JEE Physics Laws Of Motion MCQs Set 17

Answer: (d) 90°

Explanation:
1. \( \vec F=\frac{d\vec P}{dt}=-Ak\left[\hat i\sin kt+\hat j\cos kt\right] \).
2. \( \vec F\cdot\vec P=-A^2k(\sin kt\cos kt-\cos kt\sin kt)=0 \), so the angle is 90°.

Teacher's Note:
1. Force is the time derivative of momentum.
2. A zero dot product means perpendicular vectors; here the speed stays constant.

Answer: (a) 4.9 m/s²

Explanation:
1. The package just lifts when the rope tension equals its weight: \( T=15g \).
2. For the monkey climbing with upward acceleration a: \( T-10g=10a\Rightarrow a=\frac{15g-10g}{10}=\frac g2=4.9\ \text{m/s}^2 \).

Teacher's Note:
1. Lifting off the ground requires tension equal to the package's weight.
2. The monkey's acceleration is measured relative to the ground.

Answer: (d) 39 m/s

Explanation:
1. Impulse = area under the force-time graph \( =75\times5+50\times5=625\ \text{N s} \).
2. \( v=\frac{625}{16}\approx39\ \text{m/s} \).

Teacher's Note:
1. Impulse equals change in momentum.
2. The body starts from rest, so the final momentum equals the impulse.

Answer: (a) \( \frac g4 \) m/s²

Explanation:
1. Sliding down with acceleration a: \( mg-T=ma \).
2. The tension cannot exceed \( \frac34mg \), so \( a\ge g-\frac{3g}{4}=\frac g4 \).

Teacher's Note:
1. The smaller the acceleration, the larger the tension in the rope.
2. The rope limits how slowly he may slide.

Answer: (c) 40 N

Explanation:
1. For masses \( m_1=2 \) kg and \( m_2 \): \( T=\frac{2m_1m_2g}{m_1+m_2} \).
2. For a very heavy \( m_2 \) this tends to \( 2m_1g=40\ \text{N} \).

Teacher's Note:
1. A very heavy mass falls freely, so the light mass is accelerated at about g upward.
2. Then T = m1(g + g) = 2m1g.

Answer: (a) 6.0 N

Explanation:
1. Normal reaction \( N=25-9=16\ \text{N} \), so limiting friction \( =0.4\times16=6.4\ \text{N} \).
2. The applied force 6.0 N is less than 6.4 N, so the block stays at rest and static friction equals 6.0 N.

Teacher's Note:
1. Compare the applied force with the limiting friction first.
2. Static friction matches the applied force until the limit is reached.

Answer: (a) 2.0 kg

Explanation:
1. The block is at rest (limiting friction would be much larger), so friction equals the component of weight along the plane: \( f=mg\sin30^\circ \).
2. \( 10=m\times10\times\frac12\Rightarrow m=2\ \text{kg} \).

Teacher's Note:
1. For a block at rest, use equilibrium, not f = μN.
2. tan 30° is less than μ, so the block does not slide.

Answer: (c) 40 N opposite to the force

Explanation:
1. Common acceleration \( a=\frac{60}{15}=4\ \text{m/s}^2 \).
2. For the 5 kg block: \( 60-f=5\times4\Rightarrow f=40\ \text{N} \), acting backwards (it accelerates the 10 kg block).

Teacher's Note:
1. The lower block is accelerated only by friction from the upper one.
2. Check: 40 N on 10 kg gives 4 m/s².

Answer: (b) 20 N along the applied force

Explanation:
1. Common acceleration \( a=\frac{60}{15}=4\ \text{m/s}^2 \).
2. Friction on the upper block \( =5\times4=20\ \text{N} \) along the force; the limit is \( 0.5\times50=25\ \text{N} \), so no slipping.

Teacher's Note:
1. The upper block is pulled forward by friction.
2. Always check the required friction against the limiting value.

Answer: (a) 2.5 m/s²

Explanation:
1. Force on the block = rate of change of momentum of the water \( =\frac{dm}{dt}v=1\times5=5\ \text{N} \).
2. \( a=\frac{5}{2}=2.5\ \text{m/s}^2 \).

Teacher's Note:
1. The jet loses momentum at the rate (dm/dt)v, which is the force on the block.
2. Use the mass of the block alone for the initial acceleration.

Answer: (b) 0.20

Explanation:
1. \( a=\frac{8}{4}=2\ \text{m/s}^2 \).
2. \( \mu g=a\Rightarrow\mu=\frac{2}{10}=0.2 \).

Teacher's Note:
1. Friction is the only horizontal force, so a = μg.
2. The mass cancels.

Answer: (c) 0.46 m/s²

Explanation:
1. Normal reaction \( N=mg+20\sin30^\circ=50+10=60\ \text{N} \), so friction \( =0.25\times60=15\ \text{N} \).
2. \( a=\frac{20\cos30^\circ-15}{5}=\frac{17.32-15}{5}=0.46\ \text{m/s}^2 \).

Teacher's Note:
1. Pushing downward increases the normal reaction.
2. Only the horizontal component of the push drives the motion.

Answer: (c) 0.9 N s

Explanation:
1. The force vanishes at \( t=\frac{600}{2\times10^5}=3\times10^{-3}\ \text{s} \).
2. Impulse = area of the triangle \( =\frac12\times600\times3\times10^{-3}=0.9\ \text{N s} \).

Teacher's Note:
1. Impulse is the integral of force over time.
2. The force decreases linearly, so the area is a triangle.

