Topic 6: Further Mechanics

1. Impulse:

• Impulse is a fundamental concept in physics, especially in mechanics. Impulse is defined as the product of the force applied to an object and the time duration over which it is applied.
• [math] \text{Impulse (J)} = \text{Force (F)} \times \text{Time (t)} [/math]
• [math] \text{Impulse (J)} = F∆t [/math]
• Impulse is a vector quantity, and its unit is typically measured in Newton-seconds (N·s).
• We also know that
• [math] \text{Force = Rate of change of momentum (Newton’s Second Law)} \\
  F = \frac{\Delta p}{\Delta t} \\
  F = \frac{dp}{dt} [/math]
• So,
• [math] F∆t = ∆p [/math]
• [math] F∆t \, \text{is impulse so,} \\
  \text{Impulse} = \text{change in momentum} [/math]
• [math] \text{Impulse} = \text{change in momentum} \\ \text{Force} = \text{rate of change of momentum} [/math]
• Examples:
• A boy of mass 30 kg jumps off a wall 0.90 m high. On hitting the ground his impact time is 0.2 s.
  1. Determine:
     i) The speed of impact
     ii) His momentum on impact
     iii) The impulse of the ground on his feet
     iv)The force he experiences.
  2. Explain why bending his knees on impact reduces the force exerted on him.
• Given data:
• Mass of boy = m = 30 kg
• Height of wall = h = 0.90m
• Time for hitting the ground = t = 0.2 s
• Find data:
  1. i) speed of impact = v =?
     ii) momentum = p =?
     iii) impulse =?
     iv) force = F =?
  2. Explain why bending his knees on impact reduces the force exerted on him.
• Formula:
• [math] \frac{1}{2} mv^2 = mg∆h [/math]
• i) [math] v = \sqrt{2g\Delta h} [/math]
  ii) momentum = mv
  iii) impulse = change in momentum
  iv) impuls = F∆t
• v) [math] F= \frac{\text{impulse}}{∆t} [/math]
• Solution:
  1. i) 

     [math] v = \sqrt{2g\Delta h} \\
     v = \sqrt{2(10)(0.90)} \\
     v = 4.2 \, \text{m} \cdot \text{s}^{-1}[/math]

     ii)

     [math] \text{Momentum} = mv \\
     \text{Momentum} = (30)(4.2) \\
     \text{Momentum} = 126 \, \frac{\text{kgm}}{\text{s}} \, \text{or} \, 126 \, \text{N} \cdot \text{s} [/math]

     iii)

     [math] \text{Impulse} = \text{Change in momentum} \\
     \text{Impulse} = 126 \, \text{N} \cdot \text{s} \, \text{(since the object comes to rest)} [/math]

     iv)

     [math] F = \frac{\text{Impulse}}{\Delta t} \\
     F = \frac{126}{0.2} \\
     F = 630 \, \text{N} [/math]

     (This is just over twice his weight.)

  2. If he bends his knees on impact, he will increase the time of impact as he will come to rest more slowly. If the time of impact is increased, the force of impact will reduces.

2. CORE PRACTICAL 9: Investigate the relationship between the force exerted on an object and its change of momentum.

• This experiment uses the same apparatus and is similar to the experiment described earlier in this book to investigate force and acceleration.
• Figure 1 investigating change in momentum
• Safety note:
  – Place a padded box beneath the
  – load so that the floor and masses don’t get damaged. Also, to avoid injuries to feet, don’t stand under falling masses!
  – The light gates are connected to a suitable data logger and the results can either be interpreted manually or can be fed into a suitable computer programme. The following data is required:
• The length / of the card (200 mm in this experiment)
• The times for the card to pass through each light gate, 1, and 12
• The time for the car to travel from the first light gate to the second light gate ∆t.
• The mass of the trolley [math] m_T [/math](450g) in this experiments).
• A typical print-out is shown in Figure 2.
• The velocity through the first gate [math] v_1 = \frac{1}{t_1} [/math]and the velocity through the second gate [math] v_2 = \frac{1}{t_2} [/math].
• The change in momentum is then [math] ∆p = m(v_2 – v_1) =m ∆v [/math] where m is the mass being accelerated.
• Figure 2 print-out time graph
• We need to think carefully about m.
• The force, which is the weight of the masses hanging over the pulley (labelled ‘load’), has to accelerate the combined mass of the trolley, the masses on the trolley and the masses hanging over the pulley.
• In order to vary the force and at the same time keep m constant we have to take one of the masses off the trolley and add it to the load each time.
• Time [math] t_1 ,t_2 \text{ and } \Delta t [/math] obtained for a range of values for the load, keeping m constant using the technique as described above.
• The numbers can then be ‘crunched using a spreadsheet.
• If a graph of FΔt on the y-axis against Δv on the x-axis is plotted, a straight line through the origin would show that Δv ∝ F.
• If the gradient is found to be m, then we can say that
• [math] F \Delta t = m \Delta v \, \text{or} \, F = \frac{m \Delta v}{\Delta t} = \text{rate of change of momentum} [/math]

