Dynamics

3 Dynamics Learners should be able to demonstrate and apply their knowledge and understanding of:
a)	The concept of force and Newton’s 3rd law of motion
b)	How free body diagrams can be used to represent forces on a particle or body
c)	The use of the relationship ∑F = ma in situations where mass is constant
d)	The idea that linear momentum is the product of mass and velocity
e)	The concept that force is the rate of change of momentum, applying this in situations where mass is constant
f)	The principle of conservation of momentum and use it to solve problems in one dimension involving elastic collisions (where there is no loss of kinetic energy) and inelastic collisions (where there is a loss of kinetic energy)
Specified Practical Work o Investigation of Newton’s 2nd law

Learners should be able to demonstrate and apply their knowledge and understanding of:
a) The Concept of Force and Newton’s Third Law of Motion
⇒ Force
Definition:
– Force is a vector quantity that describes an interaction that causes a change in an object’s motion, either by accelerating it, decelerating it, or changing its direction. It is measured in newtons (N) and is represented mathematically as:
[math]F = ma [/math]
Where F is the force, mmm is the mass of the object, and a is its acceleration.
Types of Forces:
– Contact Forces: These arise when objects are physically in contact. Examples include friction, tension, normal force, and applied force.
– Non-contact Forces: These act at a distance. Examples include gravitational force, magnetic force, and electrostatic force.
– Vector Nature of Force: Since force has both magnitude and direction, it is represented as a vector. The resultant force acting on a body is the vector sum of all forces acting on it.
⇒ Newton’s Third Law of Motion
Statement: “For every action, there is an equal and opposite reaction.”

Figure 1 Rocket Propulsion
This means that if one body exerts a force on a second body, the second body simultaneously exerts a force of equal magnitude but in the opposite direction on the first body.
Examples:
– Rocket Propulsion: The rocket expels gases backward (action), and the gases push the rocket forward (reaction).
– Walking: When you push the ground backward with your foot, the ground pushes you forward with an equal and opposite force.
Implications:
– Forces always occur in pairs.
– Action and reaction forces act on different objects and never cancel out.
b) Free Body Diagrams (FBDs) to Represent Forces
⇒ Free Body:
A free body diagram is a simplified representation of an object (treated as a particle or rigid body) and all the forces acting on it. It helps in analyzing the forces to solve problems involving motion or equilibrium.
Figure 2 Free body diagram (FBD)
⇒ Steps to Draw an FBD:
Identify the object of interest: Represent it as a point or shape.
Isolate the object: Remove any surroundings or connections.
Identify all forces: Represent each force with a vector arrow starting at the object. Label the forces clearly.
Choose a coordinate system: Determine the axes (e.g., horizontal and vertical) for resolving forces.
⇒ Forces to Include in FBDs:
Weight ([math]W =mg [/math]): Acts downward due to gravity.
Normal Force (N): Acts perpendicular to the surface.
Friction (f): Acts parallel to the surface, opposing motion.
Tension (T): Acts along a string, rope, or cable.
Applied Force ([math]F_{applied}[/math]): Any external force applied to the object.
Air Resistance ( [math]F_{drag}[/math]): Acts opposite to motion in fluid environments.
⇒ Examples of FBDs:
Block on a horizontal surface:
– Weight (W) acts downward.
– Normal force (N) acts upward.
– Friction (f) opposes motion, parallel to the surface.
Inclined Plane:
– Resolve weight into components parallel ([math]mgsinθ[/math]) and perpendicular ([math]mgcosθ[/math] ) to the plane.
– Include normal force and friction.

