JEE Physics Motion In A Plane MCQs Set 18

Answer: (b) 5.6 m/s²

Explanation:
1. In the frame of the wedge, the pseudo force \( ma \) acts horizontally on the cube, along with weight \( mg \) and the normal reaction.
2. For equilibrium along the incline of angle 30°: \( ma\cos30^\circ=mg\sin30^\circ \).
3. \( a=g\tan30^\circ=9.8\times0.577\approx5.6\ \text{m/s}^2 \).

Teacher's Note:
1. The cube stays at rest on the incline only if the wedge's acceleration balances the component of gravity along the face.
2. The masses cancel, so the data M = 9 kg and m = 1 kg are not needed.

Answer: (b) 240 N

Explanation:
1. Treat both bodies as one system of mass 20 kg.
2. Newton's second law: \( T_1-(20)(9.8)=20\times2.2 \).
3. \( T_1=20(9.8+2.2)=240\ \text{N} \).

Teacher's Note:
1. The upper string supports the weight of both bodies and also gives them the acceleration.
2. Use the combined mass when the tension acts on the whole system.

Answer: (b) 90 N

Explanation:
1. Whole system (16 kg): \( a=\frac{180-160}{16}=1.25\ \text{m/s}^2 \).
2. Below P hang the rope (3 kg) and the 5 kg block, total 8 kg.
3. Tension at P: \( T=8(g+a)=8(11.25)=90\ \text{N} \).

Teacher's Note:
1. First find the common acceleration of the whole chain.
2. The tension at any point supports everything hanging below it.

Answer: (a) 5 m/s²

Explanation:
1. The rate of change of speed is the tangential acceleration.
2. Along the arc, the tangential component of gravity is \( mg\sin\theta \).
3. \( a_t=g\sin30^\circ=10\times0.5=5\ \text{m/s}^2 \).

Teacher's Note:
1. Tension acts along the string and changes only the direction of velocity.
2. Only the component of weight along the path changes the speed.

Answer: (d) 4 m/s

Explanation:
1. The angle subtended at the centre is twice the angle subtended at a point on the circumference.
2. So the angular velocity about the centre is \( 2\omega=4\ \text{rad/s} \).
3. \( v=R(2\omega)=1\times4=4\ \text{m/s} \).

Teacher's Note:
1. Use the inscribed angle theorem: central angle = 2 × inscribed angle.
2. Linear speed is radius times angular velocity about the centre.

Answer: (c) 1 m/s

Explanation:
1. Quarter circle: \( \theta=\frac\pi2=\frac12\alpha t^2=\frac12\cdot\frac\pi4t^2\Rightarrow t=2\ \text{s} \).
2. Displacement is the chord: \( \sqrt2R=\sqrt2\times\sqrt2=2\ \text{m} \).
3. Average velocity \( =\frac{2}{2}=1\ \text{m/s} \).

Teacher's Note:
1. Average velocity uses displacement (the chord), not the arc length.
2. Find the time from the angular displacement formula.

Answer: (a) distance travelled by the particle is 4 m

Explanation:
1. Distance \( =\int_0^2 2t\,dt=4\ \text{m} \) (a is correct).
2. Average speed = 4/2 = 2 m/s, so (c) is wrong.
3. The arc 4 m subtends 4 rad at the centre (R = 1 m), so the chord is \( 2\sin2 \) m, not 4 sin 2, and since 4 rad is not a full circle the average velocity is not zero. So (b) and (d) are wrong.

Teacher's Note:
1. Distance is the integral of speed; displacement is the chord for the angle turned.
2. Average velocity is zero only for a whole number of complete circles.

Answer: (b) \( 2\tan^{-1}\left(\frac12\right) \)

Explanation:
1. At the extreme position the speed is zero, so the acceleration is tangential: \( g\sin\theta \).
2. At the lowest point the acceleration is centripetal: \( \frac{v^2}{L}=2g(1-\cos\theta) \).
3. Equate: \( \sin\theta=2(1-\cos\theta)\Rightarrow2\sin\tfrac\theta2\cos\tfrac\theta2=4\sin^2\tfrac\theta2\Rightarrow\tan\tfrac\theta2=\tfrac12 \).
4. So \( \theta=2\tan^{-1}\left(\frac12\right) \).

Teacher's Note:
1. Energy conservation gives the lowest-point speed.
2. Use half-angle formulas to simplify.

