Pearson Edexcel Physics

Unit 4: Further Mechanics, Fields and Particles

4.3 Further Mechanics

Pearson Edexcel Physics

Unit 4: Further Mechanics, Fields and Particles

4.3 Further Mechanics

Candidates will be assessed on their ability to::
81. Understand how to use the equation impulse [math]= F∆t = ∆p[/math](Newton’s second law of motion)
82. CORE PRACTICAL 9: Investigate the relationship between the force exerted on an object and its change of momentum
83. Understand how to apply conservation of linear momentum to problems in two dimensions
84. CORE PRACTICAL 10: Use ICT to analyze collisions between small spheres, e.g. ball bearings on a table top
85. Understand how to determine whether a collision is elastic or inelastic
86. Be able to derive and use the equation [math]E_k = \frac{P^2}{2m}[/math] for the kinetic energy of a nonrelativistic particle
87. Be able to express angular displacement in radians and in degrees, and convert between these units
88. Understand what is meant by angular velocity and be able to use the equations [math]v =\omega r \text{ and } T=\frac{2\pi}{\omega}[/math]
89. Be able to use vector diagrams to derive the equations for centripetal acceleration [math]a = \frac{v^2}{r} = r\omega^2[/math] and understand how to use these equations
90. Understand that a resultant force (centripetal force) is required to produce and maintain circular motion
91. Be able to use the equations for centripetal force [math]F = ma = \frac{mv^2}{r} = mr\omega^2[/math]

  • 81) Understand how to use the equation impulse = F∆t = ∆p (Newton’s second law of motion)

  • ⇒ Impulse:
  • The product of a force applied for a known time (F× At) is known as the impulse, and this is equal to the change in momentum:
  • [math]\begin{gather}
    \text{Impulse (N·s)} = \text{force (N)} \times \text{time (s)} \\
    \text{Impulse} = \text{change in momentum (kg·m·s}^{-1}\text{)} \\
    \text{Impulse} = F \times \Delta t \\
    \text{Impulse} = \Delta p
    \end{gather}[/math]
  • To stop something moving, we need to remove all of its momentum.
  • The impulse needed to stop an object moving. If we know how long a force is applied, we could work out the size of that force.
  • Example:
  • What is the impulse needed to accelerate a 1000 kg car from rest to 25 [math]m.s^{-1}[/math]?
  • Given data:
  • Mass = 1000 kg
  • Velocity = [math]∆v = 25 m.s^{-1}[/math]
  • Formula:
  • [math]\begin{gather}
    \Delta p = m \times \Delta v \\
    \Delta p = 1000 \times 25 \\
    \text{Impulse} = \Delta p = 25000 \ \text{kg·m·s}^{-1}
    \end{gather}[/math]
  • If the car needed to stop in 3.8 seconds, what force would the brakes need to apply?
  • At the end, the car is at rest, so has no momentum. Therefore, the change in momentum will equal its initial momentum:
  • [math]\begin{gather}
    Ft = m \Delta v \\
    m \Delta v = 25000 \ \text{kg·m·s}^{-1} \\
    F = \frac{m \Delta v}{t} \\
    F = \frac{25000}{3.8} \\
    F = 6600 \ \text{N to 2 significant figures (sf)}
    \end{gather}[/math]

  • 82) CORE PRACTICAL 9: Investigate the relationship between the force exerted on an object and its change of momentum

  • ⇒ Investigating momentum change:
  • Figure 1 Measuring how a force changes the momentum of a trolley.
  • can investigate the rate of change of momentum in the school laboratory.
  • A trolley starts from rest and as a force act upon it its velocity increases.
  • If you record the trolley’s movement over time, you can find the velocity each second.
  • If you then calculate the momentum each second, you will be able to plot a graph of momentum against time.
  • It should be a straight line. As p = Ft, the gradient of this line will be equal to the accelerating force.
  • Figure 2 Accelerating from rest, momentum will be proportional to time.
  • Safety Note:
  • Place trolleys and runways so they cannot fall and cause injuries. For large masses, place a `catch box’ filled with crumpled paper or bubbled plastic in the `drop zone’ to avoid injury to feet.

