krit.club logo

Force and Laws of Motion - Newton’s Third Law of Motion

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Newton’s Third Law of Motion states that to every action, there is always an equal and opposite reaction.

•

Action and reaction forces always act on two different bodies. Therefore, they do not cancel each other out.

•

If object AA exerts a force FABF_{AB} on object BB, then object BB simultaneously exerts a force FBAF_{BA} on object AA, such that FAB=−FBAF_{AB} = -F_{BA}.

•

Though the action and reaction forces are equal in magnitude, they may produce accelerations of different magnitudes because they act on objects with different masses, according to a=Fma = \frac{F}{m}.

•

The Law of Conservation of Momentum states that the total momentum of an isolated system remains constant if no external unbalanced force acts on it.

•

Recoil of a gun is a common application: when a bullet is fired, it exerts an equal and opposite force on the gun, causing it to move backwards.

📐Formulae

FAB=−FBAF_{AB} = -F_{BA}

m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Recoil Velocity of Gun (vg)=−mbvbmg\text{Recoil Velocity of Gun } (v_g) = -\frac{m_b v_b}{m_g}

Momentum (p)=m×v\text{Momentum } (p) = m \times v

💡Examples

Problem 1:

A bullet of mass 20 g20\text{ g} is horizontally fired with a velocity 150 m/s150\text{ m/s} from a pistol of mass 2 kg2\text{ kg}. What is the recoil velocity of the pistol?

Solution:

Given: Mass of bullet m1=20 g=0.02 kgm_1 = 20\text{ g} = 0.02\text{ kg} Velocity of bullet v1=150 m/sv_1 = 150\text{ m/s} Mass of pistol m2=2 kgm_2 = 2\text{ kg} Recoil velocity of pistol = v2v_2

According to the Law of Conservation of Momentum (initial momentum = final momentum): m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 Since initial velocities u1u_1 and u2u_2 are 00: 0=(0.02×150)+(2×v2)0 = (0.02 \times 150) + (2 \times v_2) 0=3+2v20 = 3 + 2v_2 2v2=−32v_2 = -3 v2=−1.5 m/sv_2 = -1.5\text{ m/s}

The recoil velocity is −1.5 m/s-1.5\text{ m/s}.

Explanation:

The negative sign indicates that the direction in which the pistol recoils is opposite to the direction of the bullet. This is a direct consequence of Newton's Third Law and conservation of momentum.

Problem 2:

Calculate the total momentum of two objects before collision if object A (2 kg2\text{ kg}) is moving at 5 m/s5\text{ m/s} and object B (3 kg3\text{ kg}) is moving at 2 m/s2\text{ m/s} in the same direction.

Solution:

Total initial momentum Ptotal=mAuA+mBuBP_{total} = m_A u_A + m_B u_B Ptotal=(2×5)+(3×2)P_{total} = (2 \times 5) + (3 \times 2) 10+616\begin{array}{r} 10 \\ + 6 \\ \hline 16 \end{array} Ptotal=16 kg m/sP_{total} = 16\text{ kg m/s}

Explanation:

The momentum of the system is the sum of the individual momenta of the objects. Since they are moving in the same direction, both velocities are positive.