Answer: (a) 120°

Explanation:
1. Thread goes slack when T = 0. With θ measured from the lowest position (θ greater than 90°): \( mg\cos(180^\circ-\theta)=\frac{mv^2}{L}\Rightarrow v^2=-gL\cos\theta \).
2. Energy: \( v^2=u^2-2gL(1-\cos\theta)=\frac72gL-2gL+2gL\cos\theta \).
3. Equate: \( \frac32gL+2gL\cos\theta=-gL\cos\theta\Rightarrow\cos\theta=-\frac12 \), so \( \theta=120^\circ \).

Teacher's Note:
1. At slackening the tension is zero and gravity alone provides the centripetal force.
2. Combine this with energy conservation.

Answer: (d) \( \frac{u^2\cos^2\theta}{g} \)

Explanation:
1. At the highest point the velocity is horizontal: \( v=u\cos\theta \).
2. Gravity is perpendicular to it, so it supplies the centripetal acceleration: \( g=\frac{v^2}{r} \).
3. \( r=\frac{u^2\cos^2\theta}{g} \).

Teacher's Note:
1. At the top the whole acceleration g is centripetal.
2. The speed there is only the horizontal component.

Answer: (a) Lowest: mg − T1; Highest: mg + T2

Explanation:
1. At the lowest point the forces are weight mg downwards and tension T1 upwards, so the net downward force is \( mg-T_1 \).
2. At the highest point both weight and tension act downwards: \( mg+T_2 \).

Teacher's Note:
1. Net force is just the vector sum of the real forces; do not add mv²/R separately.
2. The centripetal force is the net force, not an extra force.

Answer: (c) \( \lt mv^2/\ell \)

Explanation:
1. Under tension the rubber band stretches, so the radius of the circle r is greater than the natural length ℓ.
2. The tension supplies the centripetal force: \( T=\frac{mv^2}{r} \).
3. Since r is greater than ℓ, \( T\lt \frac{mv^2}{\ell} \).

Teacher's Note:
1. A rubber band lengthens under load, unlike an inextensible string.
2. The radius of the circle, not the natural length, enters the centripetal force formula.

Answer: (b) 19.5 m/s

Explanation:
1. \( v=\frac{ds}{dt}=18+6t-6t^2 \).
2. Maximum when \( \frac{dv}{dt}=6-12t=0\Rightarrow t=0.5\ \text{s} \).
3. \( v_{max}=18+3-1.5=19.5\ \text{m/s} \).

Teacher's Note:
1. Speed is the derivative of distance covered along the path.
2. Maximise by setting the derivative of speed to zero.

Answer: (a) \( 2KS\sqrt{1+\left(\frac SR\right)^2} \)

Explanation:
1. Tangential force from \( F_t=\frac{dT}{ds}=2Ks \) (work-energy theorem).
2. Centripetal force \( F_c=\frac{mv^2}{R}=\frac{2T}{R}=\frac{2Ks^2}{R} \).
3. Total \( F=\sqrt{F_t^2+F_c^2}=2Ks\sqrt{1+\left(\frac sR\right)^2} \).

Teacher's Note:
1. The work-energy theorem in differential form gives F_t = dT/ds.
2. The net force is the vector sum of tangential and centripetal parts.

Answer: (d) 0.8 m/s²

Explanation:
1. Tangential acceleration \( a_t=0.5\ \text{m/s}^2 \). Distance covered \( s=\frac{2\pi R}{10}=\frac{at^2}{2} \).
2. Centripetal acceleration \( a_c=\frac{v^2}{R}=\frac{a^2t^2}{R}=\frac{2as}{R}=0.4\pi a\approx0.63\ \text{m/s}^2 \).
3. Total \( \sqrt{a_t^2+a_c^2}=\sqrt{0.25+0.395}\approx0.8\ \text{m/s}^2 \).

Teacher's Note:
1. Eliminate the radius by using s = one-tenth of 2πR.
2. Add tangential and centripetal accelerations vectorially.

Answer: (c) \( \frac{\omega_0}{k}\left[1-e^{-kt}\right] \)

Explanation:
1. \( \frac{d\phi}{dt}=\omega_0-k\phi\Rightarrow\int_0^\phi\frac{d\phi}{\omega_0-k\phi}=\int_0^tdt \).
2. \( -\frac1k\ln\frac{\omega_0-k\phi}{\omega_0}=t \).
3. \( \phi=\frac{\omega_0}{k}\left(1-e^{-kt}\right) \).

Teacher's Note:
1. Separate the variables and integrate both sides.
2. Check: at t = 0, φ = 0, and as t becomes large, φ approaches ω0/k.

