NEET Previous Year Questions (2015-2024): Laws of Motion | Physics Class 11 PDF Download

Ans: Find the total mass of the system:
m_total= m_A + m_B = 2 kg + 3 kg = 5kg
Calculate the acceleration of the system: Since both blocks move together on a frictionless surface, they will have the same acceleration. Using Newton's second law:
F_total = m_total ⋅ a
10N = 5kg ⋅ a
Find the force exerted by block A on block B: Block B is being accelerated by the force exerted by block A. The force on block B can be calculated using Newton’s second law again, but this time for block B alone:
F _{on B}  = m _B ⋅a
F _{on B}  = 3kg ⋅ 2m/s ² = 6 N
Thus, the force exerted by block A on block B is 6 N.

Ans: (c)
Problem Breakdown:

• The football player is initially moving southward and then suddenly turns eastward while maintaining the same speed.
• The question asks about the direction of the force acting on the player while turning.

Concept:

• Centripetal Force: When an object changes direction, there is a force directed towards the center of the circular path, which is called centripetal force.
• The player is changing direction, meaning their velocity vector is also changing direction. In this case, the player is turning from south to east, which implies a change in velocity towards the east.

Force Direction:

• The force responsible for this change in direction would act towards the north-east because the player is transitioning between the southward and eastward directions. The force is responsible for changing both the magnitude and direction of the velocity as the player turns.

Thus, the force acts along the north-east direction.

Ans: (c)
Solution:

Problem Breakdown:

We need to calculate the maximum acceleration of a moving car such that a body lying on the floor of the car remains stationary. Given:

• Coefficient of static friction (μ) = 0.15
• Acceleration due to gravity (g) = 10 m/s²

Concept:
For the body to remain stationary relative to the car, the maximum static friction force must equal the force required to accelerate the body along with the car. The friction force is what keeps the body stationary relative to the moving car.
Static friction force:
F_friction = μ⋅N
Where N is the normal force, and for a horizontal surface, N = m ⋅ g.
Therefore:
F_friction = μ ⋅ m ⋅ g
Force required for acceleration: The force required to accelerate the body is:
F = m ⋅a
 For the maximum acceleration, the static friction force must equal the force required for acceleration:
μ ⋅ m ⋅g = m⋅a
Simplifying the equation (the mass 𝑚 m cancels out):
a = μ ⋅ g
Substitute the values:
a = 0.15 ⋅ 10 m/s ² 
a = 1.5 m/s ²
Thus, the maximum acceleration of the car is 1.5 m/s².

Ans: (d)
Solution: 
First part of the motion (from u to u/3): We know the bullet decelerates uniformly as it passes through the block. Using the kinematic equation:
v² = u² + 2as
Where:

• v = u/3 is the final velocity after traveling 24 cm,
• u is the initial velocity,
• s = 24 cm is the distance traveled,
• $\mathrm{<span\; style="background-color:\; transparent;">a\; is\; the\; deceleration.</span>}$

Substituting the values:
Simplifying:
Rearranging to find a:
2. Second part of the motion (from u/3 to 0): In this part, the bullet decelerates from u/3 to 0. Using the kinematic equation again:
Substituting a= -u²/54.
Solving for s₂:
3. Total length of the block: The total length of the block is the sum of the distances traveled in both parts of the motion:
Total length = 24 cm + 9 cm = 33 cm

Ans: (d)
Solution: Given:
1. Scalar Product (Dot Product): 
The dot product of two vectors  and  is given by:
Substitute the components of

So, the scalar product is 10.
2. Vector Product (Cross Product):
The cross product of two vectors  is given by the determinant of the following matrix:
Substitute the components of
Now, calculate the determinant:





The vector product is
Magnitude of the Vector Product:
The magnitude of

Thus, the correct answer is: (d) 10, √2

Ans: (b)

Ans: (b)
Solution: 
Given:

Maximum load (lift + passengers) = 2000 kg

Constant speed v=1.5m/s

Frictional force f = 3000N

Gravitational acceleration g = 10m/s²

Step-by-Step:
Calculate the force required to lift the load: The force required to lift the load at a constant speed is equal to the gravitational force acting on the lift:

Total force acting on the lift: The total force the motor must overcome includes both the gravitational force and the frictional force:

Calculate the power: Power is given by the product of force and velocity:
Substituting the values:
P = 23000N × 1.5m/s = 34500W
Thus, the minimum power delivered by the motor to the lift is 34500 W.

Ans: (d)
Solution:
Given: 

Mass of the ball, m = 0.15kg 

Initial height, h = 10m 

Gravitational acceleration, g = 10m/s²  
Velocity just before impact: The ball falls from a height of 10 meters, and we can use the kinematic equation to calculate the velocity just before it hits the ground:

Substituting the values:

This is the velocity just before hitting the ground (downward).

