Test: Newton's Third Law of Motion (NCERT) - NEET MCQ

We can derive Newton’s third and first laws from the second law.

The correct option is C third and first laws from the second law
We can derive Newton's third and first laws from second law. If force on particle is zero, then according to Newton's second law

F=ma

0=ma

a=0 (Since mass cannot be zero)
So, this shows that the body moves with constant velocity or remains at rest if no external force on the system. This is the statement of first law itself.

Suppose an isolated system of two bodies A and B. Suppose A exerts force F₁ on B and B exerts F₂ on A. Now, rate of change of momentum of A is dp₁/dt and that of B is dp₂/dt. Therefore, from second law,

F₁=dp₁/dt …(i)

F₂=dp₂/dt …(ii)

Adding (��) and (����), we get

F₁+F₂=dp₁/dt+dp₂/dt

Since the system is isolated, so the total change in momentum or rate of change of momentum dp₁/dt+dp₂/dt should be zero. Thus, F₁+F₂=0 or F₁=−F₂.This is the statement of third law.

Action and reaction act on different bodies not on the same body.

By the retarding force the car can be stopped in a less distance if the driver apply brakes. This retarding force is actually a friction force.

It is easier to pull lawn mower than to push it . All other statements are true.

Newton's first law defines force. Newton's second law gives us a measure of force. Impulse gives us the effect of force. Recoiling of gun is accounted for by Newton's 3rd law.

As the weight of man increased by 5 times, so acceleration of  the rocket a = 5g = 5 x 10 = 50 m s^-2 
Force applied by rocket engine is
F = ma = 1.0 x 10⁴ x 50 = 5 x 10⁵ N

The sixth coin is under the weight of four coins baove it. Hence,
Reaction of the 6^th coin on the 7^th coin = Force on the 6^th coin due to 7^th coin
= 4mg

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Here, mass of the stone, m = 1 kg
As the stone is lying on the floor of the train, its acceleration is same as that of train.
∴ Force acting on the stone F = ma = (1kg)(1 m s^-2)
= 1 N

Here, v = 15 m s^-1
Area of cross section, A = 10^-2 m²
Density of water, p = 10³ kg m³
Mass of water hitting the wall per second
= p x A x v
=10³ kg m³ x 10^-2 m² x 15 m s^-1
= 150 kg s^-1
Force exerted on the wall = Momentum loss of water per second
= 150 kg s^-1 xg m³ x 10^-2 m² x 15 m s^-1
= 150 kg s^-1
Force exerted on the wall = Momentum loss of water per second
= 150 kg s^-1 x 15 m s^-1 = 2250 N
= 2.25 x 1000 N = 2250 N
= 2.25 x 10³ N

Here, u = 25 m s^-1, v = 0, t = 1, A = 10^-3 m²
Density of water p = 1000 kgm^-3
Mass of water gushed out per second, m
= volume x density / time = area x distance x density / time
= Area x velocity x density
= 10^-3 x 25 x 1000 = 25 kg
F =  = 625 N
F' = F = 625N

Here, m = 2 × 10⁴kg
Initial acceleration, a = 5ms^-2
Initial thrust = upthrust required to impart acceleration a + upthrust required to overcome gravitational pull
∴ F  = m(a + g) = (2 × 10⁴kg)(5 + 10)ms⁻²
= 3 × 10⁵N

Initial momentum of the ball is


Final momentum of the ball is
∴ Change in momentum, 

Impulse = Change in momentum = 
As impulse and force are in the same direction, therefore, force on the ball due to the wall is normal to the wall along the negative x-axis. Using Newton’s 3^rd law of motion the force on the wall due to the ball is normal to the wall along the positive x− direction.
∴ F = 

Acceleration of the rocket at any instant t is

Here, M = 6000kg, dm/dt = 16kgs⁻¹
Vr = 11kms^-1 = 11km s⁻¹ t = 1 min = 60s
∴ 
≈ 35 m s⁻²

Initial momentum of ball A = (0.05kg)(5ms(-1)) = 0.25kgms⁻¹
As the speed is reversed on collision,
Final momentum of the ball
A = (0.05kg)(−5ms⁻¹) = −0.25kgms⁻¹
Impulse imparted to the ball A = Change in momentum of ball A
= Final momentum - Initial momentum
= −0.25kg ms−¹ −0.25kg ms⁻¹ = −0.5kg ms⁻¹
Similarly,
Initial momentum of ball B = (0.05kg)(−5ms⁻¹) = −0.25kgms⁻¹
Final momentum of ball B = (0.05kg)(5ms⁻¹) = = +0.25kgms⁻¹
Impulse imparted to ball B = (0.25kgms⁻¹) − (−0.25kgms⁻¹) =  0.5kgms⁻¹
Impulse imparted to each ball is 0.5kgms⁻¹ in magnitude. The two impulses are opposite in direction.