JEE Physics Laws Of Motion MCQs Set 20

Answer: (b) the ground on the horse

Explanation:
1. The horse is the body whose motion we want, so only forces acting on the horse count.
2. The horse pushes the ground backwards with its hooves (action). By the third law the ground pushes the horse forward (reaction).
3. The cart pulls the horse backwards, and the horse's push on the ground acts on the ground, so neither makes the horse go forward.

Teacher's Note:
1. Ask which body the force acts on: the forward force on the horse must come from something outside the horse.
2. Action and reaction act on different bodies, which is why they do not cancel.

Answer: (a) \( 1:2 \)

Explanation:
1. Both blocks move together with \( a=\frac{F}{3m} \) in each case.
2. Case 1: the contact force is the force needed to accelerate \( m \): \( N_1=m\cdot\frac{F}{3m}=\frac F3 \).
3. Case 2: the contact force accelerates \( 2m \): \( N_2=2m\cdot\frac{F}{3m}=\frac{2F}{3} \).
4. Ratio \( N_1:N_2=1:2 \).

Teacher's Note:
1. The contact force always equals the mass of the block ahead of the push times the common acceleration.
2. Pushing the lighter block makes the contact force larger, because the heavier block is the one being pushed.

Answer: (d) 4 N

Explanation:
1. The blocks move together; net force \( 6-3=3 \) N on 3 kg gives \( a=1 \) m/s² towards the 3 N side.
2. For the 1 kg block: \( N-3=1\times1 \), so \( N=4 \) N.
3. Check with the 2 kg block: \( 6-N=2\times1 \) gives \( N=4 \) N, the same.

Teacher's Note:
1. Choose the lighter block when it simplifies the algebra, and check with the other.
2. The contact force is not the average of the two applied forces.

Answer: (c) \( \frac{F}{\sin\theta} \)

Explanation:
1. At B the three forces are tension \( T \) along BA, the horizontal pull \( F \) and the weight \( Mg \) (through the hanging mass).
2. Horizontal balance: \( T\sin\theta=F \).
3. Hence \( T=\frac{F}{\sin\theta} \) (vertical balance gives \( T\cos\theta=Mg \)).

Teacher's Note:
1. The angle is measured from the vertical, so the horizontal component of tension carries the sine.
2. Use the equation with fewer unknowns; here the horizontal one contains only T and F.

Answer: (b) infinitely large (not possible)

Explanation:
1. Let each half make angle \( \theta \) with the horizontal. Vertical balance at the midpoint: \( 2T\sin\theta=15g \).
2. For a straight rope \( \theta\to0 \), so \( T=\frac{15g}{2\sin\theta}\to\infty \).
3. So no finite tension can make the loaded rope perfectly horizontal.

Teacher's Note:
1. The vertical components of the two tensions must carry the weight, and a horizontal rope has none.
2. This is why a clothesline always sags a little under load.

Answer: (a) \( \frac g7 \) downwards, \( \frac{6g}{7} \)

Explanation:
1. The 4 kg side (B + A) outweighs the 3 kg side, so \( a=\frac{(4-3)g}{7}=\frac g7 \), with A moving downward.
2. For block A: \( mg-T=ma \), so \( T=1\cdot g-\frac g7=\frac{6g}{7} \).

Teacher's Note:
1. Treat the three blocks as one system to get the acceleration, then isolate one block for a tension.
2. The string between A and B carries only the net effect of the 1 kg block.

Answer: (d) 240 N, 96 N

Explanation:
1. Lower body (8 kg): \( T_2-8g=8a \), so \( T_2=8(9.8+2.2)=96 \) N.
2. Whole system (20 kg): \( T_1-20g=20a \), so \( T_1=20(9.8+2.2)=240 \) N.

Teacher's Note:
1. A string carries the weight and the required acceleration of everything hanging below it.
2. Take the lowest body first, then the whole system.