Answer: (b) \( \sqrt{\frac{g\ell}{3}},\ \cos^{-1}(2/3) \)

Explanation:
1. At angle θ from the downward vertical: \( T-mg\cos\theta=\frac{mv^2}{\ell} \). With T = mg: \( v^2=g\ell(1-\cos\theta) \).
2. Energy: \( v^2=g\ell-2g\ell(1-\cos\theta)=g\ell(2\cos\theta-1) \).
3. Equate: \( 1-\cos\theta=2\cos\theta-1\Rightarrow\cos\theta=\frac23 \), and \( v^2=\frac{g\ell}{3} \).

Teacher's Note:
1. Write the radial equation and the energy equation together.
2. Both unknowns v and θ follow from two equations.

Answer: (a) \( \frac54D \)

Explanation:
1. At the top of the loop the minimum speed satisfies \( v^2=gR \) with \( R=D/2 \).
2. Energy conservation from the starting height h (measured from the bottom) to the top: \( mgh=mg(2R)+\tfrac12mv^2=2mgR+\tfrac12mgR \).
3. \( h=\frac52R=\frac54D \).

Teacher's Note:
1. The condition at the top is gravity alone providing the centripetal force.
2. Height is measured from the lowest point of the loop.

Answer: (d) 10⁻² m

Explanation:
1. Spring constant \( k=10^2\ \text{N/cm}=10^4\ \text{N/m} \). Let the ball shift outwards by x.
2. The spring force supplies the centripetal force: \( kx=m\omega^2(0.1+x) \).
3. \( 10^4x=0.09\times10^4(0.1+x)\Rightarrow0.91x=0.009\Rightarrow x\approx10^{-2}\ \text{m} \).

Teacher's Note:
1. Convert N/cm to N/m and grams to kilograms.
2. The radius of the circular path grows with the shift.

Answer: (c) \( r\le\frac{\mu g}{\omega^2} \)

Explanation:
1. Friction provides the centripetal force, so we need \( m\omega^2r\le\mu mg \).
2. \( r\le\frac{\mu g}{\omega^2} \) (equality is the limiting case, where the coin is just about to slip).

Teacher's Note:
1. Static friction can take any value up to the limiting value μmg.
2. The coin stays on the record up to and including the limiting radius.

Answer: (b) 135.3 N

Explanation:
1. Momentum conservation: \( 0.1\times150=3v\Rightarrow v=5\ \text{m/s} \).
2. Speed at 60°: \( v'^2=25-2(10)(0.5)(1-\cos60^\circ)=20 \).
3. Tension: \( T=\frac{3(20)}{0.5}+3(10)\cos60^\circ=120+15=135\ \text{N} \).

Teacher's Note:
1. The collision is perfectly inelastic, so only momentum is conserved.
2. After the collision, use energy conservation and then the radial force equation.

Answer: (a) 10 m/s

Explanation:
1. The centre of gravity of the car moves on a circle of radius \( 8.9+1.1=10\ \text{m} \).
2. At the limit, the normal reaction is zero and gravity gives the centripetal force: \( mg=\frac{mv^2}{r} \).
3. \( v=\sqrt{gr}=\sqrt{10\times10}=10\ \text{m/s} \).

Teacher's Note:
1. Use the radius of the path of the centre of mass, not just the road radius.
2. Losing contact means N = 0.

Answer: (d) 640 N

Explanation:
1. A turn of 90° is a quarter circle: \( \frac{\pi R}{2}=628\Rightarrow R=400\ \text{m} \).
2. \( F=\frac{mv^2}{R}=\frac{1000\times256}{400}=640\ \text{N} \).

Teacher's Note:
1. Use the arc length of a quarter circle to find the radius.
2. Centripetal force is mv²/R.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Initially the coin stays on the table and moves in a circle.
2. The centripetal force is provided by static friction exerted by the table on the coin, so the table does provide the force.
3. Hence both statements are true and R explains A.

Teacher's Note:
1. The force is exerted by the table through friction.
2. Without it, the coin would fly off along a straight line.

Answer: (c) A is true, but R is false.

Explanation:
1. Bending inwards makes the normal reaction and friction together act along the line through the centre of gravity, so that the torque about it is zero; the horizontal component then supplies the centripetal force.
2. Bending does not primarily lower the centre of gravity, so R is false.

Teacher's Note:
1. The purpose of bending is to balance torques.
2. Angle of bending: tan θ = v²/(rg).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Limiting friction supplies the centripetal force: \( \mu mg=\frac{mv^2}{R} \), so \( v=\sqrt{\mu Rg}=\sqrt{0.3\times40\times10}\approx11\ \text{m/s} \).
2. The formula in R reproduces the value in A, so both are true and R explains A.

Teacher's Note:
1. Check the numbers: √120 is approximately 11.
2. Friction limits the speed on a level road.