JEE Physics Laws Of Motion MCQs Set 11

Answer: (b) \( \dfrac{v_0^2}{4g\sin\theta} \)

Explanation:
1. Constant velocity downward gives \( \mu = \tan\theta \).
2. Going up, the retardation is \( g\sin\theta + \mu g\cos\theta = 2g\sin\theta \).
3. Distance \( s = \dfrac{v_0^2}{2 \times 2g\sin\theta} = \dfrac{v_0^2}{4g\sin\theta} \).

Teacher's Note:
1. Find μ from the sliding condition first.
2. Up the plane, gravity and friction both slow the block.

Answer: (a) \( 1 - \dfrac{1}{n^2} \)

Explanation:
1. For the same distance, \( s = \tfrac12 a t^2 \), so \( t \propto 1/\sqrt{a} \).
2. Smooth incline: \( a_1 = g\sin45° \). Rough incline: \( a_2 = g\sin45°(1 - \mu) \) at 45°.
3. \( n^2 = a_1/a_2 = 1/(1 - \mu) \), so \( \mu = 1 - 1/n^2 \).

Teacher's Note:
1. At 45 degrees sin and cos are equal, which simplifies the algebra.
2. Equal distance means time is inversely proportional to the square root of acceleration.

Answer: (c) less than the angle of repose

Explanation:
1. The angle between the normal reaction and the vertical equals the angle of inclination θ of the plane.
2. The body does not slide by itself, so θ is less than the angle of repose.
3. It must be pushed to move.

Teacher's Note:
1. The normal reaction is perpendicular to the plane, so it is tilted by θ from the vertical.
2. The angle of repose equals the angle of friction.

Answer: (d) all three are possible

Explanation:
1. The angle at which sliding begins satisfies \( \tan\theta = \mu \).
2. It depends only on the coefficient of friction, not on the mass.
3. So the masses can be greater, smaller or equal; all three are possible.

Teacher's Note:
1. The angle of repose is independent of mass.
2. Different angles only tell us the coefficients of friction differ.

Answer: (a) c, d

Explanation:
1. For small F the block is at rest and \( f = F \), a line of slope 45°.
2. Beyond the limiting value, the friction drops slightly to the kinetic value \( \mu_k N \) and stays constant, so the graph is parallel to the F axis.
3. The drop from \( \mu_s N \) to \( \mu_k N \) is the small kink, so (c) and (d) are correct.

Teacher's Note:
1. Static friction grows with F up to the limit and then falls to the kinetic value.
2. That is the reason for the kink.

Answer: (d) 0.75

Explanation:
1. \( \sin\theta = 3/5 \), so \( \cos\theta = 4/5 \) and \( \tan\theta = 3/4 \).
2. At the point of sliding \( \mu = \tan\theta = 0.75 \).

Teacher's Note:
1. Slope 3 in 5 means the rise is 3 for a length of 5 along the plane.
2. The mass is not needed.

Answer: (d) \( \dfrac{P + Q\sin\theta}{mg - Q\cos\theta} \)

Explanation:
1. Q points upward, so the normal reaction is \( N = mg - Q\cos\theta \).
2. The horizontal driving force is \( P + Q\sin\theta \).
3. For equilibrium \( P + Q\sin\theta \le \mu N \), so the required coefficient is \( \mu \ge \dfrac{P + Q\sin\theta}{mg - Q\cos\theta} \).

Teacher's Note:
1. The angle is measured from the vertical, so the horizontal component is Q sin θ.
2. An upward component of Q lowers the normal reaction.

Answer: (c) 200 N

Explanation:
1. At the knot the horizontal string has tension \( T_1 = 346 \) N at most (equal to the friction).
2. For the inclined string, \( T\cos30° = T_1 \) and \( T\sin30° = W \).
3. \( W = T_1\tan30° = 346/\sqrt{3} \approx 200 \) N.

Teacher's Note:
1. Resolve the forces at the knot into horizontal and vertical parts.
2. The horizontal string tension equals the friction on the body.

Answer: (c) 28 N

Explanation:
1. Limiting friction between A and B is \( 0.5 \times 80 = 40 \) N, and the friction from the table on B is \( 0.2 \times 100 = 20 \) N.
2. If they move together, \( a = (60 - 20)/10 = 4 \) m/s².
3. For A: \( 60 - f = 8 \times 4 \), so \( f = 28 \) N, which is less than 40 N, so there is no slipping.

Teacher's Note:
1. First test whether the blocks move together.
2. Friction between A and B must stay below its limiting value.

Answer: (a) 30 N

Explanation:
1. The maximum acceleration of A (given by friction) is \( \mu g = 0.3 \times 10 = 3 \) m/s².
2. For no relative motion the whole system can have at most this acceleration.
3. \( F_{max} = (4 + 6) \times 3 = 30 \) N.

Teacher's Note:
1. The upper block is accelerated only by friction.
2. The floor is smooth, so F accelerates the total mass.

Answer: (b) 24 N

Explanation:
1. The component of velocity along the wall is unchanged.
2. The component normal to the wall is \( 12\sin30° = 6 \) m/s and it reverses, so \( \Delta p = 2 \times 0.5 \times 6 = 6 \) kg m/s.
3. \( F = 6/0.25 = 24 \) N.

Teacher's Note:
1. The angle is measured with the wall, so the normal component uses sin 30 degrees.
2. Only the normal component changes.

Answer: (a) \( 10\sqrt{2} \) kg

Explanation:
1. \( |\vec{F}| = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \) N.
2. \( m = F/a = 10\sqrt{2}/1 = 10\sqrt{2} \) kg.

