Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define momentum and impulse.
• Calculate the momentum of an object.
• Calculate the impulse on an object.
• Relate impulse to the change in momentum.
• Apply the impulse-momentum theorem to solve problems.

Lesson summary

Momentum and impulse are fundamental concepts in physics that describe the motion of objects and how forces change their motion. Understanding these concepts allows us to analyze collisions, explosions, and other interactions involving forces and moving objects. In South Africa, understanding momentum and impulse is relevant in various contexts, such as road safety (analyzing car crashes), sports (understanding the impact of a soccer ball), and engineering (designing safer structures). This week, we will lay the foundation for understanding these key concepts.

Lesson notes

2.1 Momentum (p) Momentum is a measure of an object's mass in motion. It is a vector quantity, meaning it has both magnitude and direction. The momentum of an object is directly proportional to its mass and velocity.

Definition: Momentum (p) is the product of an object's mass (m) and its velocity (v): ``` p = mv ``` Units: The SI unit for momentum is kilogram-meters per second (kg·m/s).

Direction: The direction of the momentum vector is the same as the direction of the velocity vector.

Example 1: A taxi with a mass of 1500 kg is traveling at a velocity of 20 m/s east. Calculate its momentum.

Solution: ``` m = 1500 kg v = 20 m/s east p = mv p = (1500 kg)(20 m/s east) p = 30000 kg·m/s east ``` Therefore, the momentum of the taxi is 30000 kg·m/s east.

Example 2: A soccer ball with a mass of 0.45 kg is kicked and reaches a velocity of 25 m/s north. Calculate its momentum.

Solution: ``` m = 0.45 kg v = 25 m/s north p = mv p = (0.45 kg)(25 m/s north) p = 11.25 kg·m/s north ``` Therefore, the momentum of the soccer ball is 11.25 kg·m/s north. 2.2 Impulse (J) Impulse is a measure of the change in momentum of an object. It is also a vector quantity. Impulse is equal to the force applied to an object multiplied by the time interval over which the force acts.

Definition: Impulse (J) is the product of the net force (F net ) acting on an object and the time interval (Δt) over which the force acts: ``` J = F net Δt ``` Units: The SI unit for impulse is Newton-seconds (N·s). Notice that 1 N⋅s = 1 kg⋅m/s, so impulse also has the same units as momentum.

Direction: The direction of the impulse vector is the same as the direction of the net force vector.

Example 3: A net force of 100 N is applied to a box for 2 seconds. Calculate the impulse on the box.

Solution: ``` F net = 100 N Δt = 2 s J = F net Δt J = (100 N)(2 s) J = 200 N·s ``` Therefore, the impulse on the box is 200 N·s. 2.3 Impulse-Momentum Theorem The impulse-momentum theorem states that the impulse acting on an object is equal to the change in momentum of the object.

Equation: ``` J = Δp ``` Since J = F net Δt and p = mv, we can write this as: ``` F net Δt = mv f - mv i ``` where: v f is the final velocity v i is the initial velocity This is a crucial equation that links force, time, and changes in velocity. It allows us to predict the final velocity of an object if we know the force applied and the duration of the application, or conversely, to determine the force required to achieve a specific change in velocity within a certain time.

Example 4: A cricket ball of mass 0.16 kg is moving at a velocity of 20 m/s towards a batsman. The batsman hits the ball, and it moves away from him at a velocity of 30 m/s in the opposite direction. If the contact time between the bat and the ball is 0.01 s, what is the average force exerted by the bat on the ball?

Solution: Choose a direction as positive (e.g., direction away from the batsman). ``` m = 0.16 kg v i = -20 m/s (towards the batsman) v f = 30 m/s (away from the batsman) Δt = 0.01 s F net Δt = mv f - mv i F net (0.01 s) = (0.16 kg)(30 m/s) - (0.16 kg)(-20 m/s) F net (0.01 s) = 4.8 kg·m/s + 3.2 kg·m/s F net (0.01 s) = 8 kg·m/s F net = 8 kg·m/s / 0.01 s F net = 800 N ``` The average force exerted by the bat on the ball is 800

N. Example 5: A trolley of mass 2 kg is pushed from rest with a force of 5 N over a distance of 1.5 m. Assuming negligible friction, calculate the final velocity of the trolley. The force is applied horizontally.

Solution: First, we need to find the acceleration using Newton's Second Law: ``` F net = ma 5 N = 2 kg * a a = 2.5 m/s 2 ``` Next, we can find the time the force is applied using kinematics (since force is constant).

The simplest formula is: ``` Δx = v i Δt + (1/2)a(Δt) 2 1.5 m = 0 + (1/2)(2.5 m/s 2 )(Δt) 2 Δt 2 = 1.2 Δt = 1.095 s ``` Finally, we can use the impulse-momentum theorem or kinematics to find the final velocity. Using the impulse-momentum theorem is straightforward in this instance: ``` F net Δt = mv f - mv i 5N 1.095 s = 2 kg v f - 2 kg * 0 m/s 5.475 Ns = 2kg * v f v f = 2.74 m/s ``` The final velocity of the trolley is 2.74 m/s. Guided Practice (With Solutions)

Question 1: A tennis ball of mass 0.06 kg is served with a velocity of 50 m/s. What is the momentum of the tennis ball?

Solution: ``` m = 0.06 kg v = 50 m/s p = mv p = (0.06 kg)(50 m/s) p = 3 kg·m/s ``` The momentum of the tennis ball is 3 kg·m/s.

Question 2: A force of 25 N is applied to a box for 0.5 seconds. What is the impulse on the box?

Solution: ``` F net = 25 N Δt = 0.5 s J = F net Δt J = (25 N)(0.5 s) J = 12.5 N·s ``` The impulse on the box is 12.5 N·s.

Question 3: A car of mass 1000 kg slows down from 25 m/s to 15 m/s in 5 seconds. Calculate the force exerted on the car.