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Force and Laws of Motion - Second Law of Motion (Momentum)

Grade 9CBSE

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

🔑Concepts

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Momentum (pp) is the measure of the quantity of motion contained in a body, defined as the product of its mass (mm) and velocity (vv).

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Momentum is a vector quantity; its direction is the same as the direction of velocity (vv).

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The SI unit of momentum is kg⋅m⋅s−1kg \cdot m \cdot s^{-1}.

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Newton's Second Law of Motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

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The force (FF) acting on an object is equal to the product of its mass (mm) and acceleration (aa).

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The SI unit of force is the Newton (NN), where 1N=1kg⋅m⋅s−21 N = 1 kg \cdot m \cdot s^{-2}.

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The change in momentum is also known as impulse, which occurs when a force acts over a specific time interval.

📐Formulae

p=mvp = mv

F∝ΔptF \propto \frac{\Delta p}{t}

F=kmv−mutF = k \frac{mv - mu}{t}

F=m(v−ut)F = m \left( \frac{v - u}{t} \right)

F=maF = ma

Δp=m(v−u)\Delta p = m(v - u)

💡Examples

Problem 1:

A constant force acts on an object of mass 5kg5 kg for a duration of 2s2 s. It increases the object's velocity from 3m⋅s−13 m \cdot s^{-1} to 7m⋅s−17 m \cdot s^{-1}. Find the magnitude of the applied force.

Solution:

Given: m=5kgm = 5 kg, u=3m⋅s−1u = 3 m \cdot s^{-1}, v=7m⋅s−1v = 7 m \cdot s^{-1}, t=2st = 2 s. \n Using the formula F=m(v−u)tF = \frac{m(v - u)}{t}: \n F=5(7−3)2F = \frac{5(7 - 3)}{2} \n F=5×42F = \frac{5 \times 4}{2} \n F=10NF = 10 N.

Explanation:

The force is determined by calculating the rate of change of momentum over the given time interval.

Problem 2:

Calculate the momentum of a bullet of mass 25g25 g moving with a velocity of 200m⋅s−1200 m \cdot s^{-1}.

Solution:

Given: m=25g=0.025kgm = 25 g = 0.025 kg, v=200m⋅s−1v = 200 m \cdot s^{-1}. \n Using p=mvp = mv: \n p=0.025×200p = 0.025 \times 200 \n p=5kg⋅m⋅s−1p = 5 kg \cdot m \cdot s^{-1}.

Explanation:

Momentum is the product of mass and velocity. Note that mass must be converted to the SI unit (kgkg) before calculation.

Problem 3:

Which would require a greater force: accelerating a 2kg2 kg mass at 5m⋅s−25 m \cdot s^{-2} or a 4kg4 kg mass at 2m⋅s−22 m \cdot s^{-2}?

Solution:

For the first case: F1=m1a1=2kg×5m⋅s−2=10NF_1 = m_1 a_1 = 2 kg \times 5 m \cdot s^{-2} = 10 N. \n For the second case: F2=m2a2=4kg×2m⋅s−2=8NF_2 = m_2 a_2 = 4 kg \times 2 m \cdot s^{-2} = 8 N. \n Since 10N>8N10 N > 8 N, the first case requires more force.

Explanation:

According to F=maF = ma, the force depends on both mass and the required acceleration.