JEE Physics Laws Of Motion MCQs Set 16

Answer: (a) \( 2av^2d\cos\theta \)

Explanation:
1. Take the normal to the wall as the direction of interest. The normal component of velocity reverses: it changes from \( v\cos\theta \) towards the wall to \( v\cos\theta \) away from it, a change of \( 2v\cos\theta \).
2. Mass hitting per second \( =adv \), so force \( =adv\times2v\cos\theta \).
3. \( F=2av^2d\cos\theta \).

Teacher's Note:
1. Force on a wall equals the rate of change of momentum perpendicular to the wall. The component parallel to the wall is unchanged for an elastic rebound.

Answer: (b) \( (2\hat i-10\hat j+7\hat k)\ \text{N} \)

Explanation:
1. The particle moves with constant velocity, so its acceleration is zero and the net force is zero: \( \vec F_1+\vec F_2+\vec F_3=0 \).
2. \( \vec F_1+\vec F_2=(3-5)\hat i+(2+8)\hat j+(-4-3)\hat k=-2\hat i+10\hat j-7\hat k \).
3. \( \vec F_3=-(\vec F_1+\vec F_2)=(2\hat i-10\hat j+7\hat k) \) N.

Teacher's Note:
1. Constant velocity means zero net force (Newton's first law). The third force is simply the negative of the sum of the other two.

Answer: (d) 39 m/s

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

Teacher's Note:
1. The impulse-momentum theorem gives the velocity directly for a piecewise-constant force. Add the areas of the rectangles.

Answer: (a) 2 N

Explanation:
1. Horizontal component of the pull on the 3 kg block: \( 10\cos60^\circ=5 \) N. The vertical component only reduces the normal reaction.
2. Common acceleration of the two blocks: \( a=\frac{5}{2+3}=1 \) m/s².
3. The string pulls only the 2 kg block: \( T=2\times1=2 \) N.

Teacher's Note:
1. For connected bodies, find the common acceleration from the whole system first. Then apply Newton's law to the block which the string pulls.

Answer: (c) \( \frac{320}{9}\ \text{N} \)

Explanation:
1. Left side: 8 kg. Right side: \( 6+4=10 \) kg. Common acceleration \( a=\frac{(10-8)g}{18}=\frac g9 \) (the 10 kg side goes down).
2. Consider the 4 kg block (moving down): \( 4g-T=4a \).
3. \( T=4\left(g-\frac g9\right)=4\times\frac{8g}{9}=\frac{320}{9} \) N.

Teacher's Note:
1. First treat the three blocks as one system to find acceleration. Then isolate the block whose string tension is asked.

Answer: (a) \( \frac{mMg}{k(m+M)} \)

Explanation:
1. The system accelerates with \( a=\frac{Mg}{m+M} \).
2. The spring pulls only the block of mass m, so \( kx=ma \).
3. \( x=\frac{m}{k}\cdot\frac{Mg}{m+M}=\frac{mMg}{k(m+M)} \).

Teacher's Note:
1. Spring force equals the force needed to accelerate the block it is attached to. Find the common acceleration first.

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

Explanation:
1. Whole system acceleration \( a=\frac{P}{M+m} \).
2. The rope exerts on the block the only horizontal force needed to accelerate it: \( T=Ma \).
3. \( T=\frac{PM}{M+m} \).

Teacher's Note:
1. A rope with mass has a different tension at each point. At the block end the tension is just the block's mass times acceleration.

Answer: (c) increased

Explanation:
1. When the parrot accelerates upward it pushes air downward with extra force, and the cage floor and walls transmit the reaction.
2. The centre of mass of the cage plus parrot accelerates upward, so the net upward force on the system exceeds the weight, which means the boy supports more than the static weight.
3. The weight of the cage felt by the boy increases.

Teacher's Note:
1. In a closed system a net upward acceleration of any part requires extra upward support from outside. The air-tight cage matters, because the extra force cannot escape.

Answer: (d) 8 m/s²

Explanation:
1. The spring pulls the 10 kg mass with a force \( 10\times4=40 \) N.
2. By Newton's third law the spring pulls back on the 20 kg mass with 40 N, opposite to the applied force.
3. \( a=\frac{200-40}{20}=8 \) m/s².

