Applications and Skills
Applying conservation of momentum in simple isolated systems including (but not limited to) collisions, explosions, or water jets
Using Newton’s second law quantitatively and qualitatively in cases where mass is not constant
Sketching and interpreting force–time graphs
Determining impulse in various contexts including (but not limited to) car safety and sports
Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions and explosions
Understandings
Newton’s second law expressed in terms of rate of change of momentum
Impulse and force–time graphs
Conservation of linear momentum
Elastic collisions, inelastic collisions and explosions

Momentum

The law of conservation of linear momentum makes it possible to predict outcomes in many physical situations.

Newton’s second law → $F_{net} = ma$. This equation is only valid when the mass of the system remains constant. In cases where mass changes, a different version of the second law must be used.

The linear momentum, p, is defined as the product of an object’s mass with its velocity:

$\bold{p = mv}$

The second law of Newton can be expressed in terms of momentum:

$F_{net} = \frac{\Delta p}{\Delta t}$

The average net force on a system is equal to the rate of change of the momentum of the system.

Examples:

  1. A ball of mass 0.25 kg is moving to the right at a speed of $7.4 ms^{–1}$. Calculate the momentum of the ball.

    $$ p = mv\\ p = 0.25*7.4\\ p = 1.85 (\text{Ns or kgms}^{-1}) $$

  2. A ball of mass 0.25 kg is moving to the right at a speed of $7.4 m s^{–1}$. It strikes a wall at 90° and rebounds from the wall leaving it with a speed of $5.8 m s^{–1}$ moving to the left. Calculate the change in momentum.

    $$ \Delta p = m\Delta v\\ \Delta p = 0.25*(-5.8-7.4)\\ \Delta p = 0.25*(-13.2) = -3.3 kgms^{-1} $$

  3. Gravel falls vertically on a conveyor belt at a rate of $\sigma kg s^{−1}$, as shown below.

    1. Determine:
      1. the force that must be applied on the belt to keep it moving at constant speed v
      2. the power that must be supplied by the motor turning the belt
      3. the rate at which the kinetic energy of the gravel is changing.

    Untitled

    $$ \bold{\text{i. }} F_{net} = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t} = \frac{v\Delta (m)}{\Delta t} = v\sigma\\{}\\\bold{\text{ii. }} P = Fv= v\sigma *v =v^2\sigma\\{}\\\bold{\text{iii. }} E_K = \frac{1}{2}mv^2 = \frac{1}{2}\sigma v^2 \downarrow \text{This is the increase in kinetic energy in a time of 1s.}\\{} $$

    b. The rate of increase in kinetic energy is less than the power supplied. This is because the power supplied by the motor goes to increase the kinetic energy of the gravel and also to provide the energy needed to accelerate the gravel from 0 to speed v in the short interval of time when the gravel slides on the belt before achieving the constant final speed v.