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Force and Laws of Motion - Third Law of Motion

Grade 9CBSE

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

🔑Concepts

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Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. If object A exerts a force F⃗AB\vec{F}_{AB} on object B, then object B exerts an equal and opposite force F⃗BA\vec{F}_{BA} on object A.

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

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The Law of Conservation of Momentum states that the total momentum of an isolated system (where no external unbalanced force acts) remains constant or conserved.

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In a collision between two objects, the sum of momenta before collision is equal to the sum of momenta after collision: m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.

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Recoil of a Gun: When a bullet is fired from a gun, the gun exerts a forward force on the bullet (action), and the bullet exerts an equal and opposite force on the gun (reaction), causing the gun to move backwards with a recoil velocity VV.

📐Formulae

F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Vrecoil=−m⋅vMV_{recoil} = -\frac{m \cdot v}{M}

Momentum (p)=m⋅v\text{Momentum } (p) = m \cdot v

💡Examples

Problem 1:

A bullet of mass 20 g20\text{ g} is horizontally fired with a velocity 150 m s−1150\text{ m s}^{-1} 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}, Initial velocity of bullet u1=0u_1 = 0. Mass of pistol m2=2 kgm_2 = 2\text{ kg}, Initial velocity of pistol u2=0u_2 = 0. Final velocity of bullet v1=150 m s−1v_1 = 150\text{ m s}^{-1}. Let recoil velocity be v2v_2. According to the law of conservation of momentum: m1u1+m2u2=m1v1+m2v2  ⟹  0=(0.02×150)+(2×v2)  ⟹  0=3+2v2  ⟹  v2=−32=−1.5 m s−1m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies 0 = (0.02 \times 150) + (2 \times v_2) \implies 0 = 3 + 2v_2 \implies v_2 = -\frac{3}{2} = -1.5\text{ m s}^{-1}.

Explanation:

The recoil velocity is 1.5 m s−11.5\text{ m s}^{-1}. The negative sign indicates that the direction in which the pistol recoils is opposite to the direction of the bullet's motion.

Problem 2:

Two objects of masses 100 g100\text{ g} and 200 g200\text{ g} are moving along the same line and direction with velocities of 2 m s−12\text{ m s}^{-1} and 1 m s−11\text{ m s}^{-1}, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s−11.67\text{ m s}^{-1}. Determine the velocity of the second object.

Solution:

m1=0.1 kgm_1 = 0.1\text{ kg}, u1=2 m s−1u_1 = 2\text{ m s}^{-1}, v1=1.67 m s−1v_1 = 1.67\text{ m s}^{-1}. m2=0.2 kgm_2 = 0.2\text{ kg}, u2=1 m s−1u_2 = 1\text{ m s}^{-1}. Using m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2: (0.1×2)+(0.2×1)=(0.1×1.67)+(0.2×v2)  ⟹  0.2+0.2=0.167+0.2v2  ⟹  0.4−0.167=0.2v2  ⟹  0.233=0.2v2  ⟹  v2=1.165 m s−1(0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_2) \implies 0.2 + 0.2 = 0.167 + 0.2v_2 \implies 0.4 - 0.167 = 0.2v_2 \implies 0.233 = 0.2v_2 \implies v_2 = 1.165\text{ m s}^{-1}.

Explanation:

The total momentum is conserved. The second object increases its velocity from 1 m s−11\text{ m s}^{-1} to 1.165 m s−11.165\text{ m s}^{-1} after being struck by the first object.