JEE Physics Laws Of Motion MCQs Set 07

Answer: (c) to maintain the same speed

Explanation:
1. By Newton's first law, a body in uniform motion continues in that state with no net force.
2. A force is needed only to change the speed or the direction, so it is not required to maintain the same speed.

Teacher's Note:
1. Force is the cause of a change in velocity, not of velocity itself.
2. Uniform velocity means zero net force.

Answer: (b) acceleration equals zero

Explanation:
1. A body is in equilibrium when the net force on it is zero.
2. Then by \( F = ma \) its acceleration is zero.
3. Velocity, momentum and kinetic energy need not be zero (a body in uniform motion is in equilibrium).

Teacher's Note:
1. Equilibrium means zero acceleration, not necessarily rest.
2. Static and dynamic equilibrium both have zero net force.

Answer: (a) acceleration

Explanation:
1. For a constant force, \( a = F/m \) is constant.
2. Velocity, speed and momentum all change with time under a constant acceleration.

Teacher's Note:
1. Constant force gives constant acceleration.
2. Do not confuse uniform acceleration with uniform velocity.

Answer: (b) 150 N

Explanation:
1. Acceleration \( a = (30 - 0)/2 = 15 \) m/s².
2. Force \( F = ma = 10 \times 15 = 150 \) N.

Teacher's Note:
1. First find acceleration from the change in velocity, then use \( F = ma \).
2. Units: kg times m/s² gives newton.

Answer: (a) zero

Explanation:
1. The velocity is constant, so the acceleration is zero.
2. On a frictionless table no horizontal force acts, so the force on the body in the direction of motion is zero.

Teacher's Note:
1. Constant velocity on a frictionless surface needs no horizontal force.
2. Do not use the weight here because the question asks about the force producing motion.

Answer: (b) 16 N

Explanation:
1. Retardation due to friction \( a = 8/3 \) m/s².
2. Friction force \( f = ma = 6 \times 8/3 = 16 \) N.
3. To move at constant velocity, an applied force equal to friction, 16 N, is needed.

Teacher's Note:
1. Constant velocity needs an applied force equal and opposite to friction.
2. Find friction from the deceleration first.

Answer: (c) \( \dfrac{1}{\sqrt{m}} \)

Explanation:
1. \( a = F/m \) and \( v^2 = 2ad = 2Fd/m \).
2. So \( v = \sqrt{2Fd/m} \propto 1/\sqrt{m} \).

Teacher's Note:
1. F and d are fixed, so only m varies.
2. Heavier bodies gain less speed for the same work.

Answer: (a) 39200 dyne

Explanation:
1. The body presses on the table with its weight, since it has no vertical acceleration.
2. Weight \( = mg = 40 \times 980 = 39200 \) dyne.

Teacher's Note:
1. The force on the table is the normal reaction pair, equal to the weight.
2. The horizontal motion adds no force on a frictionless table.

Answer: (d) \( \vec{v} \), unchanged

Explanation:
1. Forces represented by the sides of a triangle taken in order add up to zero.
2. With zero net force there is no acceleration, so the velocity remains \( \vec{v} \).

Teacher's Note:
1. A closed vector triangle means zero resultant.
2. Zero net force means no change in velocity.

Answer: (b) 10 s

Explanation:
1. \( \vec{a} = \vec{F}/m = (-0.6\hat{i} + 0.8\hat{j}) \) m/s².
2. The x-component of velocity is \( 6 - 0.6t \).
3. For the velocity to be along Y, this must be zero: \( t = 10 \) s.

Teacher's Note:
1. A velocity is along Y when its x-component vanishes.
2. Treat the two components independently.

Answer: (a) \( F_1/m \)

Explanation:
1. The particle was in equilibrium, so \( F_1 \) balances the resultant of \( F_2 \) and \( F_3 \).
2. Removing \( F_1 \) leaves a net force of magnitude \( F_1 \).
3. So \( a = F_1/m \).

Teacher's Note:
1. Removing one force from a balanced set leaves a net force equal to that force.
2. Direction of acceleration is opposite to the removed force.

Answer: (b) 1 N

Explanation:
1. The stone accelerates with the train at 1 m/s² (it does not slip).
2. Net force \( = ma = 1 \times 1 = 1 \) N.

Teacher's Note:
1. Net force depends only on mass and acceleration.
2. Weight is balanced by the normal reaction.

