JEE Physics System Of Particles and Rotational Motion MCQs Set 17

Answer: (b) OA

Explanation:
1. The remaining L-shaped plate is symmetric about the diagonal AC.
2. Removing mass from the corner C side shifts the centre of mass from O away from C, along the diagonal towards A.
3. So it lies on the line OA.

Teacher's Note:
1. Use symmetry first.
2. Removing mass from one side moves the centre of mass to the opposite side.

Answer: (b) first descends and then ascends

Explanation:
1. Initially the centre of mass is at the middle of the full can.
2. As liquid drains, the low-lying mass is removed from the bottom of the water column, but the centre of mass first moves down and then, when the can is nearly empty, returns up to the centre of the can.
3. It first descends and then ascends.

Teacher's Note:
1. At the start and at the end the centre of mass is at the same height.
2. In between it dips below that height.

Answer: (b) \( \dfrac{PQ + PR}{3} \)

Explanation:
1. Take P as the origin, so the positions are 0, PQ and PR.
2. \( x_{cm} = \dfrac{1(0) + 1(PQ) + 1(PR)}{3} = \dfrac{PQ + PR}{3} \).

Teacher's Note:
1. With equal masses the centre of mass is the average of the positions.
2. Measure all distances from P.

Answer: (d) the point of intersection of the medians of the triangle

Explanation:
1. For three equal masses the centre of mass is the centroid of the triangle formed by their centres.
2. The centroid of the triangle is the point where the medians meet.

Teacher's Note:
1. Equal masses at the vertices give the centroid.
2. For an equilateral triangle it is also the centre.

Answer: (c) \( \left(\dfrac{a}{6}, \dfrac{2a}{3}\right) \)

Explanation:
1. Each body can be replaced by its mass at its centre.
2. \( x_{cm} = \dfrac{M(0) + M(0) + M(a/2)}{3M} = \dfrac{a}{6} \).
3. \( y_{cm} = \dfrac{M(2a) + 0 + 0}{3M} = \dfrac{2a}{3} \).

Teacher's Note:
1. Use the centre of each symmetric body as its mass position.
2. Equal masses give simple averages of the coordinates.

Answer: (a) 2

Explanation:
1. The three particles (total 7 g) have their centre of mass at the origin, so they act like a 7 g mass at the origin.
2. \( \vec{r}_{cm} = \dfrac{7(0) + 7\vec{r}_4}{14} = \dfrac{\vec{r}_4}{2} \).
3. \( \dfrac{\alpha}{2}(2,3,4) = (2,3,4) \Rightarrow \alpha = 2 \).

Teacher's Note:
1. Replace the three particles by a single 7 g particle at the origin.
2. Then it is a two-body centre of mass problem.

Answer: (b) \( \dfrac{L\sqrt{3}}{4} \)

Explanation:
1. Each arm acts as a mass at its midpoint, at distance \( L/2 \) from O along the arm.
2. The centre of mass lies on the bisector, and each midpoint is at \( \dfrac{L}{2}\cos 30^\circ \) along it.
3. \( x_{cm} = \dfrac{L}{2}\cdot\dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}\,L}{4} \).

Teacher's Note:
1. Resolve each arm's midpoint along the bisector.
2. The equal arms give the same component, so the average is that value.

Answer: (d) \( \dfrac{R}{20} \) from the centre

Explanation:
1. The removed disc has radius \( R/4 \), so its mass is \( \dfrac{1}{16} \) of the whole and its centre is \( \dfrac{3R}{4} \) from the centre of the big disc.
2. Shift \( = \dfrac{m(3R/4)}{M - m} = \dfrac{(1/16)(3R/4)}{15/16} = \dfrac{R}{20} \), on the side opposite to the hole.

Teacher's Note:
1. Treat the cut-out as a negative mass.
2. Area (hence mass) scales as the square of the radius.

Answer: (b) internal forces on the particles

Explanation:
1. \( \vec{r}_{cm} = \dfrac{\sum m_i \vec{r}_i}{\sum m_i} \) depends on the masses and positions.
2. Internal forces cannot change the motion of the centre of mass, and are not part of the formula.

Teacher's Note:
1. Only external forces affect the motion of the centre of mass.
2. Internal forces cancel in pairs.

