Laws of Motion I - NEET Physics Class 11 Free MCQ Test with solutions

Given:
Mass of the car (m) = 1000 kg
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Time (t) = 4 s

Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can rearrange the equation to solve for the acceleration.

0 = 20 m/s + a * 4 s
a = -5 m/s² (negative sign indicates deceleration)

Since the car comes to a stop, its final velocity is 0 m/s. We can use Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration, to find the magnitude of the force.

F = 1000 kg * (-5 m/s²)
F = -5000 N

The magnitude of the force is 5000 N. However, since the force is acting in the opposite direction to the car's motion (deceleration), we take the magnitude as the answer.

Therefore, the correct answer is A: 5000N.

Given : Mass, m = 3 kg;

Initial Velocity, u = 2m/s;

Final velocity, v = 3.5 m/s

Time taken, t = 10 seconds

v = u+at

3.5 = 2 +a x 10

F = ma = 3 x 0.15 = 0.45 N

Since the applied force increase the speed of the body, it acts along the direction of motion.

When the acceleration of bob is horizontal, net vertical force on the bob will be zero.
T cos θ – mg = 0
The tangential force at that instant is 

Consider the whole system (painter + box) of total mass 150 kg.

Net force = total mass × acceleration = 150 × 5 = 750 N upward.

The total upward force provided by the two tensions in the rope is 2T (since the rope passes over the pulley and supports both painter and box).

Weight of the system = 150 × 10 = 1500 N downward.

Applying Newton's second law in vertical direction:

2T - 1500 = 750 ⇒ 2T = 2250 ⇒ T = 1125 N.

Now, for the painter alone:

Weight of painter = 100 × 10 = 1000 N downward.

Upward forces on painter are tension T and normal reaction N from the floor.

Applying Newton's second law for painter (mass 100 kg, acceleration 5 m/s² upward):

T + N - 1000 = 100 × 5 = 500 ⇒ N = 500 + 1000 - 1125 = 375 N.

Thus, the tension in the rope is 1125 N and the force of contact between the painter and the floor is 375 N.

Planck's Constant (h)  = 6.626176 x 10-34  m2 kg/s
So, Unit of planck constant=  m² kg/s

Dimensions =M L² T ⁻¹   ________ (1)

Angular momentum l = mvr

Where, m-mass

v-velocity

r-radius

Dimensions of angular momentum  =  M L T ⁻¹ L   = M  L²  T ⁻¹  _______________ (2)
From (1) and (2). 

Planck's constant and angular momentum have the same dimensions.

Force vector follows the principle of superposition which says all the force vectors can be vectorially added if applied on one point to get the net force vector. Hence we get
F = F₁ + F₂ + F₃ 
= (-2 + 3) i + (1 - 2) j
 F = i - j = ma
Thus we get a = (i - j) /m
= (i - j) / 2

Since impulse I = F x Δt

And also Impulse I = Δp (i.e. change in linear momentum)

Change in momentum

= 0.01 x 5 – (-5)

= 0.10 N s

Impulse = 0.10 N s

Therefore 0.10 = F x 0.01

Or F = 10 N

To conserve the momentum each time a single bullet is fired,
the reverse speed gained by the gun from one bullet is 
V = 400 X .035 / 20
= 0.7 m/s
Thus total speed gained in a second is = 0.7 X 400 = 280 m/s
As total speed is gained in one second only the acceleration produced = 280 m/s²
Thus total force applied on the gun by the bullets = 20 x 280 
= 5600 N

Given:
Mass of the block (m) = 2 kg
Force applied (F) = 8 N
Time (t) = 5 s

Since no external forces are acting on the block other than the applied force, the net force is equal to the applied force. Using Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration.

8 N = 2 kg * a
a = 4 m/s²

Now, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s), a is the acceleration, and t is the time.

v = 0 + 4 m/s² * 5 s
v = 20 m/s

Therefore, the final velocity of the block is 20 m/s.

Force of inertia = ma

Let the masses be 1x and 5x

Force of inertia for 1st body= 1x * a

Force of inertia for 2nd = 5x * a

Ratio= x * a / 5x * a = 1:5

Additional Information: The inability of a body to change it’s state of rest or uniform motion is called inertia.