3. Conservation of momentum:

• Kinetic energy is usually not conserved in collisions. These are known as inelastic collisions.
• However, elastic collisions in which kinetic energy is conserved do occur for example between molecules and between nuclei, and also sometimes (almost) between larger objects such as snooker balls.
• Elastic collisions between particles of equal mass result in special outcomes that are important to both particle physicists and snooker players.
• Can calculate the work done by a force as
• [math] ∆W= F_{av} ∆x [/math]

• Using this equation, we can calculate how much mechanical energy an object has been given. This energy might be stored as

• [math] \text{Gravitational Potential Energy} (\Delta \text{GPE} = mg\Delta h) [/math]
• Or as
• [math] \text{Elastic Potential Energy} (\Delta \text{EPE} = \frac{1}{2} k \Delta x^2) [/math]
• or as
• [math] \text{Kinetic Energy } (\text{KE} = E_k = \frac{1}{2} mv^2) [/math]
• Mechanical energy sometimes appears to be lost to the surroundings, often as a result of work done against air resistance or contact friction. When this happens, we say that there has been a transfer of mechanical energy to internal energy – thermal energy of the particles of the surroundings.

• The principle of conservation of energy states that energy is never created or destroyed, but it can be transferred one form into another. Other forms of energy include chemical energy and radiant energy. We will meet yet more forms, such as nuclear energy and electrostatic potential energy.

4. CORE PRACTICAL 10: Use ICT to analyze collisions between small spheres, e.g., ball bearings on a table top:

• Objective:
• Investigate collisions between small spheres (e.g., ball bearings) on a table top using ICT to analyze the motion and momentum transfer.
• Apparatus:
  – Ball bearings
  – Table top
  – ICT tools (e.g., motion sensors, cameras, or simulation software)
• Procedure:
  1. Set up the apparatus: Place the ball bearings on the table top and configure the ICT tools to record the motion.
  2. Collide the spheres: Release the spheres to collide with each other or a stationary object.
  3. Record and analyze data: Use ICT tools to record the motion and analyze the data to determine:
     – Momentum transfer
     – Velocity changes
     – Energy transfer
  4. Repeat and vary: Repeat the experiment with different initial conditions (e.g., velocity, mass, angle of incidence) to explore the effects on momentum transfer.
• ICT Tools:
  – Motion sensors: Measure velocity and acceleration
  – Cameras: Record the collision and analyze the motion frame-by-frame
  – Simulation software: Model the collision and vary parameters to explore the effects
• Outcomes:
  – Understand momentum transfer during collisions
  – Analyze the relationship between velocity, mass, and momentum
  – Use ICT to visualize and quantify the motion
• This practical investigation integrates physics concepts with ICT skills, allowing you to explore and analyze collisions in a engaging and interactive way.