Teacher's Note:
1. Use the magnitude of the force vector.
2. Then apply F = ma.

Answer: (d) \( \alpha \ge \dfrac{g}{\mu} \)

Explanation:
1. The wall pushes the block with normal force \( N = m\alpha \).
2. Friction must support the weight: \( \mu N \ge mg \), so \( \mu m\alpha \ge mg \).
3. Hence \( \alpha \ge g/\mu \).

Teacher's Note:
1. The normal reaction comes from the acceleration of the cart.
2. Friction balances the weight of the block.

Answer: (c) 11000 N

Explanation:
1. Total mass \( = 60 + 940 = 1000 \) kg.
2. \( T - Mg = Ma \).
3. \( T = 1000 \times (10 + 1) = 11000 \) N.

Teacher's Note:
1. The person's push on the button is an internal force.
2. Treat the lift and person as one system.

Answer: (c) 2 MV

Explanation:
1. Take the initial direction as positive.
2. Impulse \( = \Delta p = (-MV) - (MV) = -2MV \).
3. The magnitude is 2MV.

Teacher's Note:
1. Impulse equals change in momentum.
2. A rebound at the same speed doubles the change.

Answer: (b) zero

Explanation:
1. The blocks move with constant speed, so the acceleration of each block is zero.
2. By Newton's second law the net force on each block, including the block 2m, is zero.

Teacher's Note:
1. Constant velocity means zero net force on every part of the system.
2. Tensions balance the weights.

Answer: (c) \( \dfrac{g(1 - 2\mu)}{3} \)

Explanation:
1. Driving force \( = m_1 g = mg \).
2. Friction on the two table blocks \( = \mu(m_2 + m_3)g = 2\mu mg \).
3. \( a = \dfrac{mg - 2\mu mg}{3m} = \dfrac{g(1 - 2\mu)}{3} \).

Teacher's Note:
1. Both table blocks feel friction.
2. The total mass moving is 3m.

Answer: (a) \( \dfrac{2ma}{g + a} \)

Explanation:
1. Let the buoyant force be B. Descending: \( mg - B = ma \), so \( B = m(g - a) \).
2. After removing mass Δm, the new mass is \( m - \Delta m \) and it rises: \( B - (m - \Delta m)g = (m - \Delta m)a \).
3. So \( m(g - a) = (m - \Delta m)(g + a) \), which gives \( \Delta m = \dfrac{2ma}{g + a} \).

Teacher's Note:
1. The buoyant force stays constant because the volume does not change.
2. Write the second law for both cases.

Answer: (b) \( \dfrac{m_1 m_2 (1 + \mu_k)g}{(m_1 + m_2)} \)

Explanation:
1. For the system: \( a = \dfrac{m_2 g - \mu_k m_1 g}{m_1 + m_2} \).
2. For the hanging block: \( T = m_2(g - a) \).
3. \( T = m_2 g\left[1 - \dfrac{m_2 - \mu_k m_1}{m_1 + m_2}\right] = \dfrac{m_1 m_2 (1 + \mu_k)g}{m_1 + m_2} \).

Teacher's Note:
1. Find the common acceleration first.
2. The tension then follows from the hanging block.

Answer: (a) 6 N

Explanation:
1. Common acceleration \( a = 14/(4 + 2 + 1) = 2 \) m/s².
2. The contact force between A and B accelerates B and C together: \( N = (2 + 1) \times 2 = 6 \) N.

Teacher's Note:
1. The contact force accelerates all blocks beyond the contact.
2. First find the acceleration of the whole system.

Answer: (c) 0.6 and 0.5

Explanation:
1. The box just starts to slip at 30°, so \( \mu_s = \tan30° \approx 0.58 \approx 0.6 \).
2. From \( s = \tfrac12 a t^2 \): \( 4 = \tfrac12 a (16) \), so \( a = 0.5 \) m/s².
3. \( a = g(\sin30° - \mu_k\cos30°) \), so \( 0.5 = 5 - 8.66\mu_k \) and \( \mu_k \approx 0.52 \approx 0.5 \).

Teacher's Note:
1. The kinetic coefficient is always smaller than the static coefficient.
2. Use the distance and time to find the acceleration.

Answer: (d) \( a = g\tan\theta \)

Explanation:
1. In the frame of the wedge the forces on the block are mg down, the pseudo force ma horizontal and the normal reaction N.
2. Along the plane: \( mg\sin\theta = ma\cos\theta \).
3. So \( a = g\tan\theta \).

Teacher's Note:
1. Balance forces along the incline in the wedge's frame.
2. The result is the same as for a block at rest on an accelerating wedge.

Answer: (b) to the left, and the angle of inclination of the pendulum with the vertical is \( \tan^{-1}\!\left(\dfrac{a}{g}\right) \)

Explanation:
1. In the truck's frame a pseudo force ma acts to the left.
2. The bob swings back (to the left) until \( T\sin\theta = ma \) and \( T\cos\theta = mg \).
3. So \( \tan\theta = a/g \).

Teacher's Note:
1. The bob lags behind the acceleration of the truck.
2. Pseudo force is opposite to the acceleration.

Answer: (c) \( |\vec{F}| \le mg\sqrt{1 + \mu^2} \)

Explanation:
1. The normal reaction is \( N = mg \), and the static friction is \( f \le \mu mg \).
2. The resultant of two perpendicular forces is \( \sqrt{N^2 + f^2} \).
3. So \( |\vec{F}| \le mg\sqrt{1 + \mu^2} \).

Teacher's Note:
1. Normal reaction and friction are perpendicular, so add them as vectors.
2. Static friction is at most its limiting value.