Test: Laws of Motion II - JEE MCQ

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 

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)

As the moving elevator is a non inertial frame hence newton's laws can’t be applied directly to it. So to apply Newton's laws we need to add a pseudo force to the man's body equal to mass times the acceleration of lift in the opposite direction to that of acceleration. Thus the balancing normal force is equal to the weight of the man +  mass times the acceleration which is,
Reading = Normal force = 700 + 70 x 5
= 700 + 350
= 1050 N

We know that Impulse due to force, I = FΔt
As F = 100N
And Δt = 1sec
We get I = 100 Ns

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

Rocket works on the principle of conservation of momentum. Rocket ejaculates gases in backward direction which creates momentum of the gases backwards and thus by conservation of momentum, the rocket gets a momentum in the forward direction making it to move forward.

According to Newton's 3 rd law, every action and reaction have same magnitude and opposite direction. So the angle between them should be 180^o.

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

"Every action has an equal and opposite reaction” -Newton's 3rd Law

Simply by conserving the momentum of the system we get that,
0 (initial momentum) = 12 x 4 + 4 x v (final momentum)
Thus we get v = -12 m/s

We know that I = P, where P is momentum
As subtracting initial momentum from the final momentum won't affect its unit, we get unit if I is the same as that of P.

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

According to newton's first law of motion, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external net force.
Therefore The force required to keep the body moving with the same velocity is zero.

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.

• It is usually seen that during rainy days we tend to slip because rainwater acts as a lubricant between our feet and the ground.  

• This lubrication (rainwater) converts dry friction into fluid friction. 

• That is, the friction between feet and ground reduces, making us slip.

Therefore, due to less friction, we slip on a rainy day.

When the bus moves, the passenger's body comes into a state of motion but when the bus stops the lower part of the body which is in contact with the floor comes into a state of rest whereas the upper part of the body still remains in the state of motion and because of this the upper part of the body falls in the forward direction. This is due to the presence of inertia. 

Hence the correct answer is C.

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.