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

Grade 9CBSE

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

🔑Concepts

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Momentum (pp) is defined as the product of the mass (mm) of an object and its velocity (vv). It is a vector quantity and its SI unit is kg⋅m/skg \cdot m/s. The mathematical expression is p=mvp = mv.

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Newton’s Second Law of Motion states that the rate of change of momentum of an object is directly 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), given by the formula F=maF = ma.

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One Newton (1N1 N) is defined as the force that produces an acceleration of 1m/s21 m/s^2 in an object of mass 1kg1 kg. Thus, 1N=1kg⋅m/s21 N = 1 kg \cdot m/s^2.

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For a constant force, the acceleration produced in an object is inversely proportional to its mass: a∝1ma \propto \frac{1}{m}. This means a heavier object requires more force to achieve the same acceleration as a lighter object.

📐Formulae

p=m⋅vp = m \cdot v

F∝ΔptF \propto \frac{\Delta p}{t}

F=m(v−u)tF = \frac{m(v - u)}{t}

F=m⋅aF = m \cdot a

a=v−uta = \frac{v - u}{t}

💡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/s3 m/s to 7m/s7 m/s. Find the magnitude of the applied force.

Solution:

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

Explanation:

The force is calculated by finding the rate of change of momentum. The change in velocity is 4m/s4 m/s over 2s2 s, resulting in an acceleration of 2m/s22 m/s^2. Multiplying this by the mass (5kg5 kg) gives a force of 10N10 N.

Problem 2:

Calculate the change in momentum of a stone of mass 2kg2 kg when its velocity increases from 15m/s15 m/s to 40m/s40 m/s.

Solution:

Initial momentum p1=m⋅u=2⋅15=30kg⋅m/sp_1 = m \cdot u = 2 \cdot 15 = 30 kg \cdot m/s. Final momentum p2=m⋅v=2⋅40=80kg⋅m/sp_2 = m \cdot v = 2 \cdot 40 = 80 kg \cdot m/s. Change in momentum Δp=p2−p1\Delta p = p_2 - p_1: 80−3050\begin{array}{r} 80 \\ -30 \\ \hline 50 \end{array} Therefore, Δp=50kg⋅m/s\Delta p = 50 kg \cdot m/s.

Explanation:

Momentum is the product of mass and velocity. To find the change, we subtract the initial momentum from the final momentum using vertical arithmetic.

Problem 3:

Which would require a greater force: accelerating a 2kg2 kg mass at 5m/s25 m/s^2 or a 4kg4 kg mass at 2m/s22 m/s^2?

Solution:

Case 1: m1=2kgm_1 = 2 kg, a1=5m/s2a_1 = 5 m/s^2. Force F1=m1a1=2⋅5=10NF_1 = m_1 a_1 = 2 \cdot 5 = 10 N. Case 2: m2=4kgm_2 = 4 kg, a2=2m/s2a_2 = 2 m/s^2. Force F2=m2a2=4⋅2=8NF_2 = m_2 a_2 = 4 \cdot 2 = 8 N. Since 10N>8N10 N > 8 N, the first case requires more force.

Explanation:

By applying Newton's Second Law (F=maF=ma) to both scenarios, we compare the resulting magnitudes of force.