  • 83) Understand how to apply conservation of linear momentum to problems in two dimensions

  • ⇒ Conservation of linear momentum:
  • If we calculate the momentum of each object before they collide, the sum total of these momenta (accounting for their direction) will be the same as the sum total afterwards.
  • This principle depends on the condition that no external force acts on the objects in question. An external force would provide an additional acceleration, and the motion of the objects would not be dependent on the collision alone.
  • As we saw in the previous section, a resultant force will cause a change in momentum, so it makes sense that momentum is only conserved if no external force acts. Imagine if a juggler’s ball moving upward collided with one coming down.
  • Momentum conservation would suggest that the one falling down would bounce back with an upward velocity after the collision.
  • Common sense tells us that all balls will still end up back on the ground. The external force of gravity means that the principle of conservation of momentum alone cannot be used to predict the motions after the collision.
  • If a stationary object explodes, then the total momentum of all the shrapnel parts added up (taking account of the direction of their movements) must be zero.
  • The object had zero momentum at the start, so the law of conservation of linear momentum tells us this must be the same total after the explosion.
  • Figure 3 In this illustration, these trapeze artists are stationary. If they let go of each other, they will ‘explode’ – they will fly apart with equal and opposite momenta.
  • In physics, any such event is termed an explosion, although it may not be very dramatic.
  • For example, if the two trapeze artists in figure 3 simply let go their hands and swing apart, they have zero total momentum before and will have equal and opposite momenta afterwards, which when added together will total zero again.

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

  • ⇒  Practical Setup
  • – Use a smooth, level table (e.g. a glass or laminated surface).
  • – Get two or more ball bearings of equal or different masses.
  • – Mark a grid or place measuring scales for reference.
  • – Set up a high-speed camera or smartphone above the table for top-down recording.
  • ⇒  Step-by-Step ICT-Based Analysis
  • Record the masses, m, of the two spheres using a mass balance then place them on a level table top.
  • Position two metre rulers perpendicular to each other using a set square.
  • Position a video camera above the table top (bird’s eye view) and start camera recording.
  • Roll one sphere towards the stationary sphere and allow them to collide and roll.
  • Stop recording when both spheres come to rest.
  • Import video to tracking software, and calibrate distance and a 90o angle using the metre rulers.
  • Go through each frame of the video, use the rulers to calculate the distance travelled and calculate the time between each frame.
  • [math]V = \sqrt{(v_x)^2 + (v_y)^2}[/math]
  • Where [math]v_x[/math] is the horizontial component of velocity and [math]v_y[/math] is the vertical
  • Angle of travel of the two speeds calculated using trigonometry or calculated by the software
  • ⇒  Advantages of Using ICT
  • – More accurate motion tracking than manual measurement
  • – Easy to repeat and re-analyses without redoing the experiment
  • – Enhanced visualization of invisible quantities (e.g. momentum vectors)
  • – Ability to simulate ideal scenarios for comparison

  • 85) Understand how to determine whether a collision is elastic or inelastic

  • ⇒ Elastic Collisions:
  • “A collision in which kinetic energy is conserved is called an elastic collision.”
  • In a collision between one pool ball and another, the first one often stops completely and the second then moves away from the collision.
  • As both pool balls have the same mass, the principle of conservation of momentum tells us that the velocity of the second ball must be identical to the initial velocity of the first.
  • This means that the kinetic energy of this system of two balls before and after the collision must be the same.
  • Figure 4 Newton’s cradle maintains kinetic energy, as well as conserving momentum in its collisions.
  • ⇒ Inelastic Collisions:
  • “A collision in which total kinetic energy is not conserved is called an inelastic collision.”
  • In a crash between two bumper cars, the total momentum after the collision must be identical to the total momentum before the collision.
  • However, if we calculate the total kinetic energy before and after, we find that the total is reduced by the collision.
  • Some of the kinetic energy is transferred into other forms such as heat and sound.
  • Figure 5 Inelastic collusion in one dimension

  • 86) Be able to derive and use the equation [math]E_k = \frac{P^2}{2m}[/math]  for the kinetic energy of a nonrelativistic particle

  • ⇒ Kinetic Energy:
  • Kinetic energy, form of energy that an object or a particle has by reason of its motion.
  • [math]E_k = \frac{1}{2} m v^2[/math]
  • ⇒ Momentum:
  • Momentum is equal to the mass of an object multiplied by its velocityand is equivalent to the force required to bring the object to a stop in a unit length of time.
  • [math]\begin{gather}
    p = m v \\
    v = \frac{p}{m}
    \end{gather}[/math]
  • Put in kinetic energy equation
  • [math]\begin{gather}
    E_k = \frac{1}{2} m v^2 \\
    E_k = \frac{1}{2} m \left(\frac{p}{m}\right)^2 \\
    E_k = \frac{1}{2} m \frac{p^2}{m^2} \\
    E_k = \frac{1}{2} \frac{p^2}{m} \\
    E_k = \frac{p^2}{2m}
    \end{gather}[/math]
  • This formula is particularly useful for calculations involving the kinetic energy of subatomic particles travelling at non-relativistic speeds – that is, much slower than the speed of light.