Answer: (a) \( \dfrac{16}{5} \) m/s

Explanation:
1. \( v_{cm} = \dfrac{1(4) + 4(3)}{1 + 4} = \dfrac{16}{5} \) m/s.

Teacher's Note:
1. The centre of mass velocity is the total momentum divided by the total mass.
2. Use signs if the bodies move in opposite directions.

Answer: (b) zero

Explanation:
1. The only forces are internal, so the total momentum stays at its initial value, which is zero (they started at rest).
2. Hence the centre of mass velocity stays zero.

Teacher's Note:
1. Internal forces cannot change total momentum.
2. The centre of mass of a system starting at rest stays at rest.

Answer: (b) \( \dfrac{g}{16} \)

Explanation:
1. Acceleration of each mass \( a = \dfrac{5m - 3m}{5m + 3m}g = \dfrac{g}{4} \), 5m moving down and 3m moving up.
2. \( a_{cm} = \dfrac{5m(a) - 3m(a)}{8m} = \dfrac{2a}{8} = \dfrac{a}{4} = \dfrac{g}{16} \) downward.

Teacher's Note:
1. The two masses accelerate in opposite directions.
2. Subtract the accelerations weighted by mass.

Answer: (a) \( (2.4\hat{i} - 3.6\hat{j}) \) m/s\( ^2 \)

Explanation:
1. \( a = \dfrac{6g}{10} = 6 \) m/s\( ^2 \). The 4 kg block accelerates along +x and the 6 kg block along \( -y \).
2. \( a_{cm} = \dfrac{4(6)\hat{i} + 6(-6)\hat{j}}{10} = (2.4\hat{i} - 3.6\hat{j}) \) m/s\( ^2 \).

Teacher's Note:
1. The two blocks accelerate in perpendicular directions, so add the components separately.
2. The centre of mass acceleration is not along the string.

Answer: (a) \( \dfrac{Lm}{m + M} \)

Explanation:
1. No external horizontal force acts, so the centre of mass stays fixed.
2. \( m\,x_{person} = M\,x_{plank} \) with \( x_{person} + x_{plank} = L \) (relative displacement).
3. \( x_{plank} = \dfrac{mL}{m + M} \).

Teacher's Note:
1. The relative displacement between person and plank equals L.
2. The plank moves opposite to the person.

Answer: (c) third law of motion

Explanation:
1. Internal forces come in equal and opposite pairs (third law), so they cancel in the total.
2. The total momentum therefore changes only through external force.

Teacher's Note:
1. The third law cancels internal forces in the sum.
2. The second law then gives \( d\vec{P}/dt = \vec{F}_{ext} \).

Answer: (d) both A and B are false

Explanation:
1. Velocity depends on the observer's frame, so both momentum \( m\vec{v} \) and kinetic energy \( \tfrac{1}{2}mv^2 \) change from frame to frame.
2. So both statements are false.

Teacher's Note:
1. A particle at rest in one frame moves in another.
2. Both quantities then differ.

Answer: (a) Kinetic energy of a system can be changed without changing its momentum.

Explanation:
1. In an explosion of a body at rest, the total momentum stays zero while the kinetic energy rises, so (a) is true and (b) is false.
2. A body can have energy (potential, or internal energy) while at rest, so (d) is false.

Teacher's Note:
1. Momentum is a vector and kinetic energy is a scalar.
2. Many processes change one without the other.

Answer: (b) 1.8 N s

Explanation:
1. \( \Delta\vec{v} = (2-5,\,3+3,\,1-7) = (-3, 6, -6) \), so \( |\Delta\vec{v}| = \sqrt{9 + 36 + 36} = 9 \) m/s.
2. Impulse \( = m|\Delta\vec{v}| = 0.2\times 9 = 1.8 \) N s.

Teacher's Note:
1. Impulse equals the change in momentum, a vector difference.
2. Subtract component by component before taking the magnitude.

Answer: (c) 576 J

Explanation:
1. Momentum conservation (bomb at rest): \( 4(24) = 8v \Rightarrow v = 12 \) m/s.
2. \( KE = \tfrac{1}{2}(8)(12)^2 = 576 \) J.

Teacher's Note:
1. The lighter piece moves faster.
2. The total momentum after the explosion stays zero.

