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Physics (SSC, Railway, Police & All State exam)Chapter Unit

Physics: Laws of Motion

Introduction to Laws of Motion

The Laws of Motion were formulated by Sir Isaac Newton in his landmark work Philosophiæ Naturalis Principia Mathematica. These laws describe the relationship between the motion of an object and the forces acting upon it.


Key Concepts

  1. Force:

    • A push or pull on an object that can change its state of motion or shape.
    • SI Unit: Newton (NNN).
    • Formula: F=maF = maF=ma
  2. Inertia:

    • The property of an object to resist changes in its state of motion or rest.
    • Proportional to the mass of the object.
  3. Momentum (ppp):

    • The product of an object’s mass and velocity.
    • Formula: p=mvp = mvp=mv
    • SI Unit: kg m/skg \, m/skgm/s.

Newton's Laws of Motion

  1. First Law (Law of Inertia):
    • An object remains at rest or in uniform motion in a straight line unless acted upon by an external force.
    • Mathematical Expression: ∑F=0(for equilibrium)\sum F = 0 \quad \text{(for equilibrium)}∑F=0(for equilibrium)
    • Examples:
      • A book lying on a table remains stationary unless pushed.
      • A moving car comes to rest when brakes are applied due to external forces like friction.

  1. Second Law (Law of Acceleration):
    • The rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction of the force.
    • Mathematical Expression: F=dpdt=maF = \frac{dp}{dt} = maF=dtdp​=ma
      • FFF: Force, mmm: Mass, aaa: Acceleration.
    • Special Cases:
      • If m=1m = 1m=1, F=aF = aF=a, i.e., force equals acceleration for unit mass.

  1. Third Law (Action-Reaction Law):
    • For every action, there is an equal and opposite reaction.
    • Mathematical Expression: Faction=−FreactionF_{\text{action}} = -F_{\text{reaction}}Faction​=−Freaction​
    • Examples:
      • A rocket moves upward due to the downward expulsion of gases.
      • While walking, we push the ground backward, and the ground pushes us forward.

Types of Forces

  1. Contact Forces:

    • Arise due to physical contact.
    • Examples: Friction, Tension, Normal Force.
  2. Non-Contact Forces:

    • Act without physical contact.
    • Examples: Gravitational Force, Electrostatic Force, Magnetic Force.

Friction

  1. Definition:

    • A resistive force that opposes the relative motion of two surfaces in contact.
    • Types:
      • Static Friction (fsf_sfs​): Prevents motion.
      • Kinetic Friction (fkf_kfk​): Acts during motion.
    • Relation: fs≤μsN,fk=μkNf_s \leq \mu_s N, \quad f_k = \mu_k Nfs​≤μs​N,fk​=μk​N
      • μs,μk\mu_s, \mu_kμs​,μk​: Coefficients of static and kinetic friction.
      • NNN: Normal force.
  2. Applications:

    • Friction between tires and the road helps in vehicle motion.
    • Excess friction causes wear and tear.

Numerical Example

Example: A 5 kg5 \, kg5kg object is pushed with a force of 20 N20 \, N20N on a rough surface with μk=0.2\mu_k = 0.2μk​=0.2. Find its acceleration.

  • Given:
    • m=5 kgm = 5 \, kgm=5kg, Fapplied=20 NF_{\text{applied}} = 20 \, NFapplied​=20N, μk=0.2\mu_k = 0.2μk​=0.2.
  • Normal Force: N=m⋅g=5⋅9.8=49 NN = m \cdot g = 5 \cdot 9.8 = 49 \, NN=m⋅g=5⋅9.8=49N
  • Frictional Force: fk=μk⋅N=0.2⋅49=9.8 Nf_k = \mu_k \cdot N = 0.2 \cdot 49 = 9.8 \, Nfk​=μk​⋅N=0.2⋅49=9.8N
  • Net Force: Fnet=Fapplied−fk=20−9.8=10.2 NF_{\text{net}} = F_{\text{applied}} - f_k = 20 - 9.8 = 10.2 \, NFnet​=Fapplied​−fk​=20−9.8=10.2N
  • Acceleration: a=Fnetm=10.25=2.04 m/s2a = \frac{F_{\text{net}}}{m} = \frac{10.2}{5} = 2.04 \, m/s^2a=mFnet​​=510.2​=2.04m/s2

Applications of Newton’s Laws

  1. Free Body Diagrams (FBD):
    • A visual representation of all forces acting on a body.
    • Steps to draw an FBD:
      • Isolate the body of interest.
      • Represent all forces with arrows pointing in their respective directions.
      • Label the forces clearly (e.g., gravitational force, normal force, applied force).

  1. Equilibrium of Forces:
    • A body is in equilibrium if:

      • The net force acting on it is zero: ∑Fx=0and∑Fy=0\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0∑Fx​=0and∑Fy​=0
      • The net torque acting on it is zero: ∑τ=0\sum \tau = 0∑τ=0
    • Example: A book resting on a table is in equilibrium as the gravitational force downward is balanced by the normal force upward.


