Solved Examples: Newton’s Laws of Motion | Physics for JEE Main & Advanced PDF Download

Q1: A spherical shell is resting against the vertical wall which makes an angle 30° with the vertical as shown in the Figure. Determine the normal reaction at the wall and tension in the string.
Sol:
Draw the FBD of the sphere. Resolve the forces in horizontal and vertical directions. Apply Newton’s first law along the horizontal and vertical directions.
FBD of the sphere is provided. Since sphere is in equilibrium, hence,
N – T sin30 = 0               … (i)
and  ⇒ T cos 30= mg   ...(ii)


Q2: Two blocks each having mass of 20 kg rest on frictionless surfaces as shown in the Figure. Assume that the pulleys to be light and frictionless. Now, find (a) the time required for the block A to move 1 m down the plane, starting from rest and (b) the tension in the cord connecting the blocks.
Sol:
For block A apply Newton’s second law of motion along the inclined and for block B apply Newton’s second law along the horizontal.
Both the blocks A and B are considered as two independent systems. The FBDs for the blocks A and B are shown in the Figure and
Figure where T is tension in the string.
m_Ag sin θ − T = m_Aa   ......(i)
N = m_Agcos θ    ...(ii)
T = m_B a   ... (iii)
Now, by adding equations (i) and (iii), we obtain


Q3: A body of mass = 2 kg starts from rest whose force time graph is shown in the following graph.
(a) What is momentum of the body at t = 4 seconds?
(b) What is velocity of the body at t = 3 seconds
Sol: Area under the force-time graph and the time axis is equal to the change in momentum.
(a) Area under the curve from t = 0→2 sec.
A = 2 × 2= 4N. sec . Area from t = 2→3 sec.

Area from t = 3→4 sec.

Therefore, the net impulse = 4 + 1 – 1.5 = 3.5 N sec
P_f = impulse + P_i = 3.5 + 0 = 3.5 N. s or kg.m / s
(b) Impulse  from t = 0→3 sec
= A + A = 4 + 1= 5N. sec
Momentum at t = 3 sec =5N sec (at t = 0, P = 0)
 

Q4: Masses M and 2M are connected through pulleys A and B with strings as shown in the Figure. Assume that both the pulleys and the strings are light and all the surfaces are frictionless.
(a) Find the acceleration of the block of mass M.
(b) Find the tension in the string.
(c) Calculate the force exerted on the clamp.
Sol: 
To find the constraint relation between accelerations of blocks M and 2M, measure all distances from the fixed pulley A. Apply Newton’s second law in horizontal direction for block 2M and Newton’s second law in vertical direction for block M.
(a) Let L be the length of the string. Let x₁ be the length of the vertical string and x₂ be the length of each string in the horizontal direction. The constraint relation for the string of length L is x₁ + 2x₂ = L Now, by differentiating twice, a₁ + 2a₂ = 0
If a₁ is +ve, then a₂ is –ve,

Let a₁ = a be the acceleration of M and a/2 be the acceleration of 2M.

Now, by substituting for T in (i)


Q5: A rod AB rests with the end A on rough horizontal ground and the end B against smooth vertical wall. The rod is of uniform length and of weight W. If the rod is in equilibrium in the position shown in the Figure, then find:
(a) Frictional force at A
(b) Normal reaction at A
(c) Normal reaction at B.
Sol:
For translational equilibrium, the vector sum of all the forces acting on the rod is zero. Take component of forces along horizontal (x-axis) and vertical (y-axis) direction. Sum of components of forces along the x and y axes will be zero. For rotational equilibrium, the net torque of all the forces acting on the rod relative to a fixed point (say O) is zero.
Let the length of the rod be 2l. Using the three conditions of equilibrium, the anticlockwise moment is taken as positive.

Solving the above three equations, we obtain


Q6: In the adjacent Figure, masses of A, B and C are 1 kg, 3 kg and 2 kg, respectively. Find (a) the acceleration of the system and (b) tension in the string. Neglect friction (g = 10 m/ s²)
Sol:
Draw the FBD of each block and apply Newton’s second law along the incline plane for each block.
(a) In this case, the net pulling force

Therefore, the total mass being pulled = 1+3+2 = 6 kg
∴ Acceleration of the system
(b) For the tension in the string between A and B.

For the tension in the string between B and C. FBD of C, T₂ − m_cg sin30°= m_ca


Q7: In the system of two pulleys connected as shown in the figure, M₁ = 4M₂ and mass M₁ is 20 cm above the ground, whereas mass M₂ is lying on the ground. Find the distance covered by  when the system is released. (g = 10 m/ s²).
Sol: To find the constraint relation between accelerations of M₁ and M₂, measure their distances from fixed pulley A. Apply Newton’s second law in vertical direction for each block.
M₁ = 4M₂
As M₁ is heavier, itwill move down with acceleration a and M₂ will move upward with acceleration 2a because the strings around the pulley B will move through half the distance as compared to that of A.

