Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define momentum and impulse.
• Apply the impulse-momentum theorem to solve problems.
• State and apply the law of conservation of linear momentum.
• Distinguish between elastic and inelastic collisions.
• Express Newton's Second Law in terms of momentum.

Lesson summary

Momentum and impulse are fundamental concepts in physics, describing how forces affect the motion of objects. They are crucial for understanding collisions, explosions, and many other everyday phenomena. From the safety features in vehicles to the dynamics of sports, momentum and impulse are at play. In the South African context, understanding these concepts is vital for analysing road accidents, designing safer transport systems, and even improving athletic performance. The study of momentum also helps us understand the conservation laws, which are cornerstones of physics.

Lesson notes

Momentum Momentum (symbol p) is a measure of the "quantity of motion" of an object. It is a vector quantity, meaning it has both magnitude and direction. It is defined as the product of an object's mass (m) and its velocity (v): p = mv Units: kg⋅m/s (kilogram meters per second) The greater the mass or velocity of an object, the greater its momentum. A large truck moving slowly can have the same momentum as a smaller car moving very fast.

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

Solution: m = 1500 kg v = 20 m/s (east) p = mv = (1500 kg)(20 m/s east) = 30000 kg⋅m/s east The momentum of the taxi is 30000 kg⋅m/s east. Impulse Impulse (symbol J) is the change in momentum of an object. It is also a vector quantity. Impulse is defined as the product of the net force (F net ) acting on an object and the time interval (Δt) during which the force acts: J = F net Δt Units: N⋅s (Newton seconds) Since impulse is equal to the change in momentum, it also has units of kg⋅m/s. Impulse causes a change in momentum. A large force applied over a short time interval can produce the same change in momentum as a smaller force applied over a longer time interval.

Example 2: A builder hits a brick with a hammer. The hammer exerts an average force of 150 N on the brick for a time of 0.02 s. What is the impulse delivered to the brick?

Solution: F net = 150 N Δt = 0.02 s J = F net Δt = (150 N)(0.02 s) = 3 N⋅s The impulse delivered to the brick is 3 N⋅s. The Impulse-Momentum Theorem The impulse-momentum theorem states that the impulse acting on an object is equal to the change in momentum of that object: J = Δp = p f - p i = mv f - mv i Where: p f is the final momentum. p i is the initial momentum. v f is the final velocity. v i is the initial velocity. Combining the definitions of impulse and momentum, we get: F net Δt = mv f - mv i This equation is very powerful because it connects force, time, mass, and velocity.

Example 3: A soccer ball with a mass of 0.45 kg is initially at rest. A soccer player kicks the ball, applying an average force of 200 N for 0.01 s. What is the final velocity of the ball?

Solution: m = 0.45 kg v i = 0 m/s F net = 200 N Δt = 0.01 s Using the impulse-momentum theorem: F net Δt = mv f - mv i (200 N)(0.01 s) = (0.45 kg)v f - (0.45 kg)(0 m/s) 2 N⋅s = (0.45 kg)v f v f = (2 N⋅s) / (0.45 kg) = 4.44 m/s The final velocity of the ball is 4.44 m/s in the direction of the force. Conservation of Linear Momentum The law of conservation of linear momentum states that the total momentum of an isolated system remains constant if no external forces act on it. An isolated system is one in which the net external force is zero. This is a direct result of Newton's Third Law. For a system of two objects (A and B) colliding: p A(initial) + p B(initial) = p A(final) + p B(final) m A v A(initial) + m B v B(initial) = m A v A(final) + m B v B(final) This is a vector equation; you must consider the direction of the velocities. We often use a sign convention (e.g., right is positive, left is negative).

Example 4: A 2000 kg bakkie traveling east at 15 m/s collides head-on with a 1000 kg car traveling west at 20 m/s. If the vehicles lock together after the collision, what is their velocity immediately after the collision?

Solution: m bakkie = 2000 kg v bakkie(initial) = +15 m/s (east) m car = 1000 kg v car(initial) = -20 m/s (west) v bakkie(final) = v car(final) = v final (since they lock together)

Using the conservation of momentum: m bakkie v bakkie(initial) + m car v car(initial) = (m bakkie + m car )v final (2000 kg)(15 m/s) + (1000 kg)(-20 m/s) = (2000 kg + 1000 kg)v final 30000 kg⋅m/s - 20000 kg⋅m/s = (3000 kg)v final 10000 kg⋅m/s = (3000 kg)v final v final = (10000 kg⋅m/s) / (3000 kg) = 3.33 m/s The final velocity is 3.33 m/s east. Elastic and Inelastic Collisions Elastic Collision: A collision in which the total kinetic energy of the system is conserved (as well as momentum). In reality, perfectly elastic collisions are rare, but some collisions are close enough to be approximated as elastic (e.g., collisions between billiard balls).

Inelastic Collision: A collision in which the total kinetic energy of the system is not conserved. Some kinetic energy is converted into other forms of energy, such as heat, sound, or deformation. Most real-world collisions are inelastic. A perfectly inelastic collision is one where the objects stick together after colliding.

Example 5: (Perfectly Inelastic Collision) Two trolleys on a frictionless track collide. Trolley A has a mass of 2 kg and is moving to the right at 3 m/s. Trolley B has a mass of 1 kg and is moving to the left at 2 m/s. If the trolleys stick together after the collision, what is their final velocity?

Solution: This is a perfectly inelastic collision because the trolleys stick together. We use the conservation of momentum.