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Secondary Magazine - Issue 130: Sixth Sense


Created on 03 February 2016 by ncetm_administrator
Updated on 17 February 2016 by ncetm_administrator

 

Secondary Magazine Issue 130Shutterstock image 217318573
 

Sixth Sense

Back in Issue 123, we advocated an approach to teaching mechanics that begins by introducing the key principles in one dimension; in this article we discuss how you can extend the ideas into two dimensions with your M1 students. The very significant benefit of this approach is that your students have already understood all of the mechanical concepts: components are fiddly to work out correctly, and if you try to teach them components from the start they can become bogged down in the trigonometrical arguments and totally miss the mechanics that you’re trying to introduce. By separating these processes, students can “learn mechanics” first, and then later can apply the skills that they’ve been honing with their core teacher to solve more complicated mechanical problems.

So let’s assume that your students have developed a reasonable amount of procedural fluency in using Newton’s Laws to find unknown forces, and with the constant acceleration formulae to calculate further information given a set of initial conditions. It should be a small step for them to tackle the question:

An object moves with constant horizontal velocity \small 4ms^{-1} and constant vertical velocity \small 3ms^{-1}: how far has it travelled after 1 second? Where is it at this moment?

You may prefer to use “x direction” and “y direction” or simply “to the right” and “upwards” depending on the language you’ve used to date, but whatever you say students tend to be relatively comfortable with the idea that “every second the object moves 4m to the right and 3m up, so that’s 5m from its start point on a diagonal path (whose angle we can find using arctan 0.75)”. In our experience, this comes so naturally to students who are, by now, very familiar with Pythagoras and trigonometry, that it barely feels like we’ve introduced a new idea here. In fact, we’ve set up the entire process that allows all of our previous work to extend into two dimensions.

Next, let’s ask the reverse question:

An object travels with constant speed \small 8ms^{-1} at an angle of 30° above the x axis. Find its “rightwards velocity” and its “upwards velocity”.

A quick sketch enables students to use more trigonometry to find the correct answers (and if you don’t let them use a calculator, you can check that they remember their “trig special values” too!).

So, we’re claiming that it seems entirely natural and untroubling to split velocities into components, and this is the key to the next few lessons on projectile motion. By the time students have mastered all of the techniques involved they will be quite happy to use the same ideas to split forces into components too.

Time for another question, and perhaps worth starting with something like:

In playing a shot, a cricketer hits the ball horizontally and to the right so that, at the moment of impact, it has horizontal velocity \small 7ms^{-1}. If this happens 44.1cm above the ground, how far away is it when it first bounces?

It is sensible at this point to encourage students to get into the habit of always drawing a sketch of the motion; and, certainly, to summarise all of the information in the question:

  Horizontal (to the right) Vertical (downwards)
s x 44.1
u 7 0
v    
a 0 9.8
t t t

The conceptually difficult line of the table is the “a” line: why do we model there being no acceleration horizontally, but some acceleration (g, specifically) downwards? Your students will need to be confident with the Newtonian model that forces cause accelerations, and so modelling there being no force horizontally (modelling air resistance as negligible) means no acceleration in that direction. Their riposte may well be “but there is a force on the ball – it’s been hit by the bat!”, and so there will need to be discussion of the distinction between the instantaneous impulse that, at the moment of contact between bat and ball, changes the ball’s velocity (momentum, to be precise) – both its direction and magnitude (speed) – and thereafter the absence of any additional or further force acting on the ball, other than gravity. A comparison with a rocket may be helpful.

Then we can discuss -

What do we know? What do we want to know? Why is it not enough just to say “horizontal” and “vertical”? What’s the link between the two columns?

- and we’re now in a position to apply the constant acceleration formulae to complete the question:

Vertically: \small s=ut+\frac{1}{2}^{2}\Rightarrow 44.1=0+4.9t^{2}\Rightarrow t=3

Horizontally: same formula \small \Rightarrow x=7t=21m

A few practice questions like this to ensure that the tabular structure is in place (but keep asking them about the values in the “a” row) should prepare students for the next step:

A ball is kicked off the ground. It is given an initial velocity of \small 9ms^{-1} at an angle of 20°. For how long is it airborne? How far from the kicker does it land?

We can head straight for another table, but it makes sense here to think “upwards” rather than “downwards” [though either will work, if handled correctly!]:

  Horizontal (to the right) Vertical (upwards)
s x 0
u 9cos20° 9sin20°
v    
a 0 9.8
t t t

There’s often a conversation here about the “v” row: some students are quick to suggest that when the ball lands, either or both of these should be 0. Have your response prepared: we’re modelling the motion in flight with only gravity acting on the ball and our model doesn’t know about the ground. In the instant before hitting the ground, the ball must have velocity as per our model. As soon as the ball hits the ground our model is no longer valid as we have to model a new force acting on the ball (the instantaneous impulse of the ball-ground contact force) and the ball will start to behave differently.

Using the same formula as before (it gets used a lot in projectile motion – make sure your students are factorisation / quadratic formula experts!):

Vertically: \small 0=3.07818t-4.9t^{2} so \small t=0 or \small t=0.628 (3sf)

Horizontally: \small t=0 is the start of the motion, so we want to use the other value to find \small x=9cos20\times answer still on calculator \small =8.46m

The other “classic” question is about the maximum height achieved and it is worth thinking about what this means: again, students often find it difficult at first to comprehend that their “vertical v” must be 0 for an instant – but once they have thought about v as a positive quantity on the way up, and a negative quantity on the way down, they can be convinced that v must be 0 somewhere, and that that place must be where the maximum height is reached.

There are several lessons to spend here developing fluency, and procedural variation presents itself in some of the more challenging set-ups:

  • For how long is the object more than 1m above ground?
  • Does the object clear a wall that is 4m tall and 6m away?
  • An object is projected from 5m above ground, where does it land?
  • Two objects are projected towards each other, do they collide?

Having done all of this work on projectiles, the next time you discuss forces, the students shouldn’t (!) bat an eyelid when you draw a piece of rope pulling an object at an angle:

It will seem only natural, given all of our work with angles in projectiles, to suggest that the force can be split into components and from this that unknown forces and accelerations can be calculated using Newton’s 2nd Law both horizontally and vertically.

There’s very little teaching to be done now: if you’ve done all the work in one dimension, students just need lots and lots of interesting problems to solve – blocks on slopes, trains pulling trucks up hills – to ensure that they are entirely confident with the component process; further mathematicians can also revisit impulse, work done and impact at this point. Once they have had practice revisiting all the one-dimensional contexts but now with two-dimensional information, then they can look at more abstract questions such as “here is a set of forces, what’s the magnitude and direction of the resultant force?” – which is leading them to think about vectors.

 
 

 

 
 
 
 
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Comments

 


17 February 2016 13:53
Another set of clear and sensible advice to de-mystify mechanics (or rather to avoid the process of mystification that happens when teaching is not absolutely accurate and precise). Thank you.
By alantb
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