Answer: (d) the road on the car

Explanation:
1. The engine force is internal to the car and cannot accelerate it.
2. The wheels push the road backwards, and by Newton's third law the road pushes the car forward with friction.
3. That external force accelerates the car.

Teacher's Note:
1. Only an external force can change the momentum of a system.
2. Friction from the road is the external forward force.

Answer: (c) rate of change of momentum and external force

Explanation:
1. Newton's second law states \( \vec{F} = d\vec{p}/dt \).
2. The net external force equals the rate of change of momentum.

Teacher's Note:
1. Force equals rate of change of momentum.
2. For constant mass this becomes \( F = ma \).

Answer: (c) reaction

Explanation:
1. By Newton's third law the stone exerts an equal and opposite force (reaction) on the foot.
2. That reaction force is what hurts.

Teacher's Note:
1. Action and reaction act on different bodies.
2. Hurting the foot is a direct result of the reaction force.

Answer: (c) 400 kg m/s

Explanation:
1. Change in momentum equals impulse \( = Ft \).
2. \( \Delta p = 20 \times 20 = 400 \) kg m/s.

Teacher's Note:
1. Mass is not needed for the momentum change.
2. Impulse equals change in momentum.

Answer: (c) 2 N

Explanation:
1. Change in momentum \( = 0.1 \times 5 = 0.5 \) kg m/s.
2. \( F = \Delta p/t = 0.5/0.25 = 2 \) N.

Teacher's Note:
1. Convert grams to kilograms before calculating.
2. Longer catching time means a smaller force.

Answer: (c) 50 N

Explanation:
1. \( \Delta p = 0.05 \times (25 + 25) = 2.5 \) kg m/s.
2. \( F = 2.5/0.05 = 50 \) N.

Teacher's Note:
1. Rebounding with the same speed doubles the momentum change.
2. Use the contact time given.

Answer: (d) 2400 m/s²

Explanation:
1. Speed just before impact \( = \sqrt{2 \times 10 \times 10} \approx 14.1 \) m/s downward.
2. Speed just after \( = \sqrt{2 \times 10 \times 5} = 10 \) m/s upward.
3. \( a = (10 + 14.1)/0.01 \approx 2400 \) m/s².

Teacher's Note:
1. Find the speeds from free fall first.
2. The total velocity change is the sum of magnitudes.

Answer: (c) 10000 N

Explanation:
1. \( \Delta p = 0.15 \times 80 = 12 \) kg m/s.
2. \( F = 12/(1.2\times10^{-3}) = 10^4 \) N.

Teacher's Note:
1. Convert 1.20 ms to 1.2 x 10^-3 s.
2. Reversal of 40 m/s means a velocity change of 80 m/s.

Answer: (c) 2mnv

Explanation:
1. Each bullet's momentum change is \( mV - (-mV) = 2mV \).
2. Force is the momentum change per second: \( F = n \times 2mV = 2mnV \).

Teacher's Note:
1. Force on the wall equals the rate of momentum transferred.
2. Elastic rebound doubles the change in momentum.

Answer: (d) The greater the change in the momentum in a given time, the lesser is the force that needs to be applied.

Explanation:
1. \( F = \Delta p/\Delta t \), so for a given time a greater change in momentum needs a greater force.
2. Statement (d) says the opposite, so it is false.
3. The other three follow directly from the second law.

Teacher's Note:
1. Force is directly proportional to the change in momentum for a fixed time.
2. Check each statement against \( F = \Delta p/\Delta t \).

Answer: (a) \( 50t - 10t^2 \)

Explanation:
1. Impulse \( = \int_0^t F\,dt = \int_0^t (50 - 20t)\,dt \).
2. \( = 50t - 10t^2 \).

Teacher's Note:
1. Impulse is the integral of force over time.
2. Integrate term by term from 0 to t.

Answer: (a) 20 m/s²

Explanation:
1. The water stops on hitting the block, so force \( = (dm/dt)\,v = 3 \times 10 = 30 \) N.
2. \( a = F/m = 30/1.5 = 20 \) m/s².

Teacher's Note:
1. The jet loses all its momentum on hitting the block.
2. Force = rate of mass x speed.

Answer: (b) 0.75 kg/s

Explanation:
1. Thrust \( F = u\,\dfrac{dm}{dt} \).
2. \( dm/dt = 300/400 = 0.75 \) kg/s.

Teacher's Note:
1. Thrust equals exhaust speed times burn rate.
2. No gravity acts in force-free space.