Test: Newton’s Second Law of Motion - JEE MCQ

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 

For the whole system, 2T – 1500 = 150 x 5 ⇒ T = 1125 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 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

CONCEPT:

• Momentum: Momentum is the impact due to a moving object that has mass.

  Mathematically, the momentum of a moving object of mass mmm and velocity v is given as:

  Newton's second law of motion: It states that the rate of change of momentum of a body over time is directly proportional to the force applied, and occurs in the same direction as the applied force.

• The rate of change of momentum is directly proportional to the force applied by the moving body.

  EXPLANATION:

• Upon catching the ball, the velocity of the ball is suddenly forced to become zero and stop moving. Due to momentum, the impact it will have on the hands of the fielder will be very high and can hurt his hands.
• On swinging his hands backward, he increases the time in which the velocity of the ball will become zero.
• This decreases the rate of change of momentum and thus reduces the force acting on the fielder's hand. This avoids the chances of hurting hands.
• The relation between the rate of change of momentum and force applied is explained using Newton’s second law of motion.

CONCEPT:

• The weight associated with a moving lift (W), tension force on the supporting cable (T), and acceleration (a) of the lift can be categorized into 2 cases:
  

  • In case 1, the lift moves up with an acceleration of a.
  • In case 2, the lift moves down with an acceleration of a.

CALCULATION:

Given that:

• Mass of the lift, m=2000 kg
• Tension in the supporting cable, T=28000 N

Since the direction of lift motion is not mentioned, let us assume that the lift is moving upwards (case 1).

Sign convention: upward direction +ve, downward direction −ve

The weight of the lift acting downwards, W=mg=2000×10=20000 N
(Assume g=10 m/s²)

Net force acting on the lift = T−W=28000−20000=8000 N

Net force =ma
8000=2000×a

Therefore, a=8000/2000=4 m/s²

Hence, the lift is moving upwards with an acceleration of 4 m/s².

Important Points:

• The positive value of acceleration confirms that our assumption was correct, i.e., lift is moving up with an acceleration a.
• If the acceleration obtained was negative, it indicates that the assumption is wrong, i.e., lift is moving downwards with the acceleration a.

CONCEPT:

• A lift is designed to carry loads upwards easily and in a short span of time. This is achieved by accelerating the lift upwards.
  • Hence, the acceleration of the lift is positive in the upward direction and negative in the downward direction.
  • The downward motion of the lift is the stopping motion and hence it is retarding.
  • The lift motion can be simplified into two cases:

Case 1: Lift moving up with acceleration 'a'

Net force acting on the lift = 

Where F_g​ is the weight of the object of mass 'm', N₁ is the reaction force due to the mass.

Case 2: Lift moving down with acceleration 'a'

Net force acting on the lift = 

CALCULATION:

Given that the lift is moving up with acceleration 'a'.

The normal reaction by the lift floor on the mass = N₁

• Mass of object on floor = m
• Weight due to the mass m = F_g ​= mg (downwards)
• The force due to acceleration of lift = ma (upwards)

The net force = N₁​−Fg​=ma

∴ Normal reaction by the lift floor on the mass, N_​1=ma+F_g​=ma+mg=m(a+g)

From Newton’s 3rd law, the force by mass on the floor = force by the floor on the mass = m(g+a)