JEE Physics Centre of Mass MCQs Set 08

Answer: (b) (ii) and (iii) only

Explanation:
1. In the wall frame the ball approaches at \( u+v \) and leaves at the same speed; the ball's ground speed after impact is \( v+2u \).
2. Change in momentum \( =m(v+2u)+mv=2m(u+v) \), so the average force is \( \frac{2m(u+v)}{\Delta t} \).
3. \( \Delta KE=\tfrac12m[(v+2u)^2-v^2]=2mu(u+v) \).

Teacher's Note:
1. The moving wall does work on the ball, so the ball gains kinetic energy.
2. Use the wall's frame, where the collision looks like a bounce off a fixed wall.

Answer: (c) less than \( \tfrac12mv^2 \) if he walks along the rails, and equal to \( \tfrac12mv^2 \) if he walks normal to the rails

Explanation:
1. Walking along the rails, the car recoils: the man's ground speed is \( \frac{Mv}{M+m} \) and the car's \( \frac{mv}{M+m} \).
2. Total KE \( =\tfrac12\frac{mM}{M+m}v^2<\tfrac12mv^2 \).
3. Walking normal to the rails, the car does not move (rails exert sideways force), so the work is \( \tfrac12mv^2 \).

Teacher's Note:
1. Work done by the man equals the total kinetic energy given to man plus car.
2. When the car recoils, part of the relative velocity is carried by the car, reducing the man's ground speed.

Answer: (d) (i) and (iii)

Explanation:
1. The impact is instantaneous and the spring force (finite) cannot produce an impulse in that time: (i) true.
2. Speed just after: \( V=\frac{1\times3}{3}=1 \) m/s.
3. \( \tfrac12(3)(1)^2=\tfrac12(3)x^2\Rightarrow x=1 \) m: (iii) true.

Teacher's Note:
1. Only forces that are very large (impulsive) can alter momentum in a very short collision.
2. After the impact, the spring and the combined mass exchange KE for elastic potential energy.

Answer: (b) (i), (iii) and (iv) only

Explanation:
1. Let B have speed u and A speed qu. Momentum: \( qu+pu=p\,v_B\Rightarrow v_B=\frac{(p+q)u}{p} \).
2. \( e=\frac{v_B-0}{qu-u}=\frac{p+q}{p(q-1)}=\frac{p+q}{pq-p} \): (i) true, (ii) false.
3. \( e\le1\Rightarrow p+q\le pq-p\Rightarrow p\ge\frac{q}{q-2} \) (with q greater than 2), which exceeds 1, so (iv) holds as well.

Teacher's Note:
1. Use e as the ratio of speed of separation to speed of approach.
2. The condition e at most 1 gives a bound on the mass ratio.

Answer: (c) \( \sqrt{30} \) m/s

Explanation:
1. Horizontal momentum: \( mv=MV\Rightarrow V=\frac v2 \).
2. Energy: \( mgh=90=\tfrac12v^2+\tfrac12(2)\frac{v^2}{4}=\tfrac34v^2\Rightarrow v^2=120 \).
3. \( V^2=\frac{v^2}{4}=30 \), so \( V=\sqrt{30} \) m/s.

Teacher's Note:
1. The wedge recoils opposite to the block, so a share of the energy goes to the wedge.
2. Treat the block plus left wedge as an isolated horizontal system.

Answer: (a) \( \sqrt{\frac{40}{3}} \) m/s

Explanation:
1. At maximum height the block and right wedge move together with \( u=\frac v3 \).
2. \( u^2=\frac{v^2}{9}=\frac{120}{9}=\frac{40}{3} \).
3. \( u=\sqrt{\frac{40}{3}} \) m/s.

Teacher's Note:
1. Use the common-velocity condition from the previous part.
2. The speed of the block at the bottom, v squared = 120, comes from the first part.

Answer: (c) \( \frac12\left(\frac{Ft^2}{2m}+x_0\right) \)

Explanation:
1. Let the displacements be \( x_L \) and \( x_R \) with \( x_R-x_L=x_0 \).
2. Centre of mass: \( \frac{x_L+x_R}{2}=\frac{Ft^2}{4m}\Rightarrow x_L+x_R=\frac{Ft^2}{2m} \).
3. \( x_R=\frac12\left(\frac{Ft^2}{2m}+x_0\right) \).