Velocity after rebounding: Since the ball rebounds to the same height, the velocity after rebounding is also 14.14 m/s, but in the opposite direction (upward).
Change in velocity: The change in velocity Δv is the difference between the final velocity and the initial velocity:

Impulse: Impulse J is the change in momentum, which is given by:
J = m ⋅ Δv
Substituting the values:
J = 0.15kg × 28.28m/s = 4.242 kg⋅m/s

Ans: (a)
Solution:
1. Identify the forces acting on the masses:

• For the 4 kg mass (m₁): The force acting on it is F₁ = m₁g.
• For the 6 kg mass (m₂): The force acting on it is F₂ = m₂g.

2. Calculate the net force: The net force acting on the system is due to the difference in the weights of the two masses. The heavier mass (6 kg) will pull the lighter one (4 kg) upwards while it descends. 
F_net = m₂g − m₁g = (6g − 4g) = 2g
3. Total mass of the system: The total mass of the system is the sum of the masses of both bodies:
M_total = m₁ + m₂ = 4 kg + 6 kg = 10 kg
4. Calculate the acceleration: Using Newton’s second law, the net force is related to the total mass and the acceleration of the system:
F_net = M_total ⋅ a
Substituting the values:
2g = 10a
Solving for a:

Ans: (c)
Solution: 

Concept: For the block to remain stationary, the frictional force must balance the gravitational force. The frictional force is provided by the normal force, which in this case is the centrifugal force due to the rotation of the drum.
1. Gravitational Force (Downward): The gravitational force acting on the block is:
F_gravity = mg = 10 kg × 10 m/s² = 100N
2. Frictional Force (Upward): The frictional force must balance the gravitational force:
F_friction = μ N
Where N is the normal force, which is the centrifugal force in this case.
3. Centrifugal Force: The centrifugal force acting on the block due to the rotation is:
N = mω²r
Where ω is the angular velocity, and r is the radius of the cylinder.
4. Balance of Forces: For the block to remain stationary, the frictional force must balance the gravitational force:
F_friction = F_gravity 
Substituting the expressions:
μN = mg 
μ(mω²r)=mg
Simplifying:
μω²r=g
Solving for ω:

Substitute the Values: Substituting g = 10m/s², μ=0.1, and r = 1m:

Ans: (c)
Solution: 
Concept: The tension in the wire is responsible for keeping the mass moving in a circular path. At different points in the circle, the forces on the mass vary due to the direction of gravity. The wire will break when the tension in the wire is at its maximum.
1. At the highest point:

• The forces acting are the tension in the wire T top  and the weight of the mass mg, both acting downward.
• The net force provides the centripetal force to keep the mass moving in a circle.
• The tension is given by: 
• 

Comparison:

• At the highest point, the tension is lower because gravity acts in the same direction as the centripetal force, reducing the required tension.
• At the lowest point, the tension is greater because the wire must support the mass’s weight and provide the centripetal force. This is where the tension is maximum.

Since the wire is more likely to break at the point where the tension is highest, the wire is most likely to break when the mass is at the lowest point of the vertical circle.

Ans: (c)
Solution:
Concept: The object is projected with the same initial velocity on two inclined planes. 

The distances x₁ and x₂ are dependent on the components of velocity along and perpendicular to the inclined plane, as well as the effect of gravity along the plane.

1. Distance traveled on an inclined plane: The distance traveled by an object on an inclined plane is influenced by the acceleration due to gravity acting along the plane. The distance can be expressed using the kinematic equation for motion along an inclined plane:
where:

• v₀  is the initial velocity, 
• θ is the angle of inclination, 
• g is the acceleration due to gravity.

2. For the first case (angle θ = 60^∘): The distance x₁ along the plane when the angle is 60° is given by:

Simplifying the trigonometric values:

Thus:
For the second case (angle θ = 30^∘): The distance x₂ along the plane when the angle is 30° is given by:
Simplifying the trigonometric values:
Thus:
4. Finding the ratio x₁  : x₂  : Now, divide x₁  by  x₂ :
Therefore:
x₁ : x₂ = 1: √3

Ans: (d)
Solution: (a) Rolling friction is smaller than sliding friction: 

This is a correct statement. Rolling friction is generally much smaller than sliding friction because the contact area in rolling is reduced, and there is less interlocking between surfaces compared to sliding.

(b) Limiting value of static friction is directly proportional to normal reaction: 
This is also correct. The limiting value of static friction f_s is given by: f_s = μ_sN
where μ s  is the coefficient of static friction and N is the normal reaction force. So, static friction is directly proportional to the normal force.
(c) Frictional force opposes the relative motion:
This is a correct statement. The frictional force always acts in the direction opposite to the relative motion or the attempted relative motion between two surfaces.
(d) Coefficient of sliding friction has dimensions of length:
This is incorrect. The coefficient of friction, whether static or sliding, is a dimensionless quantity. It is simply a ratio of the frictional force to the normal force and does not have any physical dimensions.