Answer: (c) 49 N

Explanation:
1. Limiting friction \( =0.5\times100=50\ \text{N} \), larger than the weight \( 5\times9.8=49\ \text{N} \), so the block stays at rest.
2. Static friction equals the weight: 49 N.

Teacher's Note:
1. The block is in equilibrium, so friction balances its weight.
2. Do not use μN unless the block is about to slip.

Answer: (a) 4 N

Explanation:
1. Total friction \( =0.1\times10\times10=10\ \text{N} \), so \( a=\frac{20-10}{10}=1\ \text{m/s}^2 \).
2. For the 2 kg block: \( F-0.1\times2\times10=2\times1\Rightarrow F=4\ \text{N} \).

Teacher's Note:
1. Find the common acceleration, then isolate the last block.
2. Friction on the 2 kg block alone is 2 N.

Answer: (b) \( \frac{g}{\mu} \)

Explanation:
1. The normal reaction from the cart is \( N=ma \) and the friction must balance the weight: \( \mu N\ge mg \).
2. \( \mu ma\ge mg\Rightarrow a\ge\frac g\mu \).

Teacher's Note:
1. The block is pressed by the cart's acceleration, not by an applied force.
2. A smaller μ needs a larger acceleration.

Answer: (d) \( \frac{\mu g}{2b} \)

Explanation:
1. Plate acceleration \( a=\frac{dv}{dt}=2bt \).
2. The block slides when the required friction exceeds the limit: \( ma=\mu mg\Rightarrow 2bt=\mu g\Rightarrow t=\frac{\mu g}{2b} \).

Teacher's Note:
1. Friction accelerates the block with the plate until it reaches μmg.
2. The acceleration grows linearly with time.

Answer: (b) \( \frac{m(M+m)g}{\mu M} \)

Explanation:
1. Common acceleration \( a=\frac{F}{M+m} \).
2. Normal force between the blocks \( N=Ma=\frac{MF}{M+m} \).
3. Friction must support the weight: \( \mu N\ge mg\Rightarrow F\ge\frac{m(M+m)g}{\mu M} \).

Teacher's Note:
1. The normal reaction on m is produced by the acceleration of M.
2. The minimum F makes the friction exactly equal mg.

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

Explanation:
1. The surface exerts normal reaction \( N=Mg \) and friction f with \( 0\le f\le\mu Mg \).
2. \( F=\sqrt{N^2+f^2} \), so \( Mg\le F\le Mg\sqrt{1+\mu^2} \).

Teacher's Note:
1. The contact force is the vector sum of normal reaction and friction.
2. Its smallest value is when there is no friction.

Answer: (d) \( \rho\left[\frac{V}{v_2}\right](v_1+v_2)^2 \)

Explanation:
1. Cross-section of the jet \( A=\frac{V}{v_2} \). Relative speed of water with respect to the plate \( =v_1+v_2 \).
2. Mass hitting the plate per second \( =\rho A(v_1+v_2) \), and each kilogram loses momentum \( (v_1+v_2) \).
3. \( F=\rho\frac{V}{v_2}(v_1+v_2)^2 \).

Teacher's Note:
1. For a moving plate use the relative velocity of the jet.
2. Force = (mass per second) × (change of velocity).

Answer: (d) \( \frac{W}{2\sqrt3} \)

Explanation:
1. The centres form an equilateral triangle, so each contact force N from the top bar makes 30° with the vertical.
2. Vertical balance of the top bar: \( 2N\cos30^\circ=W\Rightarrow N=\frac{W}{\sqrt3} \).
3. Horizontal component pushing each lower bar apart: \( N\sin30^\circ=\frac{W}{2\sqrt3} \).

Teacher's Note:
1. Draw the triangle of centres; all sides equal 2r.
2. Each lower bar needs a horizontal force equal to the outward component.

Answer: (b) 2 and 3

Explanation:
1. Normal force greater than weight means the acceleration is upward.
2. Going up and speeding up has upward acceleration (2); going down and slowing down also has upward acceleration (3).
3. Cases 1 and 4 have downward acceleration.

Teacher's Note:
1. Check the direction of acceleration, not of velocity.
2. Slowing down while moving down means acceleration opposite to velocity, i.e. upward.

Answer: (b) 1, 3 and 4

Explanation:
1. Limiting friction is \( \mu N \), where the coefficient depends on the nature of the surfaces and on any foreign matter (dust, oil) between them, and N is the normal reaction.
2. It is independent of the apparent area of contact.

Teacher's Note:
1. Friction does not depend on area in the simple laws.
2. Lubricants and dirt change μ.

Answer: (b) 1 and 3

Explanation:
1. A body at rest has zero acceleration, so the net force is zero.
2. That happens if no force acts (1) or if several forces cancel (3). Mass alone does not decide whether a body is at rest (2).

Teacher's Note:
1. Rest requires zero net force, not zero force.
2. A massive body will move if a net force acts on it.

Answer: (b) 1, 2 and 4 only

Explanation:
1. The contact force is the vector sum of two perpendicular forces: \( F=\sqrt{F_N^2+f^2} \).
2. So F is larger than each of FN and f (1, 2) and lies between their difference and sum (4).
3. FN need not exceed f; for example when f is larger than the normal force, (3) fails.

Teacher's Note:
1. F is the hypotenuse of a right-angled triangle with sides FN and f.
2. The triangle inequality gives statement 4.