c) Using [math]∑F = ma[/math] in Situations Where Mass is Constant
⇒ Newton’s Second Law of Motion
Statement:
– The net force acting on an object is equal to the product of its mass and acceleration:
[math]∑F = ma[/math]
Where:
– ΣF: Net force (sum of all forces acting on the object).
– m: Mass of the object (constant).
– a: Acceleration of the object.
⇒ Steps to Apply [math]∑F = ma[/math] :
Draw the FBD: Identify all forces acting on the object.
Resolve Forces: Break forces into components along chosen axes.
Write Force Equations:
– Use
[math]∑F_x = ma_x[/math]
and
[math]∑F_y = ma_y[/math]
Where [math]a_x[/math] and [math]a_y[/math] are the acceleration components.
Solve for Unknowns: Rearrange and solve equations for unknown quantities such as acceleration, force, or tension.
⇒ Example Problem:
Scenario:
A 5-kg block is pulled across a horizontal surface with an applied force of 20 N. Frictional force is 5 N. Find the acceleration.
Solution:
– Draw the FBD:
– Identify forces:
– Applied force:[math]F_{applied} = 20 N (right) [/math]
– Friction: f=5 N (left).
– Weight:
[math]\begin{gather}
W = mg \\
W = (5)(9.8) \\
W &= 49 \, \text{N (down)}
\end{gather}[/math]
Normal force: N=49 N (up).
Apply
[math]\begin{gather}
\Sigma F_x = ma_x \\
\Sigma F_x = F_{\text{applied}} – f = ma_x \\
20 – 5 = 5a_x \\
a_x &= 3 \, \text{m/s}^2
\end{gather} [/math]
Acceleration is [math]3 m/s^2[/math].
d) Linear Momentum as the Product of Mass and Velocity
⇒ Definition of Linear Momentum
Linear momentum (p) is a vector quantity that describes the quantity of motion of an object. It is given by:
[math]p = mv[/math]
Where:
– p: Linear momentum (kgm/s),
– m: Mass of the object (kg)
– v: Velocity of the object (m/s).
⇒ Properties of Momentum:
1. Momentum depends directly on both mass and velocity. Heavier or faster objects have more momentum.
2. Momentum is a vector, so it has direction. Its direction is the same as the velocity of the object.
⇒ Examples of Momentum:
A truck moving at a certain velocity has more momentum than a car moving at the same velocity because of its larger mass.
A bullet, despite its small mass, can have high momentum due to its high velocity.

e) Force as the Rate of Change of Momentum
⇒ Newton’s Second Law and Momentum
Newton’s second law can be expressed in terms of momentum:
[math]F = \frac{∆p}{∆t}[/math]
Where:
– F: Net force acting on the object (N)
– ∆p: Change in momentum (kgm/s ),
– ∆t: Time over which the change occurs (s).
⇒ For Constant Mass:
If the mass (m) is constant, the change in momentum can be written as:
[math]Δp = mΔv [/math]
Substituting into the equation, we get:
[math]F = \frac{Δv}{Δt} = ma [/math]
This shows that force is proportional to the acceleration of the object when mass is constant.
⇒ Implications:
Force can be understood as the agent that changes the momentum of an object.
Larger forces produce faster changes in momentum (greater accelerations).
⇒ Examples:
Stopping a Car: The force applied by the brakes reduces the car’s momentum to zero over a certain time.
Rocket Propulsion: A rocket changes momentum by ejecting exhaust gases at high velocity, generating a force in the opposite direction.

