JEE Physics System Of Particles and Rotational Motion MCQs Set 12

Answer: (b) \(\frac{12}{7}\ \text{m}\)

Explanation:
1. Mass of the rod \(M = \int_0^3 (2+x)\,dx = 6 + 4.5 = 10.5\).
2. \(\int_0^3 x(2+x)\,dx = 9 + 9 = 18\).
3. \(x_{cm} = \frac{18}{10.5} = \frac{12}{7}\ \text{m}\).

Teacher's Note:
1. For a variable density rod use \(x_{cm} = \frac{\int x\,dm}{\int dm}\) with \(dm = \lambda\,dx\).
2. Check that the answer lies between 0 and the length.

Answer: (d) \(95\%\)

Explanation:
1. Momentum conservation gives \(0.05\times 10 = 1\times v\), so \(v = 0.5\ \text{m s}^{-1}\).
2. Initial KE \(= \frac12(0.05)(100) = 2.5\ \text{J}\) and final KE \(= \frac12(1)(0.25) = 0.125\ \text{J}\).
3. The loss is \(\frac{2.5 - 0.125}{2.5} = 95\%\).

Teacher's Note:
1. In a perfectly inelastic collision the fractional loss is \(\frac{M}{m+M}\) when the target is at rest.
2. Here \(M = 950\) and \(m + M = 1000\).

Answer: (c) \(5\hat i + \frac{25}{7}\hat j\)

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

Teacher's Note:
1. Treat the vector components separately.
2. Masses can stay in grams if the same unit is used in the numerator and the denominator.

Answer: (a) \(300\)

Explanation:
1. Force equals the momentum given to the bullets per second, \(F = n\,m\,v\) with \(n\) bullets per second.
2. \(200 = n\times 0.04\times 1000\), so \(n = 5\) per second.
3. In one minute \(5\times 60 = 300\) bullets are fired.

Teacher's Note:
1. Recoil force is the rate of change of momentum of the bullets.
2. Convert grams to kilograms and seconds to minutes carefully.

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

Explanation:
1. Both balls have the same initial momentum \(mv\).
2. The metal ball stops, so its change is \(mv\), while the rubber ball reverses direction and its change is nearly \(2mv\).
3. So the change in momentum is more for the rubber ball.

Teacher's Note:
1. Rebounding means a larger change of momentum and so a larger impulse.
2. Initial momenta are equal here.

Answer: (b) \(5.2\ \text{m s}^{-1}\)

Explanation:
1. The balls are identical and move symmetrically on either side of the initial direction with equal speeds \(v\).
2. Momentum along the initial direction gives \(m\times 9 = 2mv\cos 30^\circ\).
3. So \(v = \frac{9}{\sqrt3} = 5.2\ \text{m s}^{-1}\).

Teacher's Note:
1. The components perpendicular to the initial direction cancel out.
2. Use only the component along the initial direction.

Answer: (c) \(\frac35\)

Explanation:
1. The speed just before and after the bounce is proportional to \(\sqrt h\).
2. \(\frac{v_2}{v_1} = \sqrt{\frac{1.8}{5}} = \frac35\), so the ball rebounds with \(\frac35\) of its speed.
3. The same factor is the coefficient of restitution.

Teacher's Note:
1. The fraction of velocity retained equals \(e\).
2. The value of \(g\) cancels out.

Answer: (d) \(6\ \text{m}\)

Explanation:
1. After the collision \(60\%\) of the energy remains.
2. The rebound height is proportional to the energy, so \(h = 0.6\times 10 = 6\ \text{m}\).

Teacher's Note:
1. Energy at the top of the rise is \(mgh\).
2. So height is directly proportional to energy for a given mass.

Answer: (b) \(12\ \text{m s}^{-1}\) upwards

Explanation:
1. The ball hits the roof with speed \(\sqrt{2gh} = 10\ \text{m s}^{-1}\) downwards.
2. In the frame of the lift the ball approaches at \(10 + 1 = 11\ \text{m s}^{-1}\) and leaves at \(11\ \text{m s}^{-1}\) upwards because \(e = 1\).
3. In the ground frame the ball moves up with \(11 + 1 = 12\ \text{m s}^{-1}\).

Teacher's Note:
1. Work in the frame of the heavy moving body.
2. Then add the velocity of that body back.

Answer: (c) \(12\ \text{m s}^{-1}\), \(0\)

Explanation:
1. Identical bodies in a head-on elastic collision exchange their velocities.
2. The second ball moves with \(12\ \text{m s}^{-1}\) and the first one stops.

Teacher's Note:
1. Equal masses and \(e = 1\) mean the velocities are exchanged.
2. This is seen in a Newton's cradle.

Answer: (c) \(1\ \text{m s}^{-1}\)

Explanation:
1. The momenta are \(30\ \text{kg m s}^{-1}\) and \(40\ \text{kg m s}^{-1}\) at right angles.
2. The resultant momentum is \(\sqrt{30^2 + 40^2} = 50\ \text{kg m s}^{-1}\).
3. The final velocity is \(\frac{50}{50} = 1\ \text{m s}^{-1}\).

Teacher's Note:
1. Add perpendicular momenta as vectors.
2. The mass of the couple is the sum of both masses.

Answer: (d) the linear momentum is conserved

Explanation:
1. In any collision with no external force the linear momentum is conserved.
2. In an inelastic collision the kinetic energy is not conserved.