Teacher's Note:
1. A spring exerts equal and opposite forces at its two ends. Use the known acceleration of one mass to find the spring force.

Answer: (c) less than that in case (b)

Explanation:
1. Case (a): \( a=\frac{2m-m}{2m+m}g=\frac g3 \).
2. Case (b): the pull is \( 2mg \) on a mass m, so \( a=\frac{2mg-mg}{m}=g \).
3. \( \frac g3\lt g \), so the acceleration in case (a) is less than in case (b).

Teacher's Note:
1. Pulling with a force equal to the weight of 2m is not the same as hanging 2m. In (a) the hanging mass also has to be accelerated, so the acceleration is lower.

Answer: (b) 48 N

Explanation:
1. The component of velocity perpendicular to the wall is \( v\sin30^\circ=24 \) m/s (the angle is with the wall), and it reverses on rebound.
2. Change in momentum \( =2mv\sin30^\circ=2\times0.5\times48\times0.5=24 \) kg m/s.
3. Average force \( =\frac{24}{0.5}=48 \) N.

Teacher's Note:
1. Only the component of velocity perpendicular to the wall changes. Be careful whether the given angle is measured from the wall or from the normal.

Answer: (d) 4 m/s²

Explanation:
1. Just enough force to start the motion: \( F=\mu_sN=0.6\times200=120 \) N.
2. Once moving, kinetic friction \( =0.2\times200=40 \) N.
3. \( a=\frac{120-40}{20}=4 \) m/s².

Teacher's Note:
1. The force needed to start a body moving is larger than the kinetic friction, so the body accelerates once it begins to slide.

Answer: (a) 6.0 N

Explanation:
1. Normal reaction \( N=mg-P=25-9=16 \) N.
2. Maximum static friction \( =0.4\times16=6.4 \) N, which is larger than the applied 6.0 N.
3. The block stays at rest and the friction force equals the applied force, 6.0 N.

Teacher's Note:
1. Friction adjusts up to its limiting value. If the applied force is below the limit, friction equals the applied force and is not \( \mu N \).

Answer: (d) 42 N

Explanation:
1. The 4 kg block is accelerated by friction alone, so its maximum acceleration is \( \mu g=0.3\times10=3 \) m/s².
2. For the two to move together, the maximum common acceleration is 3 m/s².
3. \( F_{max}=(4+10)\times3=42 \) N.

Teacher's Note:
1. The maximum acceleration of the upper block is set by limiting friction. Apply that acceleration to the whole system to get the maximum force.

Answer: (b) 10 kg

Explanation:
1. The hanging 10 kg block pulls with \( 10g \). For the system to stay at rest the maximum friction must be at least this: \( \mu(10+m)g\ge10g \).
2. \( 0.5(10+m)\ge10\Rightarrow10+m\ge20 \).
3. The minimum \( m=10 \) kg.

Teacher's Note:
1. Adding a load on the block increases the normal reaction and therefore the maximum friction. Set limiting friction equal to the pulling weight.

Answer: (a) 42 N

Explanation:
1. The upper block A is accelerated by friction only, so the maximum acceleration is \( \mu g=0.6\times10=6 \) m/s².
2. For no relative motion the common acceleration is at most 6 m/s².
3. \( F_{max}=(2+5)\times6=42 \) N.

Teacher's Note:
1. Convert units carefully: 42 kgf would be about 412 N, so the option in newtons is correct here.

Answer: (a) \( 5\sqrt2 \) N s

Explanation:
1. The component parallel to the wall is unchanged; the normal component \( mv\cos45^\circ \) reverses.
2. \( |\Delta p|=2mv\cos45^\circ=2\times0.2\times25\times\frac1{\sqrt2} \).
3. \( |\Delta p|=5\sqrt2 \) N s.

Teacher's Note:
1. Change in momentum is a vector difference. Only the component normal to the wall reverses, so double that component.

Answer: (a) \( 1.35\times10^4 \) N s and 9000 N

Explanation:
1. Impulse = area of the triangle \( =\frac12\times18000\times(2.5-1)=1.35\times10^4 \) N s.
2. Contact time \( =2.5-1=1.5 \) s.
3. Average force \( =\frac{1.35\times10^4}{1.5}=9000 \) N.