f) Principle of Conservation of Momentum:
⇒ Statement of the Principle:
The total momentum of a system remains constant if no external forces act on it. Mathematically:
[math]ΣP_{initial} = ΣP_{final}[/math]
This principle applies to isolated systems, where the net external force is zero.
Figure 3 Law of conservation of momentum
⇒ Elastic and Inelastic Collisions
Elastic Collisions
– Definition: Collisions where total kinetic energy and total momentum are conserved.
– Characteristics:
No loss of kinetic energy.
Objects may bounce off each other.
Figure 4 Elastic and Inelastic Collisions
Equations:
– Conservation of Momentum:
[math]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 [/math]
Where:
– [math]m_1 \text{and} m_2[/math]: Masses of objects,
– [math]u_1 \text{and} u_2[/math]: Initial velocities,
– [math]v_1 \text{and} v_2[/math] : Final velocities.
Conservation of Kinetic Energy:
[math]\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 [/math]
Example Problem:
– Two objects of equal mass mmm collide elastically. Initially, one object is stationary ([math]u_2 = 0[/math]) while the other moves with velocity [math]u_1[/math] . After the collision
The moving object comes to rest ([math]v_1 = 0[/math] ),
The stationary object moves with velocity [math]v_2 = u_1[/math].
⇒ Inelastic Collisions
Definition:
– Collisions where total momentum is conserved, but some kinetic energy is lost (e.g., converted to heat, sound, or deformation).
Characteristics:
– Objects may stick together (perfectly inelastic collision).
– Total momentum is conserved, but kinetic energy decreases.
Equations:
– Conservation of Momentum:
[math]\begin{gather}
m_1 u_1 + m_2 u_2 &= m_1 v + m_2 v \\
m_1 u_1 + m_2 u_2 &= (m_1 + m_2)v
\end{gather}[/math]
Where v is the common final velocity of the combined mass after collision (perfectly inelastic collision).
Example Problem:
A 2-kg object moving at 3 m/s collides with a stationary 1-kg object. After the collision, the two objects stick together. Find the final velocity.
Solution:
– Initial Momentum:
[math]\begin{gather}
\Sigma p_{\text{initial}} &= m_1 u_1 + m_2 u_2 \\
\Sigma p_{\text{initial}} &= (2)(3) + (1)(0) \\
\Sigma p_{\text{initial}} &= 6 \, \text{kg·m/s}
\end{gather}[/math]
– Final Momentum (combined mass):
[math]\begin{gather}
\Sigma p_{\text{final}} &= (m_1 + m_2)v \\
\Sigma p_{\text{final}} &= (2 + 1)v \\
\Sigma p_{\text{final}} &= 3v
\end{gather}[/math]
Equating initial and final momentum:
[math]6 = 3v ⟹ v =2m/s[/math]
– Final velocity is 2 m/s.
Specified Practical Work
– Investigation of Newton’s 2nd law
⇒ Experiment:
Newton’s Second Law states that the net force acting on an object is directly proportional to its acceleration, and the proportionality constant is the object’s mass:
[math]F = ma[/math]
This law can be investigated practically in a controlled experiment. Below is a detailed guide for conducting such an investigation.
⇒ Objective
To investigate the relationship between force, mass, and acceleration and verify Newton’s Second Law
[math]F = ma[/math]
Apparatus
– Dynamics Trolley (or any object on wheels)
– Pulley System
– String
– Weights (slotted masses)
– Ticker Timer or Light Gate and Data Logger (to measure acceleration)
– Ruler or Measuring Tape (to measure displacement)
– Stopwatch (optional)
– Clamp Stand (to hold the pulley in place)
– Flat Track or Smooth Surface (to minimize friction)
Figure 5 Investigation of Newton’s Law
⇒ Experimental Setup
1. Set up a flat track or smooth surface for the trolley to move on.
2. Attach a string to the trolley and pass it over a pulley at the edge of the track.
3. Hang slotted masses (weights) at the free end of the string to act as the force (FFF) pulling the trolley.
4. Ensure the pulley is securely clamped and the string is taut.
5. Place the ticker timer (or light gate) along the trolley’s path to record its motion.
⇒ Procedure
1. Measure the Mass of the Trolley (m): Use a scale to measure the mass of the trolley. Record this value.
2. Apply Force:
– Attach a specific weight ([math]m_W[/math] ) to the string. This weight creates a force given by:
[math]F = m_W. g[/math]
Where [math]g = 9.8 m/s^2[/math] is the acceleration due to gravity.
3. Release the Trolley:
– Place the trolley at the starting position and release it so the force pulls the trolley along the track.
4. Measure Acceleration:
– Use the ticker timer or light gate to record the time taken and displacement of the trolley.
– Calculate acceleration (a) using kinematic equations, such as:
[math]a = \frac{2s}{t^2}[/math]
Where s is the displacement and t is the time taken.
5. Repeat the Experiment:
– Vary the pulling force (F) by changing the hanging mass ([math]m_W[/math] ) while keeping the mass of the trolley constant.
– Alternatively, vary the mass of the trolley while keeping the pulling force constant.
6. Record Observations:
– Record the force (F), trolley mass (m), and measured acceleration (a) for each trial.
Analysis
1. Plot Graphs:
– Plot F (force) on the y-axis and a (acceleration) on the x-axis. The graph should yield a straight line through the origin if Newton’s Second Law is valid.
– The slope of the line gives the mass of the system (m).
2. Verify Relationships:
– Check if acceleration is directly proportional to force ([math] a ∝ F[/math]) for constant mass.
– Check if acceleration is inversely proportional to mass ([math] a ∝ \frac{1}{m}[/math] ) for constant force.
⇒ Sources of Error
1. Friction between the trolley and the track.
2. Inaccurate measurement of acceleration.
3. Air resistance (minimal but present).
4. Uneven track or string tension.
⇒ Conclusion
The experiment demonstrates the relationship described by Newton’s Second Law:
[math]F = ma[/math]
– For a given force, acceleration is inversely proportional to mass. Conversely, for a constant mass, acceleration is directly proportional to the applied force.
⇒ Extensions
1. Investigate the effect of friction by adding frictional surfaces to the track.
2. Use different types of trolleys or objects to test the law in various conditions.