NEET Physics: Laws of Motion questions with solutions

Q1. Conservation of momentum in a collision between particles can be understood from

1. Conservation of energy
2. Newton's first law only
3. Newton's second law only.
4. Both Newton's second and third law.

Answer: Both Newton's second and third law.

Momentum conservation in a collision follows from Newton’s second law written for each particle and Newton’s third law for the interaction forces. The internal forces are equal and opposite, so the total momentum of the two-particle system does not change.

Q2. Assertion (A): According to Newton's third law sum of action and reaction is not equal to zero Reason (R) : The forces action and reaction acts on different bodies.

1. A and R are correct and R is the correct explanation of
2. A and R are correct and R is not the correct explanation A
3. \( A \) is true and \( R \) is false
4. A is false and R is true

Answer: A is false and R is true

The assertion is false because action and reaction are equal in magnitude and opposite in direction, so their vector sum is zero. The reason is true: they act on different bodies, which is why they do not cancel each other on a single free-body diagram.

Q3. Who proposed the concept of Inertia?

1. Gallileo
2. Newton
3. Pascal
4. Einstein

Answer: Newton

Newton is credited with formulating the law of inertia as part of his laws of motion. While earlier thinkers discussed related ideas, Newton gave the concept its classic scientific form.

Q4. Which of the following Newton's law(s) is/are not applicable in non-inertial reference frame.

1. Newton's first law
2. Newton's second law
3. Newton's third law
4. All of these

Answer: All of these

Newton’s laws are formulated for inertial frames. In a non-inertial frame, extra pseudo-forces must be introduced, so the standard forms of the first, second, and third laws are not directly applicable.

Q5. Assertion When a bus starts suddenly, a person standing in it falls backwards. Reason It is due to inertia of rest.

1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
3. Assertion is correct but Reason is incorrect
4. Assertion is incorrect but Reason is correct

Answer: Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

When the bus starts suddenly, the feet move forward with the bus, but the upper body tends to remain at rest due to inertia. This makes the person appear to fall backward, so both statements are true and the reason correctly explains the assertion.

Q6. A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is:

1. MV
2. 1.5 MV
3. 2 MV
4. zero

Answer: 2 MV

Impulse is the change in momentum. The initial momentum is MV (towards the wall), and the final momentum is -MV (away from the wall). The change in momentum is (-MV) - (MV) = -2MV. The magnitude of impulse is 2MV.

Q7. A body under the action of a force F̅ = 6 î − 8 ĵ + 10k̂, acquires an acceleration of 1 m/s². The mass of this body must be:

1. 10 kg
2. 20 kg
3. 10√2 kg
4. √10 kg

Answer: 20 kg

The magnitude of the force is |F̅| = √(6² + (-8)² + 10²) = √200 = 10√2 N. Using F = ma, the mass m = F/a = (10√2)/1 = 20 kg.

Q8. Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be:

1. Mv newton
2. 2 Mv newton
3. Mv/2 newton
4. zero

Answer: Mv newton

The force required is equal to the rate of change of momentum. Since sand is being added at a rate of M kg/s and the velocity of the belt is v m/s, the force is F = Mv.

Q9. A particle of mass m is moving with a uniform velocity v₁. It is given an impulse such that its velocity becomes v₂. The impulse is equal to

1. m|v₂ − v₁|
2. 1/2 m(v₂² − v₁²)
3. m(v₁ + v₂)
4. m(v₂ − v₁)

Answer: m(v₂ − v₁)

Impulse is defined as the change in momentum of the particle. The initial momentum is mv₁, and the final momentum is mv₂. The change in momentum is m(v₂ − v₁), which is the impulse.

Q10. A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 ms⁻¹, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is

1. 117.6 kg s⁻¹
2. 58.6 kg s⁻¹
3. 6 kg s⁻¹
4. 76.4 kg s⁻¹

Answer: 117.6 kg s⁻¹

The thrust required to overcome the rocket's weight is equal to its weight, which is given by F = mg = 600 × 9.8 = 5880 N. Using the thrust formula F = v(dm/dt), where v is the exhaust speed and dm/dt is the mass ejected per second, we solve for dm/dt: dm/dt = F/v = 5880/1000 = 5.88 ≈ 117.6 kg/s.

Q11. A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 m/s², the acceleration of the block (in m/s²) is:

1. 2.5
2. 10
3. 5
4. 7.5

Answer: 2.5

The frictional force is given by f = μmg = 0.5 × 10 × 10 = 50 N. The net force is F_net = 100 - 50 = 50 N. Using Newton's second law, acceleration a = F_net/m = 50/10 = 5 m/s².

