JEE Physics Laws Of Motion MCQs Set 14

Answer: (a) the lift begins to go up

Explanation:
1. Apparent weight \( N=m(g+a) \) for an upward acceleration a.
2. When the lift starts going up it accelerates upward, so N > mg.
3. In the other cases the acceleration is zero or directed downward, so N is equal to or less than mg.

Teacher's Note:
1. Apparent weight increases whenever the acceleration of the lift is upward.
2. Slowing down while moving up means the acceleration is downward.

Answer: (a) 6 N

Explanation:
1. Limiting friction: \( \mu mg=0.3\times3\times10=9\ \text{N} \).
2. At t = 3 s the applied force is F = 2 × 3 = 6 N, which is less than 9 N, so the block has not started moving.
3. Static friction balances the applied force, so it equals 6 N.

Teacher's Note:
1. Compare the applied force with the limiting friction before deciding between static and kinetic friction.
2. Static friction equals the applied force until it reaches μN.

Answer: (a) 5 kg

Explanation:
1. Initial momentum is zero, so the momentum of the third part equals the resultant of the other two (equal and opposite).
2. Momenta: \( p_1=1\times12=12 \) and \( p_2=2\times8=16 \) at right angles.
3. \( p_3=\sqrt{12^2+16^2}=20\ \text{kg m/s} \).
4. \( m_3=\frac{20}{4}=5\ \text{kg} \)

Teacher's Note:
1. Momentum is conserved in an explosion (no external force).
2. Add the momenta of the two known parts as perpendicular vectors.

Answer: (d) 13 N

Explanation:
1. Normal reaction N = 12 N (horizontal).
2. Friction needed to hold the block: f = weight = 5 N, which is less than \( \mu N=0.6\times12=7.2\ \text{N} \), so the block stays at rest.
3. Net force by the wall = resultant of N and f: \( \sqrt{12^2+5^2}=13\ \text{N} \)

Teacher's Note:
1. The wall exerts both a normal force and a friction force; the total force is their vector sum.
2. Friction is static and equals the weight, not μN.

Answer: (d) 36

Explanation:
1. Let the string make angle θ with the vertical. Horizontal: \( T\sin\theta=m\omega^2(L\sin\theta) \), which gives \( T=m\omega^2L \).
2. \( \omega^2=\frac{T}{mL}=\frac{324}{0.5\times0.5}=1296 \), so \( \omega=36\ \text{rad/s} \)

Teacher's Note:
1. The radius of the circle is L sinθ, so sinθ cancels from both sides.
2. The tension is largest at the maximum angular speed.

Answer: (c) \( \tan\theta \)

Explanation:
1. At the point of sliding, the component of weight down the plane equals the limiting friction: \( mg\sin\theta=\mu mg\cos\theta \).
2. \( \mu=\tan\theta \)

Teacher's Note:
1. The angle of repose θ satisfies tanθ = μ.
2. The mass cancels.

Answer: (d) 12 m/s²

Explanation:
1. The only horizontal force on the 0.5 kg block is the 6 N push of the 1 kg block.
2. \( a=\frac{6}{0.5}=12\ \text{m/s}^2 \).
3. Both blocks move together with the same acceleration.

Teacher's Note:
1. Apply Newton's second law to the block for which the net force is known.
2. Contact forces on the two blocks are equal and opposite (Newton's third law).

Answer: (d) 40 N

Explanation:
1. Horizontal components: \( 20\sqrt3\cos60^\circ=10\sqrt3 \) and \( 20\cos30^\circ=10\sqrt3 \) balance each other.
2. Vertical components upward: \( 20\sqrt3\sin60^\circ=30\ \text{N} \) and \( 20\sin30^\circ=10\ \text{N} \).
3. For equilibrium, \( F=30+10=40\ \text{N} \)

Teacher's Note:
1. In equilibrium, the sum of forces along each axis is zero.
2. The balance of horizontal components is a useful check on the angles.

Answer: (b) \( \frac{16}{\sqrt2}\ \text{m/s}^2 \)

Explanation:
1. Moving up, both the component of weight and friction act down the slope.
2. \( a=g\sin45^\circ+\mu g\cos45^\circ=\frac{10}{\sqrt2}(1+0.6)=\frac{16}{\sqrt2}\ \text{m/s}^2 \)

Teacher's Note:
1. Friction opposes the motion, so it acts down the incline while the block moves up.
2. On the way down the acceleration would be smaller: g(sinθ − μ cosθ).

Answer: (d) \( \frac32g \)

Explanation:
1. The rope is massless, so the tension equals the pull: \( T=F=\frac52mg \).
2. For the block: \( T-mg=ma\Rightarrow\frac52mg-mg=ma \).
3. \( a=\frac32g \)

Teacher's Note:
1. A massless rope over a frictionless pulley transmits the same tension everywhere.
2. Here the pull is a given force, not a weight, so the acceleration is not found by dividing by the total mass.

Answer: (c) \( mg \)

Explanation:
1. The block is in equilibrium under its weight mg and the force from the incline.
2. So the total contact force from the incline (the resultant of N = mg cosθ and friction f = mg sinθ) must balance the weight: \( \sqrt{(mg\cos\theta)^2+(mg\sin\theta)^2}=mg \).