Teacher's Note:
1. Momentum is conserved in every collision.
2. Kinetic energy is conserved only in elastic collisions.

Answer: (a) Sphere

Explanation:
1. The perpendicular axes theorem holds only for plane (two dimensional) laminar bodies.
2. A disc, a ring and a thin blade are planar, but a sphere is a three dimensional body.

Teacher's Note:
1. Parallel axes theorem works for any body.
2. Perpendicular axes theorem works only for flat bodies.

Answer: (a) \(5MR^2\)

Explanation:
1. About the dotted axis through the centre, perpendicular to the plane of the ring, \(I_{cm} = MR^2\).
2. AB is parallel to this axis at a distance \(d = 2R\).
3. By the parallel axes theorem \(I = MR^2 + M(2R)^2 = 5MR^2\).

Teacher's Note:
1. Use the parallel axes theorem when the axis does not pass through the centre of mass.
2. Add \(Md^2\) to the value about the parallel axis through the centre.

Answer: (b) \(\frac54 ml^2\)

Explanation:
1. The particle at A lies on the axis and contributes zero.
2. The particle at B is at a distance \(l\) and the particle at C is at a distance \(\frac l2\) from AX.
3. \(I = ml^2 + m\left(\frac l2\right)^2 = \frac54 ml^2\).

Teacher's Note:
1. Moment of inertia is the sum of \(mr^2\) with \(r\) measured perpendicular to the axis.
2. Find each perpendicular distance from the geometry.

Answer: (d) FH

Explanation:
1. Let \(AB = h\) and \(BC = 2h\).
2. For an axis through the centre along the length (FH) \(I = \frac{Mh^2}{12}\), and along the breadth (EG) \(I = \frac{M(2h)^2}{12} = \frac{Mh^2}{3}\).
3. Axes along the edges AB and BC give larger values, so FH gives the minimum.

Teacher's Note:
1. Mass lying closer to the axis gives a smaller moment of inertia.
2. An axis through the centre is smaller than a parallel axis along an edge.

Answer: (b) Mass of linear motion

Explanation:
1. Moment of inertia plays the role in rotation that mass plays in translation.
2. Compare \(\tau = I\alpha\) with \(F = ma\).

Teacher's Note:
1. Rotational analogues: torque is force, angular momentum is linear momentum.
2. Moment of inertia is mass.

Answer: (b) more angular acceleration will be produced in (b)

Explanation:
1. The torque is the same in both cases, \(F\times L\).
2. In (a) the heavier steel half is farther from the pivot, and in (b) it is nearer, so the moment of inertia in (b) is smaller.
3. \(\alpha = \tau/I\), so the angular acceleration is more in (b).

Teacher's Note:
1. For the same torque the smaller moment of inertia gives the larger angular acceleration.
2. Steel is denser than wood.

Answer: (a) \(0.6\pi\ \text{kg m}^2\text{/s}\)

Explanation:
1. \(\omega = 2\pi\times 0.5 = \pi\ \text{rad s}^{-1}\).
2. \(L = I\omega = 0.6\pi\ \text{kg m}^2\text{/s}\).

Teacher's Note:
1. Convert revolutions per second to rad per second by multiplying by \(2\pi\).
2. Then apply \(L = I\omega\).

Answer: (d) \(\frac{L^2}{mr^3}\)

Explanation:
1. \(L = mvr\), so \(v = \frac{L}{mr}\).
2. \(F = \frac{mv^2}{r} = \frac{m}{r}\cdot\frac{L^2}{m^2r^2} = \frac{L^2}{mr^3}\).

Teacher's Note:
1. Express the speed in terms of \(L\) first.
2. Then substitute in \(\frac{mv^2}{r}\).

Answer: (a) \(2\times 10^6\ \text{J}\)

Explanation:
1. \(\omega = \frac{2\pi}{60}\times\frac{3000}{\pi} = 100\ \text{rad s}^{-1}\).
2. \(KE = \frac12 I\omega^2 = \frac12\times 400\times 10^4 = 2\times 10^6\ \text{J}\).

Teacher's Note:
1. Change revolutions per minute to rad per second by multiplying by \(\frac{2\pi}{60}\).
2. The kinetic energy of rotation is \(\frac12 I\omega^2\).

Answer: (b) Sphere

Explanation:
1. For rolling without slipping \(a = \frac{g\sin\theta}{1 + k^2/R^2}\).
2. For a disc \(\frac{k^2}{R^2} = \frac12\) and for a solid sphere \(\frac25\), so the sphere has the larger acceleration.
3. The sphere reaches the bottom first, and mass does not matter.

Teacher's Note:
1. The smaller the ratio \(\frac{k^2}{R^2}\) the faster the body rolls down.
2. The order is sphere, disc, ring.

Answer: (a) \(\frac{ag}{h}\)

Explanation:
1. In the truck's frame a pseudo force \(m\alpha\) acts backwards at the centre of mass, at height \(\frac h2\).
2. The block is about to topple about its rear lower edge when the torques balance: \(m\alpha\cdot\frac h2 = mg\cdot\frac a2\).
3. So \(\alpha_{max} = \frac{ag}{h}\).

Teacher's Note:
1. Balance the torque of the pseudo force against the torque of the weight about the tipping edge.
2. A wider and shorter block is more stable.