Ans: (d)
Solution: 
Key Considerations:

• The wedge is smooth, meaning no friction acts between the block and the wedge.
• The block must remain stationary relative to the wedge, implying the net force on the block must be directed perpendicular to the wedge surface.

Forces acting on the block: 

• Gravitational force: The gravitational force acting on the block is mg, which acts vertically downward.
• Normal force: The normal force N acts perpendicular to the inclined plane. 
• Pseudo force due to acceleration: Since the wedge is accelerating to the right with acceleration a, a pseudo force ma acts on the block towards the left (relative to the wedge), parallel to the incline.

Resolving forces:

• The gravitational force can be resolved into two components:
  • Parallel to the inclined plane: mgsinθ
  • Perpendicular to the inclined plane: mgcosθ
• The pseudo force ma acts parallel to the wedge, providing a component in the opposite direction along the incline.

For the block to remain stationary: The net force along the incline must be zero. This means the force component along the incline due to the pseudo force must balance the force component along the incline due to gravity:

macosθ=mgsinθ
Simplifying for a:
a = gtanθ
Thus, the relationship between the acceleration $aa$a of the wedge and the angle of inclination $\mathrm{<br\; style="font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit;\; word-break:\; break-word;\; line-height:\; 1.8;\; font-size:\; 20px;\; color:\; black;">}\backslash theta$θ is: a = gtanθ

As a result, the acceleration of block A will be due to gravity, downward: a_A = g
Block B (mass m):

Before the string is cut, block B is held by the string and experiences the tension. Once the string is cut, block B is suspended by the spring.

Since the spring is initially at rest (no further tension due to the release of block A), block B will accelerate upward due to the spring force. The force exerted by the spring is equal to the weight of block A, which was pulling the spring before the cut.

The spring provides a restoring force equivalent to the weight of block A, which is 3mg.

This force will accelerate block B upwards. Using Newton's second law: F = ma ⟹ 3mg = ma_B 
Therefore, the acceleration of block B immediately after the cut is:Conclusion: 

• The acceleration of block A is g (downward). 
• The acceleration of block B is 3g (upward).

Ans: (d)
Solution: Key Concepts:

• When an object moves in a circular path, the net force directed towards the center (centripetal force) is responsible for keeping the object in circular motion.
• The centripetal force  F_c required to keep a particle of mass 𝑚 m moving in a circle of radius l at a speed v is given by:
  This force is directed towards the center of the circle.
• Tension in the String: The tension in the string, T, provides the necessary centripetal force to keep the particle moving in a circular path.
• Net Force Towards the Center: The net force on the particle, which is directed towards the center, must account for both the tension in the string and the centripetal force. The tension in the string must balance the centripetal force entirely. Therefore, the tension in the string T equals the centripetal force mv²/l.
• Thus, the net force acting on the particle towards the center is: T = mv²/l
  Conclusion: The net force on the particle, which is directed towards the center, is simply the tension T.

Ans: (a)
Solution: 
The question is illustrated in the figure below,

Let, the tension at point A be TA.
Using Newton's second law, we have

Energy at point C is,

At point C, using Newton's second law,

In order to complete a loop, T_c ≥ 0
so,

From equation (i) and (ii)
Using the principle of conservation of energy,

Ans: (b)
Solution:
Given, mass of particle. m = 0.01 kg
Radius of circle along which particle is moving , r = 6.4 cm
Kinetic energy of particle, K.E. = 8 x 10^-4 J

Given that, KE of particle is equal to 8 x 10^-4 J by the end of second revolution after the beginning of the motion of particle.
It means, initial velocity (u) is 0 m/s at this moment.
Now, using the Newton's 3rd equation of motion,

Ans: (c)
Solution: Given:

Mass of lighter stone = m, radius of its circular path = r.

Mass of heavier stone = 2m, radius of its circular path = r/2.

The centripetal forces on both stones are the same.

Let the tangential speed of the lighter stone be v₁ and that of the heavier stone be v₂.

1. Centripetal force formula: The centripetal force F_c acting on a body moving in a circular path is given by:
where:

• m is the mass of the stone,
• v is the tangential speed,
• r is the radius of the circular path.

2. Centripetal force for the lighter stone: For the lighter stone of mass m and radius r, the centripetal force is:

3. Centripetal force for the heavier stone: For the heavier stone of mass 2m and radius r/2, the centripetal force is:
4. Setting the centripetal forces equal: Since both stones experience the same centripetal force:
F₁ = F₂
Substituting the expressions for F₁  and F₂:
Simplifying:
Taking the square root of both sides:
v₁ = 2v₂ 
Conclusion: 
The tangential speed of the lighter stone is 2 times the tangential speed of the heavier stone. Thus, the value of n is 2.