75,\ 3v_A+8v_B=0\Rightarrow v_A=-2,\ v_B=0.75 \).
3. KE before \( =24+9=33 \) J, after \( =6+2.25=8.25 \) J, so loss \( =24.75=\tfrac{99}{4} \) J.

Teacher's Note:
1. A zero total momentum simplifies the algebra because both blocks reverse.
2. Loss of KE = initial KE minus final KE.

Answer: (a) 10 N and 20 W

Explanation:
1. Force = rate of change of momentum of the sand: \( F=v\frac{dm}{dt}=2\times5=10 \) N.
2. Power \( P=Fv=10\times2=20 \) W (half goes into the kinetic energy of the sand, half is dissipated by sliding).

Teacher's Note:
1. The belt gives each newly arrived mass element the belt speed.
2. Power is force times speed since the belt speed is constant.

Answer: (b) 0.9 cm right and 2.0 cm above the 1 kg mass

Explanation:
1. Take the 1 kg mass as the origin: 1.5 kg at (3, 0) and 2.5 kg at (0, 4); total mass 5 kg.
2. \( x_{cm}=\frac{1.5\times3}{5}=0.9 \) cm.
3. \( y_{cm}=\frac{2.5\times4}{5}=2.0 \) cm.

Teacher's Note:
1. The right-angle corner makes a convenient origin since each mass lies on an axis.
2. Only one coordinate of each outer mass is non-zero.

Answer: (c) 10 cm

Explanation:
1. The centre of mass of a solid cone lies on its axis at one quarter of the height from the base.
2. \( \frac{40}{4}=10 \) cm from the base centre (30 cm from the vertex).

Teacher's Note:
1. Standard results: solid cone h/4 from the base, hollow cone h/3 from the base.
2. Do not confuse the distance from the vertex with that from the base.

Answer: (a) \( \le r \)

Explanation:
1. The position vector of the centre of mass is a weighted average of vectors each of magnitude r.
2. The magnitude of an average of vectors cannot exceed the average of their magnitudes: \( |\vec R|\le r \), with equality only when all particles coincide in direction.

Teacher's Note:
1. Think of two equal masses at opposite ends of a diameter: the centre of mass is at the origin (distance 0).
2. A single particle gives distance exactly r.

Answer: (b) \( \frac{2r}{3(4-\pi)} \)

Explanation:
1. Take O as origin with x measured to the left. Rectangle: area \( 2r^2 \), centre at \( x=r/2 \). Semicircle: area \( \pi r^2/2 \), centre at \( x=\frac{4r}{3\pi} \).
2. \( x_{cm}=\frac{2r^2(r/2)-\frac{\pi r^2}{2}\cdot\frac{4r}{3\pi}}{2r^2-\frac{\pi r^2}{2}}=\frac{r^3-\frac{2r^3}{3}}{\frac{(4-\pi)r^2}{2}} \).
3. \( x_{cm}=\frac{2r}{3(4-\pi)} \).

Teacher's Note:
1. Subtract the cut-out semicircle as a negative mass.
2. The centre of mass shifts away from the removed part, which is to the left of O.

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

Explanation:
1. \( v_{cm}=\frac{1(2)+0.5(6)}{1.5}=\frac{5}{1.5}=\frac{10}{3} \) m/s.

Teacher's Note:
1. Same direction means the momenta add.
2. Divide the total momentum by the total mass.

Answer: (a) \( \left(\frac{n-1}{n+1}\right)^2g \)

Explanation:
1. Acceleration of each body: \( a=\frac{n-1}{n+1}g \) (one up, the other down).
2. \( a_{cm}=\frac{n\,a-1\cdot a}{n+1}=\frac{(n-1)}{(n+1)}a \).
3. \( a_{cm}=\left(\frac{n-1}{n+1}\right)^2g \).

Teacher's Note:
1. The heavier mass accelerates downward and the lighter upward, so the accelerations carry opposite signs in the centre-of-mass formula.
2. For n = 1 the system is balanced and a_cm = 0.

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

Explanation:
1. The initial momentum is zero, so the ball moving up must be accompanied by the Earth moving down.
2. The two therefore move apart, keeping the centre of mass fixed; the Earth's displacement is negligible because of its huge mass.

Teacher's Note:
1. With no external force, the centre of mass of the ball-Earth pair is at rest.
2. In practice we ignore the Earth's motion because of its enormous inertia.