JEE Physics System Of Particles and Rotational Motion MCQs Set 19

Answer: (c) \( -2\hat{i} - \hat{j} + \hat{k} \)

Explanation:
1. \( \vec{r}_{cm} = \dfrac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{4} \).
2. \( = \dfrac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k} \).

Teacher's Note:
1. Weight each position vector by its mass.
2. Work component by component.

Answer: (b) 5 kg

Explanation:
1. The momenta of the first two parts are 12 and 16 kg m/s at right angles, so the resultant is \( \sqrt{144 + 256} = 20 \) kg m/s.
2. The third part has equal and opposite momentum: \( m(4) = 20 \Rightarrow m = 5 \) kg.

Teacher's Note:
1. The total momentum after the explosion is zero.
2. Add perpendicular momenta by Pythagoras.

Answer: (b) 0, 1

Explanation:
1. Momentum: \( m(2) = m v_1 + 2m v_2 \). Restitution: \( v_2 - v_1 = 0.5(2) = 1 \).
2. Substituting: \( 2 = v_1 + 2(v_1 + 1) \Rightarrow v_1 = 0,\; v_2 = 1 \).

Teacher's Note:
1. Solve the momentum and restitution equations together.
2. Check by verifying that \( v_2 - v_1 = e \times \) approach speed.

Answer: (c) 10.1 m

Explanation:
1. Momentum conservation: \( 50\,V = 0.5(2) \Rightarrow V = 0.02 \) m/s upwards.
2. The stone takes \( \dfrac{10}{2} = 5 \) s to reach the floor.
3. The man rises \( 0.02\times 5 = 0.1 \) m, so he is at 10.1 m.

Teacher's Note:
1. The man recoils upward, away from the floor.
2. Find the time of flight of the stone first.

Answer: (a) zero

Explanation:
1. There is no external horizontal force on the (men + boat) system.
2. So the centre of mass of the system does not shift, whatever the men do.
3. The boat itself moves, but the system's centre of mass stays fixed.

Teacher's Note:
1. Internal motion cannot shift the centre of mass.
2. Do not confuse the boat's shift with the centre of mass shift.

Answer: (b) \( \theta = \tan^{-1}(-1/2) \) to the x-axis

Explanation:
1. Momentum along x: \( m_2 v = p_{Ax} \). Along y: \( 0 = m_2\dfrac{v}{2} + p_{Ay} \Rightarrow p_{Ay} = -\dfrac{m_2 v}{2} \).
2. \( \tan\theta = \dfrac{p_{Ay}}{p_{Ax}} = -\dfrac{1}{2} \).

Teacher's Note:
1. A's sideways momentum must cancel B's.
2. So A moves below the x-axis when B moves above it.

Answer: (b) \( \tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 - \varepsilon = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 \)

Explanation:
1. The excitation energy \( \varepsilon \) is taken from the kinetic energy of the particles.
2. Initial KE \( - \varepsilon = \) final KE, which is option (b).

Teacher's Note:
1. Energy absorbed by the particles reduces the kinetic energy.
2. Options (c) and (d) have wrong dimensions.

Answer: (b) 7.5R

Explanation:
1. The centres are 12R apart and touch when the separation is \( R + 2R = 3R \), so the total approach is 9R.
2. The centre of mass is fixed, so the distances moved are inversely proportional to the masses: the smaller body (M) moves \( \dfrac{5}{6} \) of 9R.
3. \( \dfrac{5}{6}\times 9R = 7.5R \).

Teacher's Note:
1. The centre of mass of the pair does not move.
2. The lighter body covers the larger share of the distance.

Answer: (c) 20 m/s

Explanation:
1. Energy per unit mass just before impact \( = \tfrac{1}{2}v_0^2 + 10(20) \).
2. Half of it is kept, and it must equal the energy needed to rise 20 m, i.e. \( 10(20) = 200 \).
3. \( \tfrac{1}{2}\left(\tfrac{1}{2}v_0^2 + 200\right) = 200 \Rightarrow v_0^2 = 400 \Rightarrow v_0 = 20 \) m/s.

Teacher's Note:
1. Find the energy at the ground, halve it, and equate it to the energy at the top of the rebound.
2. The ball starts with kinetic energy as well as height.

Answer: (b) \( \dfrac{2\sqrt{2}}{3}v \)

Explanation:
1. For equal masses in an elastic collision with one at rest, the final velocities are perpendicular.
2. \( v^2 = \left(\dfrac{v}{3}\right)^2 + v_2^2 \Rightarrow v_2^2 = \dfrac{8v^2}{9} \).
3. \( v_2 = \dfrac{2\sqrt{2}}{3}v \).