5. Elastic and inelastic collusion:

• Elastic Collision:
  1. Momentum is conserved (total initial momentum = total final momentum)
  2. Kinetic energy is conserved (total initial kinetic energy = total final kinetic energy)
  3. Objects bounce off each other (no permanent deformation)
  4. Coefficient of restitution (e) = 1 (objects rebound with same speed)
• Inelastic Collision:
  1. Momentum is conserved (total initial momentum = total final momentum)
  2. Kinetic energy is not conserved (some energy is lost as heat, sound, etc.)
  3. Objects stick together or deform permanently
  4. Coefficient of restitution (e) < 1 (objects rebound with less speed)
• Figure 3 A pin and cork collusion
• First suppose the mass of each trolly is m, i.e., [math] m_A = m_B = m [/math]. The equation expressing the conservation of momentum then becomes:
• [math] mu = 2mv [/math]
• That is
• [math] u = 2v \\ v = \frac{1}{2} u [/math]
• Total KE before collusion [math] = \frac{1}{2} mu^2 + 0 = [/math] [math] \frac{1}{2} mu^2 [/math]
• KE after collusion [math] = \frac{1}{2} (2m) v^2 = mv^2 = m \left(\frac{1}{2} u\right)^2 = [/math] [math] \frac{1}{4} mu^2 [/math]
• The total KE has fallen From [math] \frac{1}{2} mu^2 \text{ to } \frac{1}{4} mu^2, [/math]  i.e., [math] KE_{\text{after}} \text{ is half of } KE_{\text{before}} [/math].
• Momentum has, as always, been conserved but some kinetic energy has been lost – half the initial kinetic energy in this case. This result is the same for all collisions where objects of equal mass form linked collisions. Such a collision is an example of a non-elastic or an inelastic collision.
• But remember, total energy must be conserved. The loss of kinetic energy is equal to the gain in internal energy in this case as the pin enters the cork. The Sankey diagram in Figure 4 emphasises this energy conservation.
• Figure 4 Total energy is conserved
• Plastic Collision (a type of inelastic collision):
  1. Objects stick together after collision
  2. Momentum is conserved
  3. Kinetic energy is not conserved
• To determine the type of collision, analyze the collision using the above criteria. If the collision meets the criteria for an elastic collision, it is elastic. If not, it is inelastic. If the objects stick together, it is a plastic collision.
• Additionally, you can use the coefficient of restitution (e) to determine the type of collision:
  e = 1: Elastic collision
  0 < e < 1: Inelastic collision
  e = 0: Perfectly inelastic collision (objects stick together)
• By applying these criteria, you can determine whether a collision is elastic or inelastic.

6. Be able to derive and use the equation [math] E_k = \frac{p^2}{2m} [/math]for the kinetic energy of a non-relativistic particle.

• The equation for kinetic energy [math]E_k [/math] and linear momentum p:
• [math] E_k = \frac{1}{2} mv^2 \qquad (1) \\
  p = mv \qquad (2) [/math]
• For an object of mass m, we can eliminate the velocity v and get equations linking kinetic energy to momentum. Squaring the second equation and rearranging the first gives:
• [math] p^2 = m^2 v^2 \qquad (3) \\
  E_k = \frac{1}{2} mv^2 \\
  v^2 = \frac{2E_k}{m} [/math]
• Put in equation 3
• [math] p^2 = m^2 \left( \frac{2E_k}{m} \right) \\ p^2 = 2mE_k \\ p = \sqrt{2mE_k} \qquad (4) \\ \text{Or} \\ E_k = \frac{p^2}{2m} \qquad (5)  [/math]
• Equation 4 and 5 shows the relation between kinetic energy and momentum.
  Note: This equation is valid for non-relativistic particles, meaning particles moving at speeds much less than the speed of light. For relativistic particles, the equation is modified to account for the effects of special relativity.

Example:
(1)

• What is the momentum of:
• a) An electron of mass [math] 9.1 * 10^{-31} kg [/math]
• b) A proton of mass [math] 1.7 * 10^{-31} kg [/math],
• c) Each of which has a kinetic energy of 150 eV?
• Given data:
• For both (electron and proton)
• [math] E_k = 150 \, \text{eV} = 150 * (1.6 \times 10^{-19} \, \text{J}) = 2.4 * 10^{-17} \, \text{J} [/math]
• Formula:
• [math] p = \sqrt{2mE_k} [/math]
• Solution:
• [math] p = \sqrt{2mE_k} [/math]
• Putting values for electron
• [math] p_{\text{electron}} = \sqrt{2 \left( 9.1 * 10^{-31} \right) \left( 2.4 * 10^{-17} \right)} \\
  p_{\text{electron}} = 6.6 * 10^{-24} \, \text{N} \cdot \text{s} [/math]
• Putting values for proton
• [math] p = \sqrt{2mE_k} \\
  p_{\text{proton}} = \sqrt{2 \left( 1.7 * 10^{-27} \right) \left( 2.4 *10^{-17} \right)} \\
  p_{\text{proton}} = 2.9 * 10^{-22} \, \text{N} \cdot \text{s} [/math]

(2)

• Find the kinetic energy of a particle with a mass of 2 kg and a momentum of 4.
• Given data:
• Mass of particle = m = 2 kg
• Momentum = p = 4 kg. m/s
• Find data:
• Kinetic energy [math] = E_k = ? [/math]
• Formula:
• [math] E_k = \frac{p^2}{2m} [/math]
• Solution:
• [math]E_k = \frac{p^2}{2m} \\
  E_k = \frac{(4)^2}{2(2)} \\
  E_k = \frac{16}{4} \\
  E_k = 4 \, \text{J} [/math]

