JEE Physics System Of Particles and Rotational Motion MCQs Set 05

Answer: (d) Third law of motion

Explanation:
1. For two particles with no external force, \( \vec F_1 = -\vec F_2 \) at every instant.
2. So \( \dfrac{d}{dt}(\vec p_1 + \vec p_2) = 0 \), which is the third law applied to a system.

Teacher's Note:
1. Internal forces occur in equal and opposite pairs, so they cannot change the total momentum.

Answer: (d) \( \dfrac{mL}{M+m} \)

Explanation:
1. No external horizontal force acts, so the centre of mass stays fixed.
2. If the plank moves x, the person moves \( L - x \) relative to the ground in the opposite sense: \( Mx = m(L - x) \).
3. \( x = \dfrac{mL}{M+m} \).

Teacher's Note:
1. Keep the centre of mass at rest and write the moments of the two bodies about it. The plank moves less if it is heavier.

Answer: (d) 300

Explanation:
1. Momentum per bullet \( = 0.04\times 1000 = 40 \) kg m/s.
2. Force \( = \) momentum per second, so bullets per second \( = 200/40 = 5 \).
3. Per minute: \( 5\times 60 = 300 \).

Teacher's Note:
1. Average force equals the momentum transferred per second. Convert to bullets per minute at the end.

Answer: (c) Zero

Explanation:
1. The system is isolated and started from rest, so its total momentum is zero.
2. The velocity of the centre of mass therefore stays zero.

Teacher's Note:
1. Internal forces cannot change the velocity of the centre of mass. It remains at rest.

Answer: (d) \( \dfrac{12}{5} \) m/s

Explanation:
1. \( v_{cm} = \dfrac{2\times 3 + 3\times 2}{2+3} = \dfrac{12}{5} \) m/s.

Teacher's Note:
1. Weighted average of the velocities with the masses as weights.

Answer: (b) \( 5\hat i + \tfrac{25}{7}\hat j \)

Explanation:
1. \( \vec v_{cm} = \dfrac{0.2(10\hat i) + 0.5(3\hat i + 5\hat j)}{0.7} = \dfrac{3.5\hat i + 2.5\hat j}{0.7} \).
2. \( = 5\hat i + \tfrac{25}{7}\hat j \).

Teacher's Note:
1. Convert grams to kilograms (or use the ratio 2 : 5 directly). Do the i and j parts separately.

Answer: (a) 192 J

Explanation:
1. Momentum conservation (bomb at rest): \( 6v = 3\times 16 \Rightarrow v = 8 \) m/s.
2. \( K = \tfrac{1}{2}(6)(8)^2 = 192 \) J.

Teacher's Note:
1. In an explosion from rest the lighter piece moves faster and has more kinetic energy.

Answer: (b) \( -2\hat i + 2\hat j \)

Explanation:
1. Total momentum is zero, so with equal masses the velocity vectors add to zero.
2. \( \vec v_3 = -[(3-1)\hat i + (2-4)\hat j] = -(2\hat i - 2\hat j) = -2\hat i + 2\hat j \).

Teacher's Note:
1. Equal masses mean you can simply add velocities. The three vectors must close into a triangle.

Answer: (d) -30 m/s

Explanation:
1. \( 50\times 600 + 1000V = 0 \Rightarrow V = -30 \) m/s.
2. The negative sign shows that the cannon moves opposite to the bomb.

Teacher's Note:
1. Momentum before firing is zero, so the recoil momentum is equal and opposite.

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

Explanation:
1. The metal ball's momentum changes from mv to 0, a change of mv.
2. The rubber ball's momentum changes from mv to about -mv, a change of about 2mv. So the rubber ball has the larger change.

Teacher's Note:
1. Rebounding means the momentum reverses, giving a bigger impulse than simply stopping.

Answer: (c) 100 m/s upwards

Explanation:
1. Speed of the bomb just before explosion \( = 100 - 10\times 5 = 50 \) m/s upwards.
2. Momentum conservation (upward positive): \( 1\times 50 = 0.4(-25) + 0.6v \Rightarrow 0.6v = 60 \).
3. \( v = 100 \) m/s upwards.

Teacher's Note:
1. Gravity is negligible during the very short explosion, so momentum is conserved just before and after it. Find the velocity at the instant of explosion first.

Answer: (c) \( 10\sqrt2 \) m/s

Explanation:
1. Masses: 0.2 kg, 0.2 kg and 0.6 kg. The momenta of the small parts are perpendicular, each \( 0.2\times 30 = 6 \).
2. Resultant \( = 6\sqrt2 \). The big part has equal and opposite momentum: \( 0.6v = 6\sqrt2 \).
3. \( v = 10\sqrt2 \) m/s.

Teacher's Note:
1. Add the perpendicular momenta with Pythagoras, then divide by the mass of the large part.

Answer: (c) 1 : 2

Explanation:
1. Equal and opposite momenta give \( m_1v_1 = m_2v_2 \), so \( m_1 : m_2 = 1 : 8 \).
2. Nuclear matter has the same density, so \( r \propto m^{1/3} \): \( r_1 : r_2 = 1 : 2 \).

Teacher's Note:
1. Mass is proportional to the volume, so the radius is proportional to the cube root of the mass.