Teacher's Note:
1. The area under a force-time graph is the impulse. Average force is impulse divided by contact time, which is half the peak for a triangular pulse.

Answer: (a) 8 N and 18 N

Explanation:
1. \( F=(1+1.5+2)\times4=18 \) N.
2. The contact force between the 1.5 kg and 2 kg blocks accelerates only the 2 kg block: \( N_2=2\times4=8 \) N.
3. (The force between the 1 kg and 1.5 kg blocks is \( 3.5\times4=14 \) N.)

Teacher's Note:
1. The contact force at any joint is the mass beyond the joint times the common acceleration.

Answer: (a) 3.33 m/s²

Explanation:
1. Bananas just lose contact when the tension equals their weight: \( T=160 \) N.
2. For the monkey: \( T-120=12a \).
3. \( a=\frac{160-120}{12}=3.33 \) m/s².

Teacher's Note:
1. The limiting condition for the bananas is that the normal reaction from the ground becomes zero. The tension then equals their weight.

Answer: (a) 400 N and 300 N

Explanation:
1. Tension is the same throughout the rope. The lift (500 N weight) is pulled up by two rope segments and pushed down by the man with force \( N \): \( 2T=500+N \).
2. For the man (700 N weight), the rope pulls him up with \( T \) and the machine pushes up with \( N \): \( N+T=700 \).
3. Solving, \( T=400 \) N and \( N=300 \) N.

Teacher's Note:
1. Draw separate free body diagrams for the lift and the man. The rope pulls up on the lift twice (two segments) in this arrangement.

Answer: (a) \( T_1=25 \) N, \( T_2=25\sqrt3 \) N

Explanation:
1. Horizontal balance: \( T_1\cos30^\circ=T_2\cos60^\circ\Rightarrow T_1=\frac{T_2}{\sqrt3} \).
2. Vertical balance: \( T_1\sin30^\circ+T_2\sin60^\circ=50\).
3. \( \frac{T_2}{2\sqrt3}+\frac{\sqrt3T_2}{2}=50\Rightarrow T_2=25\sqrt3 \) N and \( T_1=25 \) N.

Teacher's Note:
1. Lami's theorem or component balance both work. The two strings are perpendicular here, so the tensions are 50 cos and 50 sin of the angles.

Answer: (a) \( \frac{\mu L}{1+\mu} \)

Explanation:
1. Let the hanging length be \( y \) and the mass per unit length be \( \lambda \). Hanging weight \( =\lambda yg \).
2. Limiting friction on the part on the table \( =\mu\lambda(L-y)g \).
3. Equating: \( y=\mu(L-y)\Rightarrow y=\frac{\mu L}{1+\mu} \).

Teacher's Note:
1. Balance the weight of the hanging part with the limiting friction of the part resting on the table.

Answer: (a) \( \frac{125}{3}\ \text{N} \)

Explanation:
1. The maximum friction on B is \( 0.25\times100=25 \) N, which gives B the maximum acceleration \( \frac{25}{15}=\frac53 \) m/s².
2. The two blocks move together if \( F\le(10+15)\times\frac53 \).
3. \( F_{max}=\frac{125}{3} \) N.

Teacher's Note:
1. The lower block is accelerated only by friction, so its limiting acceleration is the common limiting acceleration of the pair.

Answer: (a) 1.6 m/s² and 16 N

Explanation:
1. Friction between B and the floor: \( 0.2\times250=50 \) N. Net force on the system \( =90-50=40 \) N.
2. If A and B move together: \( a=\frac{40}{25}=1.6 \) m/s².
3. Friction needed on A: \( 10\times1.6=16 \) N, less than the limit \( 0.4\times100=40 \) N, so they do move together.

Teacher's Note:
1. Check that the friction needed is smaller than its limit before assuming that the blocks move together.

Answer: (a) 1100 N

Explanation:
1. Friction between A and B: \( 0.2\times100\times10=200 \) N. Friction between B and the floor: \( 0.3\times300\times10=900 \) N.
2. To make B move, \( F \) must overcome both limiting forces.
3. \( F_{min}=200+900=1100 \) N.

Teacher's Note:
1. When B slides, it feels friction from both surfaces. Note the floor friction uses the combined weight of A and B.