7. Angular displacement:

• Angular momentum is a fundamental concept in physics that describes the rotation of an object. It’s a measure of an object’s tendency to keep rotating, and it’s a vector quantity. Here’s a brief overview:
• Definition:
• Angular momentum (L) is the product of an object’s moment of inertia (I), angular velocity (ω), and the distance from the axis of rotation (r):
• [math] L = Iωr [/math]
• Units
• Angular momentum is typically measured in units of kilogram-meters squared per second (kg·m²/s).
• Properties:
  1. Conservation: Angular momentum is conserved in closed systems, meaning that the total angular momentum remains constant over time.
  2. Vector: Angular momentum is a vector quantity, having both magnitude and direction.
  3. Additivity: Angular momentum can be added vectorially, allowing for the calculation of total angular momentum.
• Radians (rad):
  – Defined as the ratio of the arc length to the radius of the circle
  – 1 rad = 1 (unitless)
  – Angular displacement (θ) in radians: [math] θ = Δs / r [/math]
• Degrees (°):
  – Defined as 1/360 of a full circle
  -[math] 1° = π/180 rad [/math]
  – Angular displacement (θ) in degrees:
• [math] \theta = \left( \frac{\Delta s}{r} \right) \left( \frac{180}{\pi} \right) [/math]
• Conversions:
  – Radians to degrees:
• [math] \theta^\circ = \theta_{\text{rad}} \times \frac{180}{\pi} [/math]
• – Degrees to radians:
• [math] \theta_{\text{rad}} = \theta^\circ \times \frac{\pi}{180} [/math]
• Example:
• – Convert 30° to radians:
• [math] \theta_{\text{rad}} = 30 \times \frac{\pi}{180} = \frac{\pi}{6} \, \text{rad} [/math]
• – Convert π/3 rad to degrees:
• [math] \theta^\circ = \frac{\pi}{3} \times \frac{180}{\pi} = 60^\circ [/math]
• Remember:
  – 2π rad = 360° (full circle)
  – π rad = 180° (half circle)
  – π/2 rad = 90° (quarter circle)
• Now you can easily switch between radians and degrees.

8. Angular velocity:

• Angular velocity (ω) is a measure of how fast an object rotates or revolves around a central axis. It’s a scalar quantity that describes the rate of change of angular displacement (θ) with respect to time (t).
• Units:
• Angular velocity is typically measured in units of radians per second (rad/s).
• Mathematical definition:
• [math] \omega_{\text{av}} = \frac{\Delta \theta}{\Delta t} \\
  \omega_{\text{inst}} = \frac{\delta \theta}{\delta t} \quad \text{or} \quad \frac{d\theta}{dt} [/math]
• From
• [math]\theta = \frac{s}{r} \\
  s = r\theta \\
  \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} \\
  \frac{\Delta s}{\Delta t} = v \quad \text{or} \quad \frac{\Delta \theta}{\Delta t} = \omega [/math]
• So,
• [math]v = r \omega [/math]
• Types:
  1. Constant angular velocity: When the angular velocity remains constant, the motion is uniform circular motion.
  2. Variable angular velocity: When the angular velocity changes, the motion is non-uniform circular motion.
• Another useful relationship concerns the time taken for something to complete one circle.
• [math] T = \frac{2 \pi }{ω} [/math]
• Reversing this to
• [math] ω = \frac{2π}{T} [/math]
• enables you to calculate the angular velocity of the spinning Earth, and then  lets you calculate the speed of any point on the Earth’s surface provided you know the radius of the circle in which it is moving. (It would be zero at the poles!)
• Example:
  (1)
  The city of Birmingham is at latitude 52.5° (the equator has a latitude of zero, the north pole 90°). The Earth has a radius of 6400 km.
• Figure 5 Birmingham at a latitude of 52.5 °
• Given data:
  Given time one day = T = 24 * 3600 s
  Latitude = θ = 52.5°
  Radius of Earth  [math] = R = 6400 km = 6400 * 10^3 m = 6.4 * 10^3 * 10^3 m = 6.4 * 10^6 m[/math]Finding data:
  a) Birmingham’s angular velocity as the Earth spins
  b) the radius of the circle in which Birmingham moves
  c) the speed at which Birmingham’s inhabitants are moving.
• Formula:
• i)   [math] \omega = \frac{2\pi}{T} [/math]
• ii)  [math] r = R \cos \theta [/math]
• iii) [math] v = r \omega [/math]
• Solution:
  1. The Earth spins once every 24 hours. 