Teacher's Note:
1. Add and subtract the two relations.
2. The pulled block moves farther than the centre of mass.

Answer: (d) \( \frac12\left(\frac{Ft^2}{2m}-x_0\right) \)

Explanation:
1. From \( x_R+x_L=\frac{Ft^2}{2m} \) and \( x_R-x_L=x_0 \):
2. \( x_L=\frac12\left(\frac{Ft^2}{2m}-x_0\right) \).

Teacher's Note:
1. The left block moves less than the centre of mass.
2. Subtract the extension condition from the centre-of-mass relation.

Answer: (b) \( m(3\hat i-\sqrt3\hat j) \) kg m/s

Explanation:
1. Impulse on B = change in momentum of B: \( m\,\vec v_B-0=m(3\hat i-\sqrt3\hat j) \) kg m/s.
2. It is directed along the line of centres (30 degrees below the x-axis), magnitude \( 2\sqrt3m \).

Teacher's Note:
1. Impulse on B is along the line joining the centres, away from A.
2. Check: the vector (3, -sqrt3) is parallel to (cos30, -sin30).

Answer: (c) (i) and (iii)

Explanation:
1. Elastic collision: \( v_1'=\frac{1-5}{6}u=-2\Rightarrow u=3 \) m/s, and \( v_2'=\frac{2}{6}u=1 \) m/s.
2. (i) total momentum \( =1(3)=3 \) kg m/s: true. (ii) 5 kg block: \( 5(1)=5 \) kg m/s, so false.
3. (iii) \( \tfrac12(6)(0.5)^2=0.75 \) J: true. (iv) total KE \( =\tfrac12(1)(9)=4.5 \) J, false.

Teacher's Note:
1. Find the initial speed from the reversal condition first.
2. The centre of mass moves at 3/6 = 0.5 m/s.

Answer: (a) 4

Explanation:
1. Energy: \( \tfrac12mV^2=\tfrac12kx^2+\mu mgx \).
2. \( 0.09V^2=\tfrac12(2)(0.0036)+0.1(0.18)(10)(0.06)=0.0036+0.0108=0.0144 \).
3. \( V^2=0.16\Rightarrow V=0.4 \) m/s, so N = 4.

Teacher's Note:
1. The initial kinetic energy is shared between the spring and friction.
2. Friction acts over the whole path of 0.06 m.

Answer: (b) 5

Explanation:
1. Speed of the first bob at the top: \( v=\sqrt{gl_1} \) (minimum for a full circle).
2. Equal masses in an elastic collision exchange velocities, so the second bob moves off with \( \sqrt{gl_1} \) at its lowest point.
3. Minimum speed at the bottom for a full circle: \( \sqrt{5gl_2} \). Equating, \( gl_1=5gl_2\Rightarrow\frac{l_1}{l_2}=5 \).

Teacher's Note:
1. The minimum speed at the top is sqrt(gl) and at the bottom sqrt(5gl).
2. The collision takes place with the second bob at its lowest position.

Answer: (d) (i) and (ii)

Explanation:
1. Horizontal momentum zero and energy: \( mv=MV,\ mgR=\tfrac12mv^2+\tfrac12MV^2\Rightarrow v=\sqrt{\frac{2gR}{1+m/M}} \): (i) true.
2. The point mass moves R to the right relative to the block; the block moves \( -\frac{mR}{M+m} \) (left): (ii) true.
3. Position of the mass: \( x=-R+\frac{MR}{M+m}=-\frac{mR}{M+m} \), not with a factor \( \sqrt2 \): (iii) false; and \( V=\frac mM v\ne\frac mM\sqrt{2gR} \): (iv) false.

Teacher's Note:
1. The centre of mass of block plus point mass does not move horizontally.
2. At the bottom the mass has only horizontal velocity relative to the table.