  1. Inclined Plane:
    • When an object is placed on an inclined plane, the forces acting are:

      • Gravitational Force (mgmgmg).
      • Normal Force (NNN).
      • Frictional Force (fff).
    • Components of gravitational force:

      • Parallel to the incline: mgsin⁡θmg \sin \thetamgsinθ.
      • Perpendicular to the incline: mgcos⁡θmg \cos \thetamgcosθ.
    • Acceleration on a frictionless incline: a=gsin⁡θa = g \sin \thetaa=gsinθ


Impulse and Momentum

  1. Impulse:

    • Change in momentum due to a force acting over a short time interval.
    • Formula: J=F⋅Δt=ΔpJ = F \cdot \Delta t = \Delta pJ=F⋅Δt=Δp
    • SI Unit: N⋅sN \cdot sN⋅s or kg m/skg \, m/skgm/s.
  2. Conservation of Momentum:

    • In the absence of external forces, the total momentum of a system remains constant.
    • Formula: m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2m1​u1​+m2​u2​=m1​v1​+m2​v2​
    • Example: A gun recoils backward when a bullet is fired forward.

Work Done by a Force

  1. Definition:

    • Work is done when a force is applied on an object, and the object moves in the direction of the force.
    • Formula: W=F⋅d⋅cos⁡θW = F \cdot d \cdot \cos \thetaW=F⋅d⋅cosθ
      • FFF: Force.
      • ddd: Displacement.
      • θ\thetaθ: Angle between force and displacement.
  2. Special Cases:

    • If θ=0∘\theta = 0^\circθ=0∘: W=FdW = FdW=Fd (maximum work).
    • If θ=90∘\theta = 90^\circθ=90∘: W=0W = 0W=0 (no work done).
  3. SI Unit: Joule (JJJ).


Circular Motion

  1. Uniform Circular Motion:

    • An object moves in a circular path with constant speed.
    • Velocity changes due to continuous change in direction.
  2. Centripetal Force:

    • Force required to keep an object in circular motion.
    • Formula: Fc=mv2rF_c = \frac{mv^2}{r}Fc​=rmv2​
      • mmm: Mass of the object.
      • vvv: Velocity.
      • rrr: Radius of the circle.

Numerical Examples

  1. Example 1: A block of 10 kg10 \, kg10kg is sliding down a frictionless inclined plane at an angle of 30∘30^\circ30∘. Find its acceleration.

    • Given:
      • m=10 kgm = 10 \, kgm=10kg, θ=30∘\theta = 30^\circθ=30∘, g=9.8 m/s2g = 9.8 \, m/s^2g=9.8m/s2.
    • Acceleration: a=gsin⁡θa = g \sin \thetaa=gsinθ a=9.8⋅sin⁡30∘=9.8⋅0.5=4.9 m/s2a = 9.8 \cdot \sin 30^\circ = 9.8 \cdot 0.5 = 4.9 \, m/s^2a=9.8⋅sin30∘=9.8⋅0.5=4.9m/s2
  2. Example 2: A 5 kg5 \, kg5kg object moving at 10 m/s10 \, m/s10m/s is brought to rest by a constant force in 2 s2 \, s2s. Find the impulse and the magnitude of the force.

    • Impulse: J=Δp=m⋅ΔvJ = \Delta p = m \cdot \Delta vJ=Δp=m⋅Δv J=5⋅(0−10)=−50 N⋅sJ = 5 \cdot (0 - 10) = -50 \, N \cdot sJ=5⋅(0−10)=−50N⋅s
    • Force: J=F⋅Δt  ⟹  F=JΔtJ = F \cdot \Delta t \implies F = \frac{J}{\Delta t}J=F⋅Δt⟹F=ΔtJ​ F=−502=−25 NF = \frac{-50}{2} = -25 \, NF=2−50​=−25N

Motion of Connected Bodies

  1. Pulleys and Strings:
    • Systems with connected masses are analyzed using tension (TTT) and the acceleration of the system.
    • For two masses connected by a string over a frictionless pulley:
      • Let m1m_1m1​ and m2m_2m2​ be the masses, m1>m2m_1 > m_2m1​>m2​, and aaa be the acceleration.
      • Net force for m1m_1m1​: T−m1g=m1aT - m_1 g = m_1 aT−m1​g=m1​a
      • Net force for m2m_2m2​: m2g−T=m2am_2 g - T = m_2 am2​g−T=m2​a
      • Solving for aaa: a=(m2−m1)gm1+m2a = \frac{(m_2 - m_1) g}{m_1 + m_2}a=m1​+m2​(m2​−m1​)g​
      • Tension in the string: T=2m1m2gm1+m2T = \frac{2 m_1 m_2 g}{m_1 + m_2}T=m1​+m2​2m1​m2​g​

Uniform and Non-Uniform Motion

  1. Uniform Motion:

    • Motion with constant velocity (a=0a = 0a=0).
    • Example: A car moving at a steady 60 km/h60 \, km/h60km/h on a straight road.
  2. Non-Uniform Motion:

    • Motion with changing velocity (a≠0a \neq 0a=0).
    • Example: A freely falling body.

