Momentum
Forces and Motion

Episode 222: Impulse of a force

Lesson for 16-19 IOP TAP

This episode focuses on the forces involved in changing momentum, and introduces the concept of impulse.

Lesson Summary

  • Discussion: To show that impulse equals change in momentum (10 minutes)
  • Demonstration: A qualitative comparison of the trade-off between force and collision time (10 minutes)
  • Worked example and discussion: Using the area under a force time graph (20 minutes)
  • Demonstration: Force, time and impulse (20 minutes)
  • Student questions: Testing understanding, including calculations of area under a graph (30 minutes)

Discussion: To show that impulse equals change in momentum

It is obvious that, the greater a force and the longer it acts for, the greater will be its effect. Hence, the quantity F  ×  t is important. This is known as the impulse of the force, measured in N s.

Show that it has the same units as momentum (N s = kg m s-1).

State the impulse equation, which describes the effect of a force acting on one body:

Impulse = change in momentum

Ft =  Δ mv or Ft = mv − mu

(You may need to explain that Δ mv means the change in momentum, and that Δ itself is not a quantity.)

At this point, you might return to the discussion at the end of Episode 219, and extend it slightly to deduce the impulse equation.

If appropriate to the specification you are following, introduce the alternative way of writing this expression: F = dmvd t and explain that this is an alternative way of writing Newton’s second law.

Or without using calculus notation

F = m × a

F = m (vfinal  −  vinitial)t

F = mvfinal  −  mvinitialt

F = pfinal  −  pinitialt

i.e. Ft = change in momentum

Demonstration: A qualitative comparison of the trade-off between force and collision time

The average force on an object can be reduced by increasing the time over which it acts.

Episode 222-1: Momentum demonstrations (Word, 27 KB)


RP 23: Egg and sheet demo – egg fails to break when thrown against a suspended sheet

Relate this to real world collisions such as car crashes, or jumping off a table (important safety point: never try jumping, even small heights, with stiff legs and ankles – it is remarkably easy to break a bone). NB. No-one should climb onto a stool or table in the lab. If any experimental work is undertaken, it must be done in the gym under the supervision of a trained PE teacher.

Worked example and discussion: Using the area under a force time graph

Use a simple example involving constant force acting for a known time.

Episode 222-2: Impulse examples (Word, 37 KB)


Forces between objects are not usually constant. Ask students to think about, say, a bat striking a ball, or a collision between cars or a car hitting a wall. What would the force-time graph look like? (The force will rise to a peak and then fall, being approximately triangular.)

The impulse of the force will be given by the area under the graph, since this area represents the sum of force ´ time in each time interval.

Demonstration: Force, time and impulse

The average force on a football can be calculated as a pupil led demonstration.

(An alternative approach is to use sensors, but this is more demanding to set up and to interpret.)

Episode 222-3: Kicking a football (Word, 128 KB)


Student questions: Testing understanding, including calculations of area under a graph

Look for two types of question:

Numerical questions involving impulse, including force-time graphs. These will require the student to find the area under the graph, the total momentum change and possibly the average force. This latter is a particularly difficult concept, being the equivalent force which, if applied constantly over the entire contact time would produce the same impulse. This can be related to the idea of an average speed.

Descriptive questions which take these ideas and develop them in contexts such as seat belts and airbags.

Episode 222-4: Momentum questions (Word, 77 KB)


If the students are to be stretched mathematically, look for questions which use dmvd t directly, such as the force caused by water from a hose hitting a wall, or the lift provided by the air driven down by a helicopter’s blades.

Momentum
appears in the relation p=mv F=dp/dt λ=h/p ΔxΔp>ℏ/2
has the special case Angular Momentum
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