Laws of Motion II - Free MCQ Practice Test with solutions, NEET Physics

• The stone is moving in a circular path with constant speed, this is uniform circular motion.
• In uniform circular motion, at any instant velocity (magnitude speed) acts tangentially to the circular path.
• Hence, when string is cut the stone will fly off tangentially to circular path with velocity of magnitude speed.

Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as, 
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg

Key points:

• The stone has a mass of 0.1 kg.
• The train is moving at a constant velocity (1 m/s).
• The stone is at rest relative to the train.
• We are neglecting air resistance.

Since the stone is at rest relative to the train and the train is moving with a constant accleration, the net force acting on mass .
F= ma 
F =0.1 x 1 N
F = 0.1 N

• The third law states that all forces between two objects exist in equal magnitude and opposite direction. 
• If one object A exerts a force F_A on a second object B, then B simultaneously exerts a force F_B on A, and the two forces are equal in magnitude and opposite in direction, F_A = −F_B
• Newton's third Law:
  

• Force is required to start a stationary object and to stop a moving object due to inertia. 
• Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
• This is also called law of inertia or newton's first law of motion.
• Newton's first Law:
  

• When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
• To every action there is always opposed an equal reaction or the mutual actions of two bodies upon each other are always equal and opposite in direction.

If net force act on a body then acceleration of the body will also be zero. Hence velocity will not be changed i.e. it continues in its existing state of rest or uniform motion in a straight line.

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms^-2

Force in horizontal direction 

F = ma = 0.1 x 1 = 0.1N

According to Newton's second Law of motion,
Apparent weight, R = m(a+g) 

= 70(10+10) 

= 1400N 

mass = R/g

= 1400/10 

= 140Kg

Given,
Mass of the helicopter, m_h = 1000 kg
Mass of the crew and passengers, m_p = 300 kg
Total mass of the system, m=1300 kg

Acceleration of the helicopter, a =15 m/s²

Using Newton’s second law of motion, the reaction force R:

R − m_pg = ma

R=m_p(g+a)

R=300(10+15)

R=300×25

R=7500 N

The reaction force will be directed upwards, as the helicopter is accelerating vertically upwards.

According to Newton’s third law of motion, the force on the floor by the crew and passengers =7500 N, directed downward.

• According to Newton s Second Law of Motion, also known as the Law of Force and Acceleration, a force upon an object causes it to accelerate according to the formula net force = mass x acceleration.
• It states that the force applied on a body is proportional to the rate of change of momentum of the body.
• So the acceleration of the object is directly proportional to the force and inversely proportional to the mass.

When no external forces act on a system of several interacting particles, then the total momentum of system is conserved.
Law of conservation of Momentum:

According to Newton's second Law
Force = mass x accleration
F = m.a
   = m.v/t
t  = m.v/F
   = 20x15/50
   = 6 sec

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

R = m(g + a)

= 70 (10 + 5) = 70 x 15

= 1050 N

∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg

Option A is correct.

m = 40 kg, g = 10 m/s², maximum tension = 600 N.

Taking upward as positive, the tension in the rope is given by R = m(g + a), where a is positive for upward acceleration and negative for downward acceleration.

Case (a): a = +6 m/s².

R = 40 × (10 + 6) = 40 × 16 = 640 N.

Since 640 N > 600 N, the rope will break in case (a).

Case (b): a = -4 m/s².

R = 40 × (10 - 4) = 40 × 6 = 240 N.

Since 240 N < 600 N, the rope will not break in case (b).

Case (c): a = 0 (uniform speed).

R = 40 × 10 = 400 N.

Since 400 N < 600 N, the rope will not break in case (c).

Case (d): nearly free fall ⇒ a ≈ -10 m/s².

R ≈ 40 × (10 - 10) = 0 N (tension is nearly zero in free fall).

Since 0 N < 600 N, the rope will not break in case (d).

Therefore only case (a) produces tension exceeding the limit; correct answer: A.

Newton's second law of motion gives the quantitative definition of force. The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

• According to newton's third law of motion for every action there is always equal and opposite reaction or the mutual actions of two bodies upon each other.
• Whenever two objects interact with each other, then the pair of forces acting on them are equal in magnitude and opposite in direction. Contact forces are mutual forces (action and reaction).

Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity

⇒ ma + mg = m(a+g)
⇒ 20000(5+10) = 300000 N

For some instance assume both masses as one system, thus we get that
600 = 30a 
Where a is the common acceleration of the system.
Now if we consider the 10kg block we get that
T = 10a
And a = 20m/s² 
Thus we get T = 200N

Correct option: C - 240 N.

Let m = 40 kg, g = 10 m/s², and downward acceleration a = 4 m/s².

Choose downward as the positive direction. The forces acting on the monkey are its weight mg (downward) and the tension T (upward). Applying Newton's second law along the vertical gives mg - T = m a.

Rearranging the equation gives T = m (g - a).

Substituting the values: T = 40 × (10 - 4) = 40 × 6 = 240 N.

Compare with the rope's limit: 240 N < 600 N, so the rope does not break.

When an object has several forces acting on it, the effect of force is the same as one force acting on the object in a certain direction and this overall force is called the ‘resultant force’. The resultant force is essential to change the velocity of an object.

• Friction opposes the relative motion between the surfaces in contact with each other. Friction is the force that opposes relative motion or tendency of relative motion between two surfaces in contact.
• When two surfaces move relative to each other or they have a tendency to move relative to each other, at the point (or surface) of contact, there appears a force which opposes this relative motion or tendency of relative motion between two surfaces in contact.
• It acts on both the surfaces in contact with equal magnitude and opposite directions (Newton's 3rd law). also Friction force tries to stop relative motion between two surfaces in contact.

Initial velocity of the truck, u = 0

Acceleration, a = 2 m/s²

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

v = u + at

⇒ 0 + 2 X 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

When the stone is dropped, its initial velocity is the same as the velocity of the truck at the time of release. The stone's motion can be broken into two components:

1. Horizontal motion: The stone has the same horizontal velocity as the truck when dropped, which is 20 m/s. Since there is no force acting horizontally (neglecting air resistance), its horizontal velocity remains constant at 20m/s.

2. Vertical motion: The stone experiences a downward acceleration due to gravity, which is g=9.8 m/s².

Thus, the total acceleration of the stone at any point is the combination of:

• The constant horizontal acceleration (which is 0 for the stone).
• The vertical acceleration due to gravity, which is 9.8 m/s² downward.

Therefore, at t=11 s, the total acceleration of the stone is the magnitude of the vector sum of these two accelerations:

However, since the truck is moving horizontally with an acceleration of 2.0 m/s², at t=11s, the stone will still have the same horizontal acceleration as the truck. The total acceleration of the stone is the combination of:

• The acceleration of the truck horizontally, 2.0 m/s², and
• The gravitational acceleration vertically, 9.8 m/s².

Thus, the total magnitude of the acceleration of the stone at t=11s is:

Option A is correct.

For uniform speed, the acceleration is zero, so the net force on the monkey is zero and T - mg = 0.

Therefore T = mg.

Using g = 10 m s⁻² (standard approximation used to match the given choices), T = 40 × 10 = 400 N.

Thus the tension in the rope is 400 N, which is less than the rope's maximum tension of 600 N.

Initial momentum of the ball,
P_i​ = 0
Final momentum of the ball,
P_f  = mv
⇒ 0.25×10 = 2.5 kg m/s
Impulse imparted on the ball,
I = P_f − P_i 
⇒ I = 2.5 − 0 = 2.5 N s

• It is the maximum friction, where a body just starts to move over the surface. 
• For example a large block of mass m is placed on a horizontal table. Apply a small force the block does not move due to the friction force which balances the applied force.
• Now Gradually increase the applied force until it start moving. This maximum applied force acting in opposite direction is called as Static Friction. 

At the extreme position of the oscillation, the speed of the bob is zero. So the bob is momentarily at rest. if the string is cut, the bob will fall vertically downwards.

The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).

Rolling friction is always less than static friction because in order for an object to roll the force of friction between it and surface must be large enough to keep the object from sliding. Hence rolling friction is always greater than static force.