Teacher's Note:
1. Kinetic energy conservation with the 90° rule gives a Pythagorean relation.
2. Use the speeds, not momenta, since the masses are equal.

Answer: (a) 120 m/s

Explanation:
1. Speed of the block just after: \( V = \sqrt{2gh} = \sqrt{2\times 10\times 0.1} = \sqrt{2} \approx 1.41 \) m/s.
2. Momentum: \( 0.01(400) = 0.01\,v + 2(1.41) \).
3. \( v = \dfrac{4 - 2.83}{0.01} \approx 117 \approx 120 \) m/s.

Teacher's Note:
1. Use energy conservation for the swing and momentum conservation for the impact.
2. The answer is approximate, so choose the nearest option.

Answer: (d) 0.5 m/s and \( -0.3 \) m/s

Explanation:
1. Identical balls in a one-dimensional elastic collision exchange velocities.
2. So B gets 0.5 m/s and A gets \( -0.3 \) m/s.

Teacher's Note:
1. The order in the question is B first, then A.
2. Equal masses swap velocities.

Answer: (b) \( \dfrac{8}{9} \)

Explanation:
1. \( v_A = \dfrac{4m - 2m}{4m + 2m}u = \dfrac{u}{3} \).
2. Fraction of energy retained \( = \dfrac{1}{9} \).
3. Fraction lost \( = 1 - \dfrac{1}{9} = \dfrac{8}{9} \).

Teacher's Note:
1. Energy goes as the square of the speed.
2. The remaining energy of A is one ninth.

Answer: (a) \( 4\hat{i} + 23\hat{j} - 16\hat{k} \)

Explanation:
1. Momentum conservation with masses m and 5m: \( 6m(20, 25, -12) = m(100, 35, 8) + 5m\,\vec{v} \).
2. \( 5\vec{v} = (120 - 100,\; 150 - 35,\; -72 - 8) = (20, 115, -80) \).
3. \( \vec{v} = 4\hat{i} + 23\hat{j} - 16\hat{k} \).

Teacher's Note:
1. Treat each component separately.
2. The total mass is 6m and the larger piece is 5m.

Answer: (b) the blade piece will have the larger mass

Explanation:
1. About the centre of mass the moments balance: \( m_{handle}\,d_{handle} = m_{blade}\,d_{blade} \).
2. The handle part extends farther from the centre of mass, so its mass must be smaller.
3. The blade piece therefore has the larger mass.

Teacher's Note:
1. The centre of mass lies closer to the heavy end.
2. The same pieces would be equal only if the cut was at the geometric middle of a uniform bar.

Answer: (b) 10 cm from the centre of the rod

Explanation:
1. The rod's centre of mass is at its middle (take it as the origin) and the ball's centre is at \( 25 + 5 = 30 \) cm from it.
2. \( x_{cm} = \dfrac{100(0) + 50(30)}{150} = 10 \) cm from the centre of the rod, towards the ball.

Teacher's Note:
1. The ball's centre is beyond the end of the rod by its radius.
2. The balance point is the centre of mass of the combination.

Answer: (b) \( \dfrac{6}{5} \) m

Explanation:
1. \( M = \int_0^2(3x + 2)\,dx = 6 + 4 = 10 \).
2. \( \int_0^2 x(3x + 2)\,dx = 8 + 4 = 12 \).
3. \( x_{cm} = \dfrac{12}{10} = \dfrac{6}{5} \) m.

Teacher's Note:
1. Use \( x_{cm} = \dfrac{\int x\,dm}{\int dm} \) with \( dm = \lambda\,dx \).
2. The centre lies beyond the middle because the rod is denser at the far end.

Answer: (c) 1.24 Å

Explanation:
1. Place H at the origin and Cl at 1.27 Å.
2. \( x_{cm} = \dfrac{1(0) + 35.5(1.27)}{36.5} \approx 1.24 \) Å.

Teacher's Note:
1. The centre of mass is very close to the heavy chlorine atom.
2. The answer must be less than 1.27 Å.

Answer: (c) \( \dfrac{r}{14} \)

Explanation:
1. The removed sphere has radius \( r/2 \) and mass \( \dfrac{1}{8} \) of the whole. The shift is greatest when it touches the surface, with its centre at \( r/2 \) from the centre.
2. Shift \( = \dfrac{(1/8)(r/2)}{7/8} = \dfrac{r}{14} \).

Teacher's Note:
1. Removal at the surface gives the maximum shift.
2. The remaining mass is 7/8 of the original.

