### Archive

Posts Tagged ‘M1’

## All you need is sine

Today I was going through an M1 question with a year 13 student and was surprised to see the method he had used. The question involved finding an angle in a right angled triangle given the opposite and adjacent sides. The learner had used Pythagoras’s Theorem to find the hypotenuse then used the sine ratio to find the angle.

Puzzled I questioned further, thinking he may have instinctively found the hypotenuse without fully reading the question then having all 3 sides so going with the first. This turned out not to be the case:

“I know sine equals opposite over hypotenuse innit sir, I have trouble remembering the other ones so I just always use sine.”

This was extra interesting as earlier I had come across a markscheme which suggested the way to resolve a force at an angle of 30 degrees was to use Fsin30 for the vertical and Fsin60 for the horizontal! Further checking showed this learner did that too.

I wasn’t too sure what to make of it. It’s mathematically correct, so there’s no issue there. The learner has a grasp of the other ratios but is more confident with sine so I can see why he would default to that position, although I hope the extra time it takes isn’t an issue tomorrow. I can’t fathom, however, why the markscheme would show it this way in the first instance. (Not the only time a markscheme has confused me recently!)

What do you think? Have you got any quirky methods like this? Have any if your students? Do you have an idea why a markscheme would default to this position? I’d love to hear your response.

## Dumbed down markschemes?

Today I was working through the June 2014 Edexcel M1 paper for my year 13s, noting down thought processes etc and I arrived at this question: It’s quite a nice, general bouncing ball question that takes into account many of the rules of mechanics. When I finished the question I checked it against the markscheme so I could jot down where each mark came from and on part E I received a bit if a shock.

The weight of the ball is given as is the distance it’s dropped from and the distance it bounces to. In earlier parts of the question you have been asked to calculate the final speed of the bit before the first bounce and the initial speed after. Part E asks for the time between the ball being dropped and the second bounce. I split it into to, used s = ut + (1/2)at^2 (u=0, s=2,a=9.8) to find the time it took to drop, then the same equation for the time between bounces (this time s=0, u= the value already calculated, a=(-9.8)). A simple question, a simple solution and a nice answer. The markscheme, however, says this: I don’t understand why they have felt the need to work out the time between the first bounce and the top then double it. It’s a valid mrthod, granted, but surely looking at the whole motion is sensible as acceleration is constant?

Worryingly, in the notes there’s no other solutions offered and I was left wondering if examiner’s might miss that thus method is not only correct,, but actually more sensible.

Which way would you have done it? Do you think I’m correct thinking mines the mire logical sensible way?