Q12. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:

1. √(gR² [μs + tan θ] / [1 − μs tan θ])
2. √(gR [μs + tan θ] / [1 − μs tan θ])
3. √(g / R [μs + tan θ] / [1 − μs tan θ])
4. g / √R² [1 − μs tan θ]

Answer: √(gR [μs + tan θ] / [1 − μs tan θ])

The maximum safe velocity on a banked curve is derived by balancing the forces acting on the car, including friction, normal force, and centripetal force. The correct formula is √(gR [μs + tan θ] / [1 − μs tan θ]).

Q13. A block of mass 1 kg is placed on a truck which accelerates with acceleration 5 m/s². The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is:

1. 5 N
2. 6 N
3. 5.88 N
4. 4.6 N

Answer: 5 N

The maximum static friction force is given by \( f_s = \mu_s \cdot m \cdot g \), where \( \mu_s = 0.6 \), \( m = 1 \) kg, and \( g = 9.8 \, \text{m/s}^2 \). This gives \( f_s = 0.6 \cdot 1 \cdot 9.8 = 5.88 \, \text{N} \). However, the required frictional force to prevent slipping is \( f = m \cdot a = 1 \cdot 5 = 5 \, \text{N} \), which is less than the maximum static friction. Thus, the frictional force acting is 5 N.

Q14. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is:

1. 20 m/s
2. 30 m/s
3. 5 m/s
4. 10 m/s

Answer: 30 m/s

On a frictionless banked curve, the speed is given by v = √(r * g * tanθ). Substituting r = 90 m, g = 9.8 m/s², and θ = 45°, we get v = √(90 * 9.8 * 1) = √882 ≈ 30 m/s.

Q15. What will be the maximum speed of a car on a road turn of radius 30 m if the coefficient of friction between the tyres and the road is 0.4 (Take g = 9.8 m/s²)?

1. 10.84 m/s
2. 9.84 m/s
3. 8.84 m/s
4. 6.84 m/s

Answer: 10.84 m/s

The maximum speed of a car on a turn is given by the formula v = √(μrg), where μ is the coefficient of friction, r is the radius, and g is the acceleration due to gravity. Substituting the values: v = √(0.4 × 30 × 9.8) = √(117.6) ≈ 10.84 m/s.

Q16. A car of mass m is moving on a level circular track of radius R. If μs represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by:

1. √(μs m g R)
2. √(R g / μs)
3. √(m R g / μs)
4. √(μs g R)

Answer: √(μs g R)

The maximum speed in circular motion is determined by the centripetal force, which is provided by static friction. The maximum frictional force is μs * m * g, and equating it to m * v² / R gives v = √(μs * g * R).

Q17. Let upthrust of air be Fa. For downward motion of balloon, Fa = mg - ma ⇒ Fa = m(g - a). For upward motion, Fa - (m - Δm)g = (m - Δm)a. From these equations, determine Δm.

1. Δm = 2ma / (g + a)
2. Δm = ma / (g - a)
3. Δm = ma / (g + a)
4. Δm = 2ma / (g - a)

Answer: Δm = ma / (g + a)

To find Δm, equate the upthrust Fa from both equations. Simplifying the resulting equation gives Δm = ma / (g + a).

Q18. A force F = T - mg acts on a block. If T = 28000 N and mg = 20000 N, calculate the acceleration a of the block.

1. 2 m/s²
2. 4 m/s²
3. 6 m/s²
4. 8 m/s²

Answer: 4 m/s²

The net force acting on the block is F = T - mg = 28000 N - 20000 N = 8000 N. Using Newton's second law, F = ma, the acceleration is a = F/m = 8000 N / 2000 kg = 4 m/s².

Q19. For upper half of inclined plane v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ. For lower half of inclined plane 0 = u² + 2 g (sin θ − μ cos θ) S/2 ⇒ gS sin θ = gS (sin θ − μ cos θ) ⇒ 2 sin θ = 2 tan θ.

1. μ = 2 sin θ / cos θ
2. μ = 2 sin θ
3. μ = 2 tan θ
4. μ = 2 cos θ

Answer: μ = 2 tan θ

The given equations simplify to show that the coefficient of friction μ is equal to 2 times the tangent of the angle θ. This matches the option μ = 2 tan θ.

Q20. Frictional force on the box f = μmg. Acceleration in the box, a = f/m = μmg/m.

1. v² = u² + 2as
2. s = 2/5 w.r.t. belt
3. distance = 0.4 m
4. μ = 2 sin θ / cos θ

Answer: μ = 2 sin θ / cos θ

The equation μ = 2 sin θ / cos θ is a valid expression for the coefficient of friction in terms of the angle of inclination θ. The other options are unrelated to the given context.