Answer: (c) Statement-I is false, Statement-II is true

Explanation:
1. Maximum loss occurs in a perfectly inelastic collision (Statement-II true): \( V=\frac{mv}{m+M} \).
2. Energy lost \( =\tfrac12mv^2-\tfrac12(m+M)V^2=\tfrac12mv^2\cdot\frac{M}{m+M} \).
3. So \( f=\frac{M}{M+m} \), not \( \frac m{M+m} \): Statement-I is false.

Teacher's Note:
1. The fraction lost is the target mass divided by the total mass.
2. If M is very large the whole kinetic energy is lost, so f tends to 1.

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

Explanation:
1. The centre of mass of a solid cone lies on the axis at h/4 from the base.
2. Hence from the vertex: \( z_0=h-\frac h4=\frac{3h}{4} \).

Teacher's Note:
1. The answer depends only on h, not on R.
2. Do not mix up the distance from the base with the distance from the vertex.

Answer: (d) \( \sqrt2\,v_0 \)

Explanation:
1. Momentum: \( v_1+v_2=v_0 \). Energy: \( v_1^2+v_2^2=1.5v_0^2 \).
2. \( 2v_1v_2=v_0^2-1.5v_0^2=-0.5v_0^2 \).
3. \( (v_1-v_2)^2=1.5v_0^2+0.5v_0^2=2v_0^2 \), so the relative speed is \( \sqrt2v_0 \).

Teacher's Note:
1. Use the identity (a - b) squared = (a squared + b squared) - 2ab.
2. The energy increase means energy was released during the collision (explosive collision).

Answer: (a) \( v_4-v_2 \)

Explanation:
1. Momentum: \( m_1v_1+0.5m_1v_2=m_1(0.5v_1)+0.5m_1v_4 \).
2. \( v_1+0.5v_2=0.5v_1+0.5v_4\Rightarrow0.5v_1=0.5(v_4-v_2) \).
3. \( v_1=v_4-v_2 \).

Teacher's Note:
1. Cancel \( m_1 \) from every term.
2. The result is independent of the collision type; only momentum conservation is needed.

Answer: (b) 1

Explanation:
1. Momentum: \( \vec v_B'=3\hat i+5\hat j-4\hat i-4\hat j=-\hat i+\hat j \) m/s.
2. \( KE_B=\tfrac12(0.1)(1+1)=0.1\ \text{J}=\frac{1}{10} \).
3. So \( x=1 \). (Check: total KE before \( =0.05(9+25)=1.7 \) J, after \( =0.05(32+2)=1.7 \) J.)

Teacher's Note:
1. Equal masses make the velocity of B straightforward to find from the momentum equation.
2. The energy check confirms that the collision is elastic.

Answer: (c) \( \sqrt{\frac32} \)

Explanation:
1. Relative speed \( \sqrt{2gh} \), so they meet after \( t=\frac{h}{\sqrt{2gh}}=\sqrt{\frac{h}{2g}} \) at height \( h-\tfrac12gt^2=\frac{3h}{4} \).
2. Speeds at that instant: first \( gt=\sqrt{gh/2} \) down, second \( \sqrt{2gh}-gt=\sqrt{gh/2} \) up; equal and opposite, so the combined mass has zero velocity.
3. It then falls freely from \( 3h/4 \): \( t'=\sqrt{\frac{2(3h/4)}{g}}=\sqrt{\frac32}\sqrt{\frac hg} \).

Teacher's Note:
1. Equal masses with equal and opposite velocities stick together at rest.
2. The question asks only for the fall after the collision.

Answer: (d) \( \frac34\left(\frac{2a+b}{3a+b}\right)L \)

Explanation:
1. Total mass: \( \int_0^L\rho\,dx=aL+\frac{bL}{3} \).
2. First moment: \( \int_0^Lx\rho\,dx=\frac{aL^2}{2}+\frac{bL^2}{4} \).
3. \( x_{cm}=\frac{L^2(2a+b)/4}{L(3a+b)/3}=\frac34\frac{(2a+b)}{(3a+b)}L \).

Teacher's Note:
1. For b = 0 the result is L/2 and for a = 0 it is 3L/4, as expected.
2. Compute both integrals with the density written as a polynomial.