Answer: (c) \( -30 \) m/s

Explanation:
1. Momentum conservation: \( 50(600) + 1000\,V = 0 \).
2. \( V = -30 \) m/s, the negative sign meaning opposite to the bomb.

Teacher's Note:
1. The cannon recoils opposite to the shell.
2. Its speed is small because its mass is large.

Answer: (b) \( 2mv^2 \)

Explanation:
1. The two pieces carry momenta mv at right angles, so the resultant is \( \sqrt{2}\,mv \).
2. The third piece (mass m) must have speed \( \sqrt{2}\,v \) opposite to it.
3. \( KE = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 + \tfrac{1}{2}m(2v^2) = 2mv^2 \).

Teacher's Note:
1. Balance the vector momentum first.
2. Then add the three kinetic energies.

Answer: (b) 6 kJ

Explanation:
1. \( 2(60) = 3v \Rightarrow v = 40 \) m/s.
2. \( KE = \tfrac{1}{2}(2)(60)^2 + \tfrac{1}{2}(3)(40)^2 = 3600 + 2400 = 6000 \) J.

Teacher's Note:
1. Find the speed of the heavier piece from momentum conservation.
2. Add both kinetic energies.

Answer: (b) 1089 J

Explanation:
1. \( 22(9) = 18v \Rightarrow v = 11 \) m/s.
2. \( KE = \tfrac{1}{2}(18)(121) = 1089 \) J.

Teacher's Note:
1. Equal and opposite momenta, so the lighter piece has the higher speed.
2. Square the speed carefully.

Answer: (a) \( \dfrac{m_2}{m_1} \)

Explanation:
1. With equal momenta \( p \), \( E = \dfrac{p^2}{2m} \).
2. So \( \dfrac{E_1}{E_2} = \dfrac{m_2}{m_1} \).

Teacher's Note:
1. For equal momentum, kinetic energy is inversely proportional to mass.
2. The lighter fragment carries more energy.

Answer: (b) 1 : 3

Explanation:
1. Equal momenta give mass ratio \( m_1 : m_2 = 1 : 27 \) for speed ratio 27 : 1.
2. Nuclear density is constant, so \( R \propto m^{1/3} \).
3. The radii are in the ratio \( 1 : 3 \).

Teacher's Note:
1. Mass is proportional to volume, so radius goes as the cube root of mass.
2. The faster fragment is the lighter one.

Answer: (c) \( \dfrac{d}{m}\sqrt{kM} \)

Explanation:
1. The gun's recoil energy is stored in the spring: \( \tfrac{1}{2}MV^2 = \tfrac{1}{2}kd^2 \Rightarrow V = d\sqrt{k/M} \).
2. Momentum conservation: \( mv = MV \Rightarrow v = \dfrac{M}{m}\,d\sqrt{\dfrac{k}{M}} = \dfrac{d}{m}\sqrt{kM} \).

Teacher's Note:
1. Two steps: energy for the gun, then momentum for the bullet.
2. Check the units of the final expression.

Answer: (b) change in momentum in the rubber ball is more

Explanation:
1. The metal ball goes from mv to 0, so \( \Delta p = mv \).
2. The rubber ball goes from mv to \( -mv' \), so \( \Delta p = m(v + v') \gt mv \).

Teacher's Note:
1. Rebounding reverses the momentum, which increases the change.
2. Both balls have the same initial momentum.

Answer: (b) \( -3\hat{i} - \hat{j} \)

Explanation:
1. Total momentum is zero, and the masses are equal, so the sum of the velocities is zero.
2. \( \vec{v}_3 = -[(2+1)\hat{i} + (3-2)\hat{j}] = -3\hat{i} - \hat{j} \).

Teacher's Note:
1. For equal masses the velocities add to zero.
2. Keep track of the signs of the components.

Answer: (b) Both pieces have zero vertical velocity immediately after the breakup, and therefore they hit the ground at the same time.

Explanation:
1. At the top the vertical velocity is zero, and the explosion forces are internal, so the vertical momentum stays zero.
2. A is at rest, so B moves horizontally with speed \( \tfrac{3}{2}v\cos\theta \) and has no vertical velocity.
3. Both fall from the same height with zero initial vertical velocity, so they land together.

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. Time of fall depends only on height and initial vertical velocity.