  • 87) Be able to express angular displacement in radians and in degrees, and convert between these units

  • Angles measured in degrees are used extensively in navigation to locate places.
  • To describe the difference between moving from one starting point to two possible destinations. This is measuring angular displacement on the surface of the Earth.
  • Each degree is subdivided into 60 ‘arcminutes’ and each of those arcminutes into 60 ‘arcseconds’.
  • When we are measuring rotation, we often use the alternative unit to measure angles – the radian.
  • The circle itself defines this. Imagine an object moves around part of the circumference of a circle. The angle through which it moves, measured in radians, is defined as the distance it travels, divided by its distance from the centre of the circle (the radius).
  • If the radius of the circle were one metre, then the distance the object travels around the circumference (also in metres) would be equal to the angle swept out in radians.
  • [math]\begin{gather}
    \text{angle (in radians)} = \frac{\text{length of arc}}{\text{radius of arc}} \\
    \theta = \frac{s}{r}
    \end{gather}[/math]
  • For a complete circle, in which the circumference is equal to , the angle swept out would be:
  • [math]\begin{gather}
    \theta = \frac{2 \pi r}{r} \\
    \theta = 2 \pi \text{ radians}
    \end{gather}[/math]
  • This means that the angle will be 1 radian (rad) if the distance moved around the circle is the same as the radius; just over of the distance around the circumference.
  • Figure 6 Angle measure in radian
  • ⇒ Angular Displacement:
  • Angular displacement is the vector measurement of the angle through which something has moved.
  • The standard convention is that anticlockwise rotation is a positive number and clockwise rotation is a negative number.
Angle in Radians Angle in degrees
0 0
[math]\frac{π}{4}[/math] 45
[math]\frac{π}{2}[/math] 90
[math]\frac{3π}{4}[/math] 135
[math]π[/math] 180
[math]\frac{5π}{4}[/math] 225
[math]\frac{3π}{2}[/math] 270
[math]\frac{7π}{4}[/math] 315
[math]2π[/math] 360

  •  88) Understand what is meant by angular velocity and be able to use the equations [math]v =\omega r \text{ and } T=\frac{2\pi}{\omega}[/math]

  • ⇒ Angular Velocity:
  • An object moving in a circle sweeps out a particular angle in a particular time, depending upon how fast it is moving.
  • The rate at which the angular displacement changes is called the angular velocity, ω. So, angular velocity is measured in [math]rad.s^{-1}[/math], and is defined mathematically by:
  • [math]\begin{gather}
    \omega = \frac{2\pi}{T} \\
    \omega = 2\pi \left(\frac{1}{T}\right) \\
    T = \frac{2\pi}{\omega}
    \end{gather}[/math]
  • The frequency of rotation is the reciprocal of the time period.
  • [math]\begin{gather}
    f = \frac{1}{T} \\
    \omega = 2 \pi f
    \end{gather}[/math]
  • ⇒ Instantaneous Velocity:
  • Instead of thinking about the angular movement, let us consider the actual velocity of the moving object (sometimes called the ‘instantaneous velocity’).
  • We know that
  • [math]v = \frac{s}{t}[/math]
  • and from the definition of the angle in radians
  • [math]θ = \frac{s}{r}[/math]
  • So, that
  • [math]s = rθ[/math]
  • Thus:
  • [math]\begin{gather}
    v = \frac{r \theta}{t} \\
    v = r \omega
    \end{gather}[/math]
  • Example:
  • In figure 4 we can see a geostationary satellite orbiting the Earth. What is its angular velocity?
  • To find the angular velocity, remember that it completes an orbit at the same rate as the Earth revolves, so one full circle every 24 hours.
  • Figure 7 How quickly does a satellite rotate through a certain angle?
  • [math]\begin{gather}
    \omega = \frac{2\pi}{T} \\
    \omega = \frac{2 \times 3.14}{24 \times 60 \times 60} \\
    \omega = \frac{2 \times 3.14}{86400} \\
    \omega = 7.27 \times 10^{-5} \ \text{rad.s}^{-1}
    \end{gather}[/math]
  • If the radius of the Earth is 6400 km and the satellite in figure 4 is in orbit 35 600 km above the Earth’s surface, what is the velocity of the satellite?
  • Form before
  • [math]\begin{gather}
    \omega = 7.27 \times 10^{-5} \ \text{rad.s}^{-1} \\
    v = r \omega \\
    v = (6400 + 35600)(7.27 \times 10^{-5}) \\
    v = 3050 \ \text{m.s}^{-1}
    \end{gather}[/math]

  • 89) Be able to use vector diagrams to derive the equations for centripetal acceleration [math]a = \frac{v^2}{r} = r\omega^2[/math]  and understand how to use these equations