Answer: (c) \( 4mg \)

Explanation:
1. For masses \( M \) and \( m \): \( T=\frac{2mM}{M+m}g \).
2. For \( M\gg m \), \( T\approx2mg \) (the small mass accelerates upward at about \( g \)).
3. The string pulls the pulley down with a tension on each side, so the force is \( 2T\approx4mg \).

Teacher's Note:
1. The pulley feels the tension twice, once from each side of the string.
2. Do not stop at T: the question asks for the force on the pulley.

Answer: (b) \( \frac{U}{\cos\theta} \)

Explanation:
1. By symmetry M moves only vertically, with speed \( v \).
2. The string is inextensible, so the velocity components along a segment are equal at its two ends: at the pulley end the string moves at \( U \), and M's velocity component along the segment is \( v\cos\theta \).
3. So \( v\cos\theta=U \) and \( v=\frac{U}{\cos\theta} \).

Teacher's Note:
1. Resolve M's velocity along the string, because only that component must match the string's own speed.
2. As the angle grows, the mass moves faster than the string for the same pull.

Answer: (a) 2 m/s

Explanation:
1. B moves on a circle about A, so its velocity is perpendicular to AB. Its horizontal component equals the block's speed, \( \sqrt3 \) m/s.
2. The rod makes 60° with the ground, so a velocity perpendicular to it makes 60° with the vertical and its horizontal component is \( v\sin60^\circ \).
3. Setting \( v\sin60^\circ=\sqrt3 \) gives \( v=2 \) m/s, with vertical component \( v\cos60^\circ=1 \) m/s downward.

Teacher's Note:
1. The end of a rod pivoted at the other end moves perpendicular to the rod.
2. Match the component that the contact forces you to match, then use the direction to find the speed.

Answer: (d) keep it moving with uniform velocity

Explanation:
1. Uniform motion in a straight line means zero acceleration, so the net force needed is zero (first law).
2. Increasing speed, decreasing momentum and changing direction all change the velocity, so each needs a force.

Teacher's Note:
1. Force is needed to change velocity, not to maintain it.
2. Changing only the direction is still an acceleration.

Answer: (c) the resultant force on it is zero

Explanation:
1. By \( F=ma \), acceleration continues as long as the resultant force is non-zero.
2. A decreasing force still accelerates the body, a reversal of velocity does not need a zero force (a ball thrown up), and a force at right angles changes the direction of velocity, which is still acceleration.

Teacher's Note:
1. Acceleration stops only when the net force vanishes.
2. Do not confuse a smaller force with a zero force.

Answer: (b) A ball freely falling from a height

Explanation:
1. Net force is non-zero only when the velocity changes.
2. The kite is at rest, the aeroplane has constant velocity (constant speed along a straight line) and the cork is at rest, so net force is zero in each.
3. The falling ball accelerates at \( g \), so a net force (its weight) acts.

Teacher's Note:
1. Constant velocity in a straight line means zero net force.
2. Free fall is the cleanest example of non-zero net force.

Answer: (a) \( \left(\frac{M_2\sin\beta-M_1\sin\alpha}{M_1+M_2}\right)g \)

Explanation:
1. \( M_2\sin\beta>M_1\sin\alpha \), so \( M_2 \) moves down its face.
2. Along the incline for \( M_2 \): \( M_2g\sin\beta-T=M_2a \); for \( M_1 \): \( T-M_1g\sin\alpha=M_1a \).
3. Adding: \( a=\frac{M_2\sin\beta-M_1\sin\alpha}{M_1+M_2}g \).

Teacher's Note:
1. On each incline use the component of weight along the plane.
2. The driving term is the difference of the two components.

Answer: (d) less than that in case II

Explanation:
1. Case I: \( T-mg=ma_1 \) and \( 2mg-T=2ma_1 \), so \( a_1=\frac g3 \).
2. Case II: the rope tension is \( 2mg \), so \( 2mg-mg=ma_2 \) and \( a_2=g \).
3. Hence \( a_1<a_2 \).