F = F1 + F2 + F3 = ( 20i + 30j) N+(8i - 50j) N+(2i +2 j) N

=(20+8+2)i+(30-50+2)j

F=30i - 18j

F = ma

Substituting F and m = 6 kg (given) in the above equation we get,

30i - 18j = 6*a

A = 5i - 3j

Hence (b) is correct choice

To calculate the acceleration of the object, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Given:
Mass of the object (m) = 5 kg
Initial velocity (u) = 10 m/s
Force applied (F) = -30 N (opposite direction of motion)

Using Newton's second law, F = ma,
we can rearrange the equation to solve for acceleration (a).
-30 N = 5 kg * a
a = -6 m/s² (negative sign indicates the opposite direction)

Therefore, the acceleration of the object is 6 m/s² in the opposite direction of its initial motion.

The correct answer is a) 6 m/s².

Given: 

Speed of the particle after a time 2T, 

• Newton’s first law of inertia states that an object at rest or in motion tends to stay at rest or moving with constant motion unless enacted upon by an unbalanced force. 

• The law of inertia relates to the sport of cycling due to the fact that the cyclist is in constant motion when on the cycle. 

• The motion of the bike is caused by the rider pressing down upon the pedals, which, in turn, enables the rider to accelerate on the bike. 

• As well, the cyclist will continue to move unless enacted upon by an opposing unbalanced force, such as the force of friction on the bike tires, when the bike decelerates to a stop. 

• The Law of Inertia is also displayed when the cyclist begins to move on the bicycle from a stop. 

• The action is related to the concept of Inertia for the reason that the cyclist is applying an unbalanced force to the bicycle pedals when not in motion. As a result, the bicycle would gain acceleration and therefore begin to move.

Therefore the correct answer is B. 

Let us take the cart and the blocks together as system.

Acceleration of the system,

The cart moves because of the normal reaction applied by the block ' D ' on the cart. Now considering the F.B.D. of the cart (considering horizontal forces only)
∴ N_D,can = m_car⋅ a = 4 × 6 = 24 N

The velocity of the sphere is same as that of the cube, which is given as 
Hence, acceleration of the sphere:

The spring will exert maximum force when the ball is at its lowest position. If the ball has descended through a distance x to reach the position.
mgx = 1/2Kx² or x = 2mg/K (i)
For the block B to leave contact spring force
Kx=Mg (ii)
Comparing equations (i) and (ii), m=M/2

Three forces are acting on a particle of mass m initially in equilibrium. If the first two forces R₁ and R₂ perpendicular to each other, R₃ has to balance the resultant of the other two forces. So,

Now, the force R₃ is removed
So, the particle moves with the resultant force R₁ and R₂
Now, the acceleration is

Hence, the acceleration of the particle is R₃/m.

Method I: Considering equilibrium of each part of system The whole system is in equilibrium; therefore, for each part 
From the free-body diagram of block C,T_C = 300 N.

Now consider the equilibrium of point O


From Eq. (i), we get

By Lami's theorem, we have

R + Psin60 = Mg
R = Mg − Psin60^∘
Frictional force

The body just moves when

In the free body diagram of B [Fig.]


In the free body diagram of the bob (b)
Tsinθ = ma (iii)
Tcosθ = mg (iv)
From equations (iii) and (iv), we get

For a frictionless surface,

In the presence of friction,

Dividing eq. (2) by eq. (1), we get

Put the value of N from eqs (i) and (v)

Now the minimum Force F to Just start the motion is 12 N
Now From eqs (i) and (iv) weget 75 N (by putting N₂⋅0)

The free body diagrams of two blocks are as shown in Fig,

Let the coefficient of friction be μ. Since block A slides down with constant velocity, the net acceleration is zero. From the free body diagrams:

For block B:

• Normal reaction from A on B, N₁ = mg cos 37°
• Frictional force between A and B, f₁ = μN₁ = μmg cos 37°
• Since B rests on A without slipping, frictional force f₁ acts upward along the incline, opposing motion.
• The component of weight of B down the incline is mg sin 37°.
• For B to remain at rest relative to A: Tension or frictional forces balance the component down the incline, so T = mg sin 37° + f₁

For block A:

• Normal reaction from incline on A, N₂ = N₁ + mg cos 37° = 2mg cos 37°
• Frictional force between A and incline, f₂ = μN₂ = μ × 2mg cos 37°
• Since A slides down with constant velocity, the net force along the incline is zero:

mg sin 37° = f₁ + f₂

Substitute f₁ and f₂:

mg sin 37° = μmg cos 37° + μ × 2mg cos 37° = 3μ mg cos 37°

Divide both sides by mg cos 37°:

tan 37° = 3μ

We know tan 37° ≈ 3/4, so

3μ = 3/4 ⇒ μ = 1/4

Hence, the coefficient of friction is 1/4.

for any value of M (M should be greater than zero.

The free body diagrams of two blocks are shown in Fig. Under the assumption that both the blocks are moving together,