  • ⇒ Centripetal acceleration:
  • Velocity is a vector, and so it is correctly described by both its magnitude and direction. An acceleration can change either of these, or both.
  • An object moving in a circle may travel at a constant speed (and a constant angular velocity) but the direction it is moving in must constantly change. This means it is constantly accelerating. As this acceleration represents the changes in direction around the circle, it is called the centripetal acceleration, a.
  • To calculate the centripetal acceleration, we must consider how quickly the direction, and therefore the velocity, is changing.
  • Figure 8 Vector components of velocity lead us to the centripetal acceleration equation.
  • At the arbitrary positions of the rotating object, A, and at a time later, B, we consider the components of the object’s velocity, in the x and y directions.
  • As A and B are equal distances above and below point C, the vertical velocity component, [math]V_y[/math] is the same in both cases:
  • [math]v_y = v cosθ[/math]
  • So, the vertical acceleration is zero:
  • [math]a_y = 0[/math]
  • Horizontally, the magnitude of the velocity is equal at both points, but in opposite directions:
  • [math]\begin{gather}
    \text{at A:} \quad v_x = v \sin \theta \\
    \text{at B:} \quad v_x = -v \sin \theta
    \end{gather}[/math]
  • So, the acceleration is just the horizontal acceleration, calculated as change in velocity divided by time:
  • [math]a_x = \frac{2 v \sin \theta}{t}[/math]
  • Figure 9 Uniform Circular motion
  • From the definition of angular velocity above:
  • [math]\begin{gather}
    v = \frac{r \theta}{t} \\
    t = \frac{r \theta}{v}
    \end{gather}[/math]
  • Here, the angle moved in time t is labelled as 2θ​ , so:
  • [math]\begin{gather}
    t = \frac{r 2 \theta}{v} \\
    a_x = \frac{2 v (v) \sin \theta}{r 2 \theta} \\
    a_x = \frac{v^2 \sin \theta}{r \theta} \\
    a_x = \frac{v^2}{r} \cdot \frac{\sin \theta}{\theta}
    \end{gather}[/math]
  • This must be true for all values of θ, and as we want to find the instantaneous acceleration at any point on the circumference, we must consider the general answer as we reduce θ to zero. In the limit, as θ tends to zero:
  • [math]\frac{\sin \theta}{\theta} = 1[/math]
  • Centrpetal acceleration,[math]a = \frac{v^2}{r}[/math]
  • From the definition of the instantaneous velocity above:
  • [math]\begin{gather}
    v = r \omega \\
    a = \frac{(r \omega)^2}{r} \\
    \text{Centripetal acceleration, } a = r \omega^2
    \end{gather}[/math]
  • The centripetal acceleration in this case is just the horizontal acceleration, as we considered the object in a position along a horizontal radius.
  • Following a similar derivation at any point around the circle will always have identical components of velocity that are perpendicular to the radius on either side of the point being considered.
  • Example:
  • What is the centripetal acceleration of the satellite in figure 5
  • [math]\begin{gather}
    a = \frac{v^2}{r} \\
    a = \frac{(3050)^2}{(6400 + 35600) \times 10^3} \\
    a = 0.22 \ \text{m.s}^{-2}
    \end{gather}[/math]
  • Or
  • [math]\begin{gather}
    a = r \omega^2 \\
    a = \big((6400 + 35600) \times 10^3\big) \left(7.27 \times 10^{-5}\right)^2 \\
    a = 0.22 \ \text{m.s}^{-2}
    \end{gather}[/math]

  • 90) Understand that a resultant force (centripetal force) is required to produce and maintain circular motion

  • ⇒ Circular motion:
  • When a hammer thrower moves athletics hammer around in a circle, the hammer has an angular velocity.
  • When the thrower lets the hammer go, it will fly off, following the straight line in the direction that it was moving at the instant of release. This direction is always along the edge of the circle (a tangent) at the point when it was released.
  • The hammer is moved in a circle at a constant speed and so the magnitude of the velocity is always the same.
  • For any object moving in a circle, there must be a resultant force to cause this acceleration, and it is called the centripetal force.
  • Figure 10 The instantaneous velocity of an object moving in a circle is tangential to the circle. When there is no resultant force, velocity will be constant, so it moves in a straight line, as this hammer will do when the athlete lets go.

  • 91) Be able to use the equations for centripetal force

  • [math]F = ma = \frac{mv^2}{r} = mr \omega^2[/math]

  • ⇒ Centripetal Force:
  • The mathematical formula for the centripetal force on an object moving in a circle can be found from Newton’s second law, and the equation we already have for the centripetal acceleration:
  • [math]\begin{gather}
    F = ma \quad \text{and} \quad a = \frac{v^2}{r} \\
    F = \frac{mv^2}{r} \\
    \text{Centripetal force} = \frac{\text{mass} \times (\text{velocity})^2}{\text{radius}} \\
    v = r \omega
    \end{gather}[/math]
  • So,
  • [math]\begin{gather}
    F = \frac{m (r \omega)^2}{r} \\
    F = \frac{m r^2 \omega^2}{r} \\
    F = m r \omega^2
    \end{gather}[/math]
  • The resultant centripetal force needed will be larger if:
  • – The rotating object has more mass
  • – The object rotates faster
  • – The object is closer to the center of the circle.
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