Answer: (d) \( \dfrac{L}{3} \)

Explanation:
1. The explosion is internal, so the centre of mass of the system (box + bomb) stays fixed.
2. After the parts stick to the walls, the lighter part m is \( 2L \) to one side of the box's centre and 2m is \( 2L \) to the other.
3. The box's centre moves a distance x with \( 3m(x) + m(x - 2L) + 2m(x + 2L) = 0 \Rightarrow 6mx + 2mL = 0 \Rightarrow |x| = \dfrac{L}{3} \).

Teacher's Note:
1. The centre of mass of the system does not shift.
2. The box moves towards the side of the lighter fragment.

Answer: (b) B will be accelerated but A will be retarded

Explanation:
1. A packet thrown from A has A's high forward speed, and one thrown from B has B's lower speed.
2. A loses packets with high momentum and receives packets with lower momentum, so A slows down.
3. B gains the high-momentum packets, so B speeds up.

Teacher's Note:
1. Think of the packets as carrying momentum from one train to the other.
2. The faster train loses momentum and the slower gains it.

Answer: (a) \( \dfrac{4v}{A - 4} \)

Explanation:
1. The daughter nucleus has mass number \( A - 4 \).
2. Momentum conservation: \( 4v = (A - 4)V \).
3. \( V = \dfrac{4v}{A - 4} \).

Teacher's Note:
1. Use mass numbers as the masses.
2. The daughter is lighter than the parent by 4 units.

Answer: (b) \( \dfrac{2}{3} \) kg and \( \dfrac{6}{8}v \)

Explanation:
1. \( \dfrac{v}{2} = \dfrac{2 - m}{2 + m}v \Rightarrow 2 + m = 4 - 2m \Rightarrow m = \dfrac{2}{3} \) kg.
2. \( v_{cm} = \dfrac{2v}{2 + 2/3} = \dfrac{6}{8}v \).

Teacher's Note:
1. Use the elastic collision formula for the first body's final speed.
2. Centre of mass speed is total momentum over total mass.

Answer: (b) \( -4.6 \), 1.4

Explanation:
1. Mass is proportional to \( r^3 \), so the masses are in ratio \( 1 : 8 \).
2. \( v_1 = \dfrac{1 - 8}{9}(6) = -4.67 \) and \( v_2 = \dfrac{2(1)}{9}(6) = 1.33 \).
3. So the velocities are about \( -4.6 \) and 1.4 cm/s (the small ball bounces back).

Teacher's Note:
1. Steel balls have the same density, so mass goes as radius cubed.
2. The lighter ball rebounds.

Answer: (b) \( \dfrac{1}{4} \)

Explanation:
1. B must start up the incline with \( \sqrt{2gh} = \sqrt{2\times 10\times 5} = 10 \) m/s.
2. For equal masses, \( v_B = \dfrac{(1 + e)}{2}\,v = \dfrac{(1 + e)}{2}(16) = 10 \).
3. \( 1 + e = 1.25 \Rightarrow e = \dfrac{1}{4} \).

Teacher's Note:
1. The speed at the foot of the incline decides the height reached.
2. Use the equal mass relation for the struck ball's speed.

Answer: (b) 8/9

Explanation:
1. Take the neutron mass m and the deuteron mass 2m.
2. \( v_n = \dfrac{m - 2m}{3m}u = -\dfrac{u}{3} \).
3. Fraction retained \( = \dfrac{1}{9} \), so the fraction lost is \( \dfrac{8}{9} \).

Teacher's Note:
1. Energy fraction retained is the square of the speed fraction.
2. The standard formula is \( \dfrac{4m_1m_2}{(m_1 + m_2)^2} \).

Answer: (a) 11 m/s upwards

Explanation:
1. The ball hits with \( v = \sqrt{2(9.8)(4.9)} = 9.8 \) m/s downwards.
2. In the lift's frame it approaches at \( 9.8 + 0.6 = 10.4 \) m/s and rebounds at 10.4 m/s upwards.
3. In the ground frame this is \( 10.4 + 0.6 = 11 \) m/s upwards.

Teacher's Note:
1. Switch to the heavy lift's frame, where the speed is unchanged for e = 1.
2. Then return to the ground frame.

Answer: (c) 0.25 m

Explanation:
1. At maximum compression both move with a common velocity: \( V = \dfrac{3(5) + 5(3)}{8} = 3.75 \) m/s.
2. Kinetic energy lost \( = \tfrac{1}{2}\cdot\dfrac{15}{8}(5 - 3)^2 = 3.75 \) J, which is stored in the spring.
3. \( \tfrac{1}{2}(120)x^2 = 3.75 \Rightarrow x = 0.25 \) m.

Teacher's Note:
1. Maximum compression occurs when the relative velocity is zero.
2. The energy stored equals the loss in kinetic energy.