Physics Study Notes on Force and Motion and Related Principles

Physics Study Notes

Chapter: Force and Motion

Projectile Motion

• Definition: Projectile motion is two-dimensional motion observed under constant acceleration due to gravity.

  • Examples include:
  • A football kicked by a player.
  • A ball thrown by a cricketer.
  • A missile fired from a launching pad.
  • All these projectiles are projected at angles with the horizontal.

• Horizontal and Vertical Components:

  • The motion of a projectile can be broken down into independent horizontal and vertical components.
  • If a projectile is fired at an angle $ heta$ with an initial velocity $v$:
  • Horizontal component: $v_x = v imes ext{cos}( heta)$
  • Vertical component: $v_y = v imes ext{sin}( heta)$
  • Horizontal acceleration ($a_x$) is 0 assuming no air resistance.
  • Vertical acceleration ($ay$) is given by $a</em>y = -g$ (where $g$ is the acceleration due to gravity).
  • Therefore, the horizontal component of velocity $v_x$ remains constant.

• Equations of Motion:

  • At any time $t$, for horizontal motion:

  • $V<em>x = V = V</em>x imes ext{cos}( heta)$ (Equation 2.18)

  • For vertical motion:

  • Initially, the vertical velocity is $v<em>{y0} = v imes ext{sin}( heta)$ and at any time t, the vertical component $vy$ can be calculated as:

  • $v<em>y = v</em>{y0} - gt$ (Equation 2.19)

• Magnitude of Velocity:

  • The resultant velocity (magnitude) at any instant can be found using:
  • y = ext{tan}^{-1}igg(rac{V{tx}}{V{ty}}igg) (Equation 2.21)

Determining Projectile Parameters

• Maximum Height (h):

  • To find the maximum height:

  • Initial vertical velocity when at the zenith is 0.

  • Using the equation of motion, with acceleration $a = -g$ and initial vertical velocity $v_{y0}$:

  • h = rac{v_{y0}^2 imes ext{sin}^2( heta)}{2g} (Equation 2.22)

  • Effect of air resistance:

  • The presence of air reduces the height by decreasing the upward velocity of the projectile over time.

• Time of Flight (T):

  • Defined as the total time taken for the projectile to rise and then descend back to its original launch height.
  • For projectiles launched from the ground with an angle $ heta$:
  • Vertical displacement is zero $S = h = 0$.
  • The equation becomes:
  • 0 = v_{y0}t - rac{1}{2}gt^2
  • Rearranging yields:
  • t = rac{2v imes ext{sin}( heta)}{g} (Equation 2.23)

• Range (R):

  • Defined as the maximum horizontal distance covered by the projectile.
  • The formula for range is:
  • $R = v_{x0} imes T$
  • Which can also be expressed as:
  • R = rac{v^2 imes ext{sin}(2 heta)}{g} (Equation 2.24)
  • This implies for maximum range, the angle $ heta = 45^{ ext{o}}$ (since $ ext{sin}(90^{ ext{o}})=1$).

• Effect of Air Resistance on Range:

  • Air resistance decreases velocity and subsequently reduces both the height and range of the projectile.

Example Problem:

• A ball thrown with a speed of 30 m/s at an angle of 30° above the horizon:
  • Calculate:
  • Horizontal velocity: $v_{x0} = v imes ext{cos}(30^{ ext{o}}) = 30 imes ext{cos}(30^{ ext{o}}) = 25.98 ext{ m/s}$
  • Vertical velocity: $v_{y0} = v imes ext{sin}(30^{ ext{o}}) = 30 imes ext{sin}(30^{ ext{o}}) = 15 ext{ m/s}$
  • **Maximum Height (
    **): h = rac{v_{y0}^2}{2g} = rac{(15)^2}{2 imes 9.8} = 11.5 ext{ m}
  • Time of Flight: T = rac{2 imes 15}{9.8} ext{s} ext{ or approximately 3.1s}
  • Range (R): R = rac{v^2 imes ext{sin}(60^{ ext{o}})}{g}= 19.6m

Chapter 5: Work and Energy

Elastic Potential Energy

• Definition: The amount of energy stored in a material when it is stretched or compressed.
  • Strained energy stored in a spring is given by:
  • E = F imes x = rac{1}{2}kx^2 (Equation 5.8)
• Work Done: The work done in stretching a material equals the area under the force-extension graph.

Archimedes' Principle and Floatation

• Definition: Archimedes' principle states that:

  • An object wholly or partly immersed in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced.

• Calculation of Upthrust:

  • For a solid cylinder of height $h$, with cross-sectional area $A$, immersed in a liquid:
  • Let the depths at the top and bottom of the cylinder be $h1$ and $h2$. The pressure at these depths can be written as:
    • $P<em>1 = pgh</em>1$
    • $P<em>2 = pgh</em>2$
  • Therefore, the net upthrust $F_u$ acting on the body is:
    • $F<em>u = P</em>2A - P<em>1A = pgh</em>2A - pgh<em>1A = pgA(h</em>2 - h_1)$
    • Simplifying gives us:
    • $F_u = pghA$

• Example Problem:

  • A wooden cube with side lengths of 10 cm is dipped in water:
  • Volume of cube: $V = L^3 = (0.1 m)^3 = 1 imes 10^{-3} m^3$
  • Upthrust is calculated as:
    • $F_u = p imes g imes V = 1000 ext{kg/m}^3 imes 9.8 ext{m/s}^2 imes 1 imes 10^{-3} m^3 = 9.8N$

Applications of Archimedes' Principle

• Explains why balloons rise in the air and why objects feel lighter in water.
• Relevant in understanding the buoyancy of boats and their design.

Conclusion

• Projectile motion can be analyzed through its components and corresponding equations of motion. Understanding Archimedes' principle is crucial in fluid mechanics, providing insights into buoyancy and displacement.