Dynamics of Rotational Motion

  1. Torque (τ\tauτ):

    • Rotational equivalent of force.
    • Formula: τ=r⋅F⋅sin⁡θ\tau = r \cdot F \cdot \sin \thetaτ=r⋅F⋅sinθ
      • rrr: Distance from the axis of rotation.
      • FFF: Force applied.
      • θ\thetaθ: Angle between rrr and FFF.
  2. Moment of Inertia (III):

    • Rotational equivalent of mass.
    • Formula for a point mass: I=mr2I = m r^2I=mr2
    • Examples:
      • For a solid sphere: I=25mr2I = \frac{2}{5} m r^2I=52​mr2
      • For a hollow cylinder: I=mr2I = m r^2I=mr2
  3. Angular Momentum (LLL):

    • Rotational equivalent of linear momentum.
    • Formula: L=I⋅ωL = I \cdot \omegaL=I⋅ω
      • III: Moment of inertia.
      • ω\omegaω: Angular velocity.
  4. Rotational Dynamics:

    • Newton’s second law in rotational motion: τ=I⋅α\tau = I \cdot \alphaτ=I⋅α
      • α\alphaα: Angular acceleration.

Work-Energy Principle

  1. Statement:

    • The work done on a body is equal to the change in its kinetic energy.
    • Formula: W=ΔKE=12mv2−12mu2W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2W=ΔKE=21​mv2−21​mu2
  2. Power:

    • Rate at which work is done.
    • Formula: P=Wt=F⋅vP = \frac{W}{t} = F \cdot vP=tW​=F⋅v
    • SI Unit: Watt (WWW).

Conservation Laws

  1. Conservation of Mechanical Energy:

    • In the absence of non-conservative forces: KE+PE=constantKE + PE = \text{constant}KE+PE=constant
      • Example: A pendulum converting potential energy at the highest point to kinetic energy at the lowest point.
  2. Conservation of Linear Momentum:

    • Total momentum remains constant in an isolated system.
    • Example: Explosion of a bomb where fragments move in opposite directions.
  3. Conservation of Angular Momentum:

    • Total angular momentum remains constant if no external torque acts on a system: I1ω1=I2ω2I_1 \omega_1 = I_2 \omega_2I1​ω1​=I2​ω2​
    • Example: A figure skater spinning faster by pulling in their arms.

Numerical Examples

  1. Example 1: Two masses, 5 kg5 \, kg5kg and 3 kg3 \, kg3kg, are connected by a string over a frictionless pulley. Find the acceleration and tension in the string.

    • Given:
      • m1=5 kgm_1 = 5 \, kgm1​=5kg, m2=3 kgm_2 = 3 \, kgm2​=3kg, g=9.8 m/s2g = 9.8 \, m/s^2g=9.8m/s2.
    • Acceleration: a=(m1−m2)gm1+m2a = \frac{(m_1 - m_2) g}{m_1 + m_2}a=m1​+m2​(m1​−m2​)g​ a=(5−3)⋅9.85+3=2⋅9.88=2.45 m/s2a = \frac{(5 - 3) \cdot 9.8}{5 + 3} = \frac{2 \cdot 9.8}{8} = 2.45 \, m/s^2a=5+3(5−3)⋅9.8​=82⋅9.8​=2.45m/s2
    • Tension: T=2m1m2gm1+m2T = \frac{2 m_1 m_2 g}{m_1 + m_2}T=m1​+m2​2m1​m2​g​ T=2⋅5⋅3⋅9.85+3=2948=36.75 NT = \frac{2 \cdot 5 \cdot 3 \cdot 9.8}{5 + 3} = \frac{294}{8} = 36.75 \, NT=5+32⋅5⋅3⋅9.8​=8294​=36.75N
  2. Example 2: A rotating disc has a moment of inertia of 2 kg m22 \, kg \, m^22kgm2 and an angular acceleration of 5 rad/s25 \, rad/s^25rad/s2. Find the torque acting on it.

    • Formula: τ=I⋅α\tau = I \cdot \alphaτ=I⋅α
    • Substituting values: τ=2⋅5=10 N⋅m\tau = 2 \cdot 5 = 10 \, N \cdot mτ=2⋅5=10N⋅m

Recap: Key Points to Remember

  • Newton's laws describe the motion of objects under the influence of forces.
  • Free body diagrams are essential tools for solving force-related problems.
  • Conservation laws are fundamental principles in mechanics.
  • Rotational motion has parallels to linear motion, with torque, moment of inertia, and angular momentum.

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