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Laws of Motion - Newton's Laws of Motion

Grade 11ICSEPhysics

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

🔑Concepts

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Newton's First Law of Motion: An object remains in its state of rest or of uniform motion in a straight line unless compelled by an external force to change that state. This is also known as the Law of Inertia.

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Linear Momentum: The product of the mass mm and velocity v⃗\vec{v} of an object. It is a vector quantity given by p⃗=mv⃗\vec{p} = m\vec{v}.

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Newton's Second Law of Motion: The rate of change of linear momentum of a body is directly proportional to the applied force and takes place in the direction of the force. Mathematically, F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}.

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Newton's Third Law of Motion: To every action, there is always an equal and opposite reaction. If body AA exerts a force F⃗AB\vec{F}_{AB} on body BB, then BB exerts a force F⃗BA=−F⃗AB\vec{F}_{BA} = -\vec{F}_{AB} on AA.

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Impulse: The product of the average force and the time interval for which it acts, equal to the change in momentum. J⃗=F⃗avgΔt=Δp⃗\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}.

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Law of Conservation of Linear Momentum: In the absence of an external force, the total linear momentum of a system remains constant (sump⃗i=constant\\sum \vec{p}_i = \text{constant}).

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Apparent Weight in a Lift: When a lift accelerates upwards with acceleration aa, the apparent weight RR is m(g+a)m(g + a). When it accelerates downwards, R=m(g−a)R = m(g - a).

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Friction: The opposing force that comes into play when one body slides or tends to slide over another. Static friction fs≤μsRf_s \le \mu_s R and kinetic friction fk=μkRf_k = \mu_k R.

📐Formulae

p⃗=mv⃗\vec{p} = m\vec{v}

F⃗=ma⃗\vec{F} = m\vec{a}

J⃗=∫t1t2F⃗dt=p⃗2−p⃗1\vec{J} = \int_{t_1}^{t_2} \vec{F} dt = \vec{p}_2 - \vec{p}_1

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

vrecoil=−mbulletvbulletMgunv_{recoil} = -\frac{m_{bullet} v_{bullet}}{M_{gun}}

R=m(g±a)R = m(g \pm a) (Apparent weight in a lift)

fk=μkRf_k = \mu_k R

Acceleration of connected masses (Pulley):a=(m2−m1)gm1+m2\text{Acceleration of connected masses (Pulley)}: a = \frac{(m_2 - m_1)g}{m_1 + m_2}

💡Examples

Problem 1:

A bullet of mass 20 g20\text{ g} is fired from a gun of mass 5 kg5\text{ kg} with a velocity of 400 m/s400\text{ m/s}. Calculate the recoil velocity of the gun.

Solution:

Using the Law of Conservation of Momentum: mgvg+mbvb=0m_g v_g + m_b v_b = 0. Given mb=0.02 kgm_b = 0.02\text{ kg}, vb=400 m/sv_b = 400\text{ m/s}, and mg=5 kgm_g = 5\text{ kg}. Therefore, 5×vg+(0.02×400)=0  ⟹  5vg=−8  ⟹  vg=−1.6 m/s5 \times v_g + (0.02 \times 400) = 0 \implies 5 v_g = -8 \implies v_g = -1.6\text{ m/s}.

Explanation:

The negative sign indicates that the gun moves in the direction opposite to the bullet. This is a direct application of Newton's Third Law and the conservation of momentum.

Problem 2:

A block of mass 10 kg10\text{ kg} is kept on a horizontal surface where the coefficient of static friction μs=0.4\mu_s = 0.4. Find the force of friction if a horizontal force of 30 N30\text{ N} is applied.

Solution:

First, calculate the maximum static friction (limiting friction): fms=μsR=μsmg=0.4×10×9.8=39.2 Nf_{ms} = \mu_s R = \mu_s mg = 0.4 \times 10 \times 9.8 = 39.2\text{ N}. Since the applied force F=30 NF = 30\text{ N} is less than fmsf_{ms}, the body does not move.

Explanation:

In the static region, the force of friction is self-adjusting and equal to the applied force until the limiting friction is reached. Therefore, the force of friction is 30 N30\text{ N}.

Problem 3:

A man of mass 60 kg60\text{ kg} stands on a weighing scale in a lift which is accelerating upwards at 2 m/s22\text{ m/s}^2. What will be the reading on the scale? (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

For upward acceleration, the apparent weight R=m(g+a)R = m(g + a). Substituting values: R=60(10+2)=60×12=720 NR = 60(10 + 2) = 60 \times 12 = 720\text{ N}. The reading in kgf would be R/g=72 kgfR/g = 72\text{ kgf}.

Explanation:

When the lift accelerates upwards, the pseudo force acts downwards, increasing the normal reaction force recorded by the scale.

Newton's Laws of Motion - Revision Notes & Key Formulas | ICSE Class 11 Physics