Therefore, the time taken for M₁ to reach the ground at

Velocity of M₂ after 0.4 seconds =v=u+2at

Distance covered by M₂ with velocity 2ms^-1 upwards before coming to rest

Distance covered by M₂ before coming to rest
= x = x1 + x = 0.4 + 0.2 = 0.6 m

Q8: A pendulum is hanging from the ceiling of a car having acceleration a₀ with respect to the ground. Find the angle made by the string with the vertical.
Sol: In the reference frame of the car the pendulum bob will experience a pseudo force. For the bob to be in equilibrium, the vector sum of all the forces acting on it in the frame of the car should be zero.
The situation is shown in the Figure. Suppose that the mass of the bob is m and string makes an angle θ with the vertical. We shall now proceed based on the car frame. This frame is non-inertial as it has acceleration  with respect to an inertial frame (the road). Hence, if we use Newton’s second law we shall have to include a pseudo force.
Now, consider the bob as the complete system.
Then, the forces acting on it are:
(a) T along the string, by the string
(b) mg downward, by the earth
(c) m a₀ towards left (pseudo force).
The FBD is shown in the Figure. As the bob is at rest (remember we are discussing the motion with respect to car) the force in (a), (b) and (c) should add to zero. Take the X-axis along the forward horizontal direction and the Y-axis along the upward vertical direction.
The components of the forces along the X-axis give
T sin θ− ma₀ = 0 or , T sin θ = ma₀ .....(i)  
And the components along the Y-axis give
T cos θ− mg = 0 or , T cos θ = mg ......(ii)
Dividing (i) by (ii) tan θ= a₀ / g.
Thus, the string makes an angle tan⁻¹ (a₀ / g) with the vertical

Q9: A man is standing on a weighing machine placed in a lift, when stationary, his weight is recorded as 40 kg. If the lift is accelerated upwards with an acceleration of 2m/s², then the weight recorded in the machine will be
(a) 48 kg
(b) 32 kg
(c) 64 kg
(d) 80 kg
Ans: (a)
We have given that the man is standing on the weighing machine placed in the lift which is accelerated upwards with acceleration a = 2m/s². We know that the weight of the person is determined by the normal force provided by the ground. Let us draw the free body diagram of the accelerated frame of lift as follows,
From the above free body diagram, we can write,
N − mg = ma
⇒ N = m(a+g)
Here, m is the mass of the person, a is the acceleration of the lift and g is the acceleration due to gravity.
Substituting m = 40kg , a =2m/s² and g = 10m/s² in the above equation, we get,
N = (40)(2+10)
⇒ N = 480N
Thus, the weight of the person is 480 N. But the weight recorded in the machine will be,

∴ m = 48kg
Thus, the weight recorded in the weighing machine will be 48 kg.

Q10:  A bullet of mass 0.1 kg moving horizontally with speed 400 ms⁻¹ hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is __________.
(Given g=10 m/s²)
(a) 0.65
(b) 0.25
(c) 0.50
(d) 0.90
Ans: (b)
First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.
The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.
Setting initial momentum equal to final momentum:

Solving for (v_final)

Substituting the given values:

Next, we know the block comes to rest after moving 20 m due to friction. The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0). The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).
So, setting the work done by friction equal to the initial kinetic energy of the block:



Q11: At any instant the velocity of a particle of mass 500 g is  ms^-1. If the force acting on the particle at t = 1 is  N. Then the value of x will be:
(a) 2
(b) 4
(c) 6
(d) 3
Ans: (d)
Given the velocity vector of a particle , the acceleration a is the derivative of the velocity vector with respect to time.
So, we have:

At t = 1 s, the acceleration a is

According to Newton's second law, the force F is equal to the mass m times acceleration a. The mass m is given as 500 g, or equivalently, 0.5 kg.
Therefore, the force F on the particle at t = 1 s is:

So, the force acting on the particle at  N, where x — 3.
Therefore, the answer is x = 3.

Q12: As shown in figure, a 70 kg garden roller is pushed with a force of F → = 200 N at an angle of 30^∘ with horizontal. The normal reaction on the roller is
(Given g=10 m s⁻²)
(a) 800 N
(b) 600 N
(c) 200√3 N
(d) 800 √2N
Ans: (a)
Normal reaction = 70 g + Fcos 60
= 700 + 100
= 800 N

Q13: An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s²)
(a) 90 N
(b) 240 N
(c) 30 N
(d) 300 N
Ans: (d)
Torque balance about  'O'




Q14: As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s²)
(a)
(b)
(c)
(d)
Ans: (d)
Let's consider T tension in string and a acceleration of blocks.
FBD of 4 kg
4g sinθ — T = 4a

FBD of 1 kg

From (1) and (2), we get




Q15: Given below are two statements:
Statement I: An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable.
Statement II: Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed.
In the light of the above statements, choose the correct answer from the options given below:
(a) Both Statement I and Statement II are false
(b) Both Statement I and Statement II are true
(c) Statement I is false but Statement II is true
(d) Statement I is true but Statement II is false
Ans: (d)
Statement I says that when the weight of an elevator is balanced with the tension of its cable, it can move up or down with a uniform speed. This is true because the weight of the elevator is balanced by the tension in the cable, which allows it to move smoothly and at a constant speed.
Statement II says that the force exerted by the floor of an elevator on a person's foot is greater than their weight when the elevator goes down with increasing speed. This is false because the force exerted by the floor on a person's foot is equal to their weight, regardless of the speed of the elevator. The person's weight is a constant force and does not change with the speed of the elevator. The apparent weight of a person may change with the speed of the elevator, but this is due to the effects of acceleration and not an increase in the force exerted by the floor.