     [math] \omega = \frac{2\pi}{T} \\ \omega = \frac{2\pi \, \text{(rad)}}{24 \times 3600 \, \text{s}} \\ \omega = 7.27 * 10^{-5} \, \text{rad/s} [/math]

  2. At a latitude of 52.5°, 

     [math] r = (6.4 * 10^6) \cos(52.5^\circ) = 3.9 * 10^6 \, \text{m} [/math]

  3. Using

     [math] v = r \omega \\ v = (3.9 * 10^6) (7.3 * 10^{-5}) \\ v = 280 \, \text{m/s}[/math]

9. Centripetal acceleration:

• Centripetal acceleration ([math] a_c [/math]) is the acceleration of an object moving in a circular path, directed towards the center of the circle. It’s a fundamental concept in physics, essential for understanding circular motion.
• Mathematical definition:
• [math] a_c = \frac{v^2}{r} \\
  v = r \omega \\
  a_c = \frac{(r \omega)^2}{r} \\
  a_c = \frac{r^2 \omega^2}{r} \\
  a_c = r \omega^2 [/math]
• where:
  – [math] a_c [/math] = centripetal acceleration (m/s²)
  – v = velocity (m/s)
  – r = radius of the circular path (m)
  – ω = angular velocity (m/s)
• Figure 6 Vector diagram for centripetal acceleration
• Examples
• (1)
• Find the centripetal acceleration of a car turning a corner with a radius of 50 m at a speed of 20 m/s.
• Given data:
• Radius = r = 50m
• Angular speed = ω = 20 m/s
• Find data:
• Centripetal acceleration = [math] a_c [/math]  =?
• Formula:
• [math] a_c  = r ω^2 [/math]
• Solution:
• [math] a_c = rω^2 \\ a_c =(50)(20)^2 \\ a_c = 8 m.s^{-2} [/math]

10. Centripetal force:

• Centripetal force ([math] F_c [/math]) is the resultant force that acts on an object moving in a circular path, directed towards the center of the circle. It’s responsible for keeping the object on its circular trajectory.
• Types of centripetal forces:
  1. Frictional force (e.g., car turning)
  2. Normal force (e.g., satellite orbiting)
  3. Tension force (e.g., spinning top)
  4. Gravitational force (e.g., planet orbiting)
• Mathematical representation:
• [math] F_c = m a_c = m \left(\frac{v^2}{r}\right) \\
  F_c = m \left(\frac{(r \omega)^2}{r}\right) \\
  F_c = m \left(\frac{r^2 \omega^2}{r}\right) = m r \omega^2 [/math]
• Where:
  – [math] F_c [/math] = centripetal force
  – m = mass
  – [math] a_c [/math] = centripetal acceleration (v²/r or rω²)
  – v = velocity
  – r = radius
  – ω = angular velocity
• Figure 7 Uniform circular motion
• Characteristics:
• – Acts towards the center of the circle
  – Perpendicular to the velocity vector
  – Necessary for circular motion
  – Resultant force (net force)
• Examples:
  – A car turning a corner (frictional force)
  – A satellite orbiting the Earth (gravitational force)
  – A spinning top (tension force)
  – A planet orbiting a star (gravitational force)
• Examples:
• (1)
• “A 1500 kg car is traveling at a speed of 30 m/s around a circular track with a radius of 200 m. What is the centripetal force acting on the car?”
• Given data:
  Mass of car (m) = 1500 kg
  Velocity of moving car (v) = 30 m/s
  Radius of circular tack (r) = 200 m
  Centripetal force = [math] F_c [/math]=?
• Formula:
• [math]  F_c = \frac{m v^2}{r} [/math]
• Solution:
• [math] F_c = \frac{m v^2}{r} \\
  F_c = \frac{(1500)(30^2)}{200} \\
  F_c = \frac{(1500)(900)}{200} \\
  F_c = 6750 \, \text{N} [/math]
• Therefore, the centripetal force acting on the car is 6750 Newtons.
• The centripetal force acting on the car is 6750 N, which is provided by the frictional force between the tires and the track, keeping the car moving in a circular path.