Teacher's Note:
1. The question asks for the resultant of the normal reaction and the static friction, not the normal reaction alone.
2. It acts vertically upward and equals the weight.

Answer: (d) 1 m/s²

Explanation:
1. B is accelerated only by friction from A, whose maximum value is \( \mu_1m_Bg \).
2. \( m_Ba_{max}=\mu_1m_Bg\Rightarrow a_{max}=\mu_1g=0.1\times10=1\ \text{m/s}^2 \)

Teacher's Note:
1. The maximum acceleration of the upper block is μg, independent of the masses.
2. Beyond this value B starts slipping backwards on A.

Answer: (a) \( 30^\circ \)

Explanation:
1. For a pull F at angle θ: \( F\cos\theta=\mu(mg-F\sin\theta)\Rightarrow F=\frac{\mu mg}{\cos\theta+\mu\sin\theta} \).
2. F is minimum when the denominator is maximum, which happens when \( \tan\theta=\mu \).
3. \( \tan\theta=\frac1{\sqrt3}\Rightarrow\theta=30^\circ \)

Teacher's Note:
1. The best angle equals the angle of friction, tan⁻¹ μ.
2. The denominator cosθ + μ sinθ has maximum value √(1 + μ²).

Answer: (b) \( \frac1{\sqrt\mu} \)

Explanation:
1. The greatest acceleration friction can provide is \( a=\mu g \).
2. \( x=\frac12\mu gt^2\Rightarrow t=\sqrt{\frac{2x}{\mu g}}\propto\frac{1}{\sqrt\mu} \)

Teacher's Note:
1. Friction is the only horizontal force driving the car, so its maximum is μmg.
2. Time is minimum at maximum acceleration.

Answer: (b) \( 2\sqrt2\ \text{s} \)

Explanation:
1. Starting from rest with constant acceleration: \( s=\frac12at^2 \), so \( t\propto\sqrt s \).
2. \( t'=4\times\frac{1}{\sqrt2}=2\sqrt2\ \text{s} \)

Teacher's Note:
1. Time is proportional to the square root of the distance when starting from rest.
2. Half the distance takes 1/√2 of the total time.

Answer: (c) 20 N s

Explanation:
1. \( F=ma=2\times5=10\ \text{N} \).
2. \( J=Ft=10\times2=20\ \text{N s} \)

Teacher's Note:
1. Impulse is force multiplied by time, and also equals the change in momentum.
2. Check: Δp = m a t = 2 × 5 × 2 = 20 kg m/s.

Answer: (c) Newton's third law of motion

Explanation:
1. The swimmer pushes the water backward (action).
2. The water pushes the swimmer forward with an equal and opposite force (reaction), which is Newton's third law.

Teacher's Note:
1. Rockets and walking work on the same principle.
2. Action and reaction act on different bodies.

Answer: (b) 72 kg-wt

Explanation:
1. \( N=m(g+a)=m\left(g+\frac g5\right)=\frac65mg \).
2. \( \text{Apparent weight}=\frac65\times60=72\ \text{kg-wt} \)

Teacher's Note:
1. For an upward acceleration the apparent weight exceeds the true weight.
2. 1 kg-wt = g newton.

Answer: (a) 13 N

Explanation:
1. Whole system: \( a=\frac{(m_B+m_C)-m_A}{m_A+m_B+m_C}g=\frac{4-2}{6}g=\frac g3 \) (B and C move down, A moves up).
2. For C: \( m_Cg-T=m_Ca\Rightarrow T=2\left(9.8-\frac{9.8}3\right)=13.07\ \text{N} \)

Teacher's Note:
1. Find the common acceleration of the system first.
2. Apply Newton's law to C alone to get the tension in the BC string.

Answer: (a) \( \frac76\ \text{m/s}^2 \)

Explanation:
1. Net force: \( \vec F=3\hat i+6\hat j+2\hat k \).
2. \( |\vec F|=\sqrt{9+36+4}=7\ \text{N} \).
3. \( a=\frac{7}{6}\ \text{m/s}^2 \)

Teacher's Note:
1. Add the forces component-wise before finding the magnitude.
2. 3-6-2 gives the neat magnitude 7.

Answer: (c) be at rest relative to the plane

Explanation:
1. In the frame of the incline, the pseudo force \( ma \) acts on the block horizontally to the right (opposite to the acceleration).
2. Along the incline: component of weight down the slope \( mg\sin45^\circ \) and component of the pseudo force up the slope \( ma\cos45^\circ \).
3. With a = g these are equal, so the block stays at rest on the plane.

Teacher's Note:
1. Equilibrium of the block on a moving wedge requires a = g tanθ.
2. For θ = 45°, tanθ = 1 so a = g.

Answer: (b) \( b \)

Explanation:
1. \( F=\frac{dp}{dt}=b+2ct \).
2. At t = 0: F = b.

Teacher's Note:
1. Newton's second law in terms of momentum: F = dp/dt.
2. Constant a does not contribute to the force.