Teacher's Note:
1. A hanging mass also has to be accelerated, which wastes part of the pull.
2. A fixed pull of 2mg beats a hanging mass of 2m.

Answer: (c) 2.4 m/s²

Explanation:
1. \( m_1=\frac F4,\ m_2=\frac F6 \).
2. Combined mass \( F\left(\frac14+\frac16\right)=\frac{5F}{12} \), so \( a=\frac{12}{5}=2.4 \) m/s².

Teacher's Note:
1. For the same force, \( \frac1a \) adds like mass.
2. The combined acceleration is less than either individual one.

Answer: (b) zero

Explanation:
1. Three equal vectors at 120° form a closed equilateral triangle.
2. Their vector sum is zero, so the net force and the acceleration are zero.

Teacher's Note:
1. Equal forces at equal angles in a plane always cancel.
2. Check by components: \( F(1+\cos120^\circ+\cos240^\circ)=0 \).

Answer: (a) \( \frac g4 \)

Explanation:
1. Sliding down with acceleration \( a \): \( mg-T=ma \).
2. With \( T_{max}=\frac34mg \): \( a=g-\frac34g=\frac g4 \).
3. A smaller acceleration would need a larger tension.

Teacher's Note:
1. To keep tension low the fireman must accelerate downward.
2. Set the tension to its limit to find the extreme acceleration.

Answer: (d) \( 10\sqrt2 \) kg

Explanation:
1. Mass \( =\frac{|\vec F|}{a} \).
2. The magnitude is \( \sqrt{6^2+8^2+10^2}=\sqrt{200}=10\sqrt2 \) N, so \( m=10\sqrt2 \) kg.

Teacher's Note:
1. Mass is a scalar, so use the magnitude of the vector force.
2. Make sure to take the square root of the sum of squares.

Answer: (c) \( 9F \)

Explanation:
1. Stopping needs \( v^2=2ax \), so \( a=\frac{v^2}{2x} \).
2. The force is \( F=ma \) so \( F\propto v^2 \).
3. Tripling the speed multiplies the force by \( 3^2=9 \).

Teacher's Note:
1. For a fixed stopping distance, force scales with the square of speed.
2. This is the work-energy theorem seen through the second law.

Answer: (b) \( \frac{Mg\sin\theta}{2} \)

Explanation:
1. Incline block: \( Mg\sin\theta-T=Ma \); table block: \( T=Ma \).
2. Adding: \( a=\frac{g\sin\theta}{2} \).
3. So \( T=Ma=\frac{Mg\sin\theta}{2} \).

Teacher's Note:
1. The block on the table is pulled only by the string, so its equation is simply T = Ma.
2. The system's driving force is the weight component along the incline.

Answer: (a) 10 N

Explanation:
1. Each block is in equilibrium, so each string has tension \( T=1\times10=10 \) N.
2. A spring balance reads the force on one of its ends, which is the tension, not the sum of the two pulls.
3. So the reading is 10 N.

Teacher's Note:
1. A balance hung between two equal pulls reads one of them.
2. The other end could be a fixed wall and the reading would be the same.

Answer: (d) \( \frac{F-M_1a_1}{M_2} \)

Explanation:
1. Spring force \( f \) on \( M_1 \): \( F-f=M_1a_1 \), so \( f=F-M_1a_1 \).
2. The same force acts on \( M_2 \): \( a_2=\frac{f}{M_2}=\frac{F-M_1a_1}{M_2} \).

Teacher's Note:
1. The spring transmits the same force to both ends, but the accelerations differ while the spring is changing length.
2. The accelerations equal each other only after the spring settles to a fixed length.

Answer: (c) 49 m/s²

Explanation:
1. 10 kg-wt means \( 10\times9.8=98 \) N, the tension in the balance.
2. This is the only horizontal force on the trolley: \( 98=2a \).
3. So \( a=49 \) m/s².

Teacher's Note:
1. Convert kg-wt to newtons before using the second law.
2. A massless spring balance transmits the force it reads to the trolley.