Answer: (b) \( m_2/m_1 \)

Explanation:
1. Equal momenta, so \( K = p^2/2m \), and \( \dfrac{E_1}{E_2} = \dfrac{m_2}{m_1} \).

Teacher's Note:
1. In an explosion from rest, the lighter fragment gets the greater share of the energy.

Answer: (a) \( \tfrac{3}{2}mv^2 \)

Explanation:
1. The two momenta combine to \( \sqrt2\,mv \), so the third piece (mass 2m) has speed \( v_3 = \dfrac{\sqrt2\,mv}{2m} = \dfrac{v}{\sqrt2} \).
2. \( K = 2\times\tfrac{1}{2}mv^2 + \tfrac{1}{2}(2m)\dfrac{v^2}{2} = mv^2 + \tfrac{1}{2}mv^2 = \tfrac{3}{2}mv^2 \).

Teacher's Note:
1. The mass of the third piece is 4m - 2m = 2m. Add the three kinetic energies.

Answer: (a) 4.8 kJ

Explanation:
1. The 2 kg piece moves with \( \dfrac{1\times 80}{2} = 40 \) m/s in the opposite direction.
2. \( K = \tfrac{1}{2}(1)(80)^2 + \tfrac{1}{2}(2)(40)^2 = 3200 + 1600 = 4800 \) J = 4.8 kJ.

Teacher's Note:
1. The total energy released equals the sum of the kinetic energies, because the bomb was at rest.

Answer: (a) 486 J

Explanation:
1. \( 12v = 18\times 6 \Rightarrow v = 9 \) m/s.
2. \( K = \tfrac{1}{2}(12)(81) = 486 \) J.

Teacher's Note:
1. Momentum conservation gives the speed of the lighter piece. Then use K = ½mv².

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

Explanation:
1. The gun's kinetic energy goes into the spring: \( \tfrac{1}{2}MV^2 = \tfrac{1}{2}kd^2 \Rightarrow V = d\sqrt{k/M} \).
2. Momentum: \( mv = MV \Rightarrow v = \dfrac{M}{m}d\sqrt{\dfrac{k}{M}} = \dfrac{d}{m}\sqrt{kM} \).

Teacher's Note:
1. Combine energy conservation for the spring with momentum conservation for the firing.

Answer: (d) 4 m/s

Explanation:
1. Momentum conservation: \( 10\times 2 = 5v \Rightarrow v = 4 \) m/s.

Teacher's Note:
1. Spring forces are internal, so the total momentum stays zero.

Answer: (c) South by 4 m

Explanation:
1. The centre of mass of the system stays at rest. Let the plank move x towards the south, then the man moves \( 6 - x \) towards the north relative to the ground.
2. \( 80(6 - x) = 40x \Rightarrow x = 4 \) m.

Teacher's Note:
1. The man and plank move in opposite directions. The lighter plank moves farther.

Answer: (c) \( \left(1 + \dfrac{m_2}{m_1}\right)h \)

Explanation:
1. The ball takes \( t = h/u \) to reach the floor. The man recoils upward with \( v = \dfrac{m_2u}{m_1} \).
2. He rises \( vt = \dfrac{m_2}{m_1}h \), so he is at \( h + \dfrac{m_2h}{m_1} = \left(1 + \dfrac{m_2}{m_1}\right)h \).

Teacher's Note:
1. There is no gravity, so both the ball and the man move with constant speed after the throw.

Answer: (a) \( 3V\cos\theta \)

Explanation:
1. At the highest point the shell moves horizontally with speed \( V\cos\theta \).
2. One half retraces its path, so its velocity is \( -V\cos\theta \). Momentum: \( 2m(V\cos\theta) = m(-V\cos\theta) + mv \).
3. \( v = 3V\cos\theta \).

Teacher's Note:
1. At the top, only the horizontal component of velocity is left. Apply momentum conservation in the horizontal direction.

Answer: (a) 3.2 m

Explanation:
1. Relative to the trolley the man walks 4 m. The man gets a fraction \( \dfrac{M_{trolley}}{M_{trolley} + m} = \dfrac{320}{400} \) of the relative displacement.
2. \( 4\times 0.8 = 3.2 \) m.

Teacher's Note:
1. The man moves forward and the trolley recoils backward, so the ground displacement is less than the relative displacement.

Answer: (d) 10.1 m

Explanation:
1. The stone reaches the floor after \( t = 10/2 = 5 \) s.
2. Man's recoil speed \( = \dfrac{0.5\times 2}{50} = 0.02 \) m/s upwards, so he rises 0.1 m.
3. Height \( = 10 + 0.1 = 10.1 \) m.

Teacher's Note:
1. Compute the recoil speed from momentum conservation, then multiply by the time the stone takes to land.

Answer: (a) 0.2 m

Explanation:
1. No external horizontal force acts, so the horizontal position of the centre of mass is fixed.
2. If the wedge moves x, the block moves \( 2.2 - x \) the other way: \( M(2.2 - x) = 10Mx \).
3. \( x = 0.2 \) m.

Teacher's Note:
1. Use the horizontal centre-of-mass argument; there is no need to find the accelerations.