Christian Lawson-Perfect

Christian Lawson-Perfect commented on Resolve a force into $x$ and $y$ components 9 years, 5 months ago

Use the LaTeX commands \sin and \cos to get he correct rendering $\sin$ and $\cos$.

For part $b$, I'd say $F \sin \theta$ instead of $F \cos (90-\theta)$. I use SOHCAHTOA to remember which function to use.

Christian Lawson-Perfect commented on Resolve a force into $x$ and $y$ components 9 years, 5 months ago

Need units in the statement: $F \, \mathrm{N}$ and $\theta °$.

Christian Lawson-Perfect commented on Particle acted on by gravity and a constant force against its motion 9 years, 5 months ago

For the first time, we have a negative sign for an acceleration, in part a! You must say what direction to resolve in.

Christian Lawson-Perfect on Ball in a box 9 years, 5 months ago

Saved a checkpoint:

Fixed "madnitude" to "magnitude" in part a.

Christian Lawson-Perfect commented on Ball in a box 9 years, 5 months ago

A diagram would really help here!

The statement could be made easier to understand. Maybe something like

A box is being lifted upwards by a crane. It is accelerating at a rate of $a \, \mathrm{ms^{-2}}$ upwards.

A ball of mass $m\, \mathrm{kg}$ is resting inside the box.

Christian Lawson-Perfect commented on Constant deceleration 9 years, 5 months ago

Statement: you decelerate at a rate, not a speed.

a)

I would say "after it has travelled ...", just to make things clearer.

b)

I think you mean "tension" instead of "thrust".

c)

This part is trivial - just subtract the given resistance from your answer to b. What's the point of it? It's not used in the following part, and in fact students might think they need to recalculate the acceleration for the new force.

d)

I'd use $s = ut + \frac{1}{2}at^2$ instead of reusing the speed I worked out in part a, to avoid a rounding error. Maybe mention both methods in the advice, and allow wiggle room in the marking to tolerate any rounding error.

Christian Lawson-Perfect commented on Find tension in rope attached to mass 9 years, 5 months ago

The statement could very easily have an unlabelled image of a mass hanging off a rope, just for illustration.

I wouldn't say "the block of mass moves upward", I'd just say "the block moves upward".

Again, say what precision you want.

In parts a and b you resolve in different directions. Don't you find this confusing? I'd prefer to always resolve in the same direction, but change the sign of the acceleration, so the equations always look the same.

In the advice for part d, you need to give the units for the acceleration $a$.I found it hard to parse the prompt; something like "the block is moving downwards but decelerating at a rate of ..." would be clearer to me.

Christian Lawson-Perfect commented on Find distance travelled and reaction force given mass and force 9 years, 5 months ago

Say what precision you want.

State what units you're using. You could just put it after the input.

I didn't round off the acceleration when working out the distance travelled, which made my answer to part b wrong. Consider adding a bit of wiggle room to the marking.

Again, state $g = 9.8$ for part c.

How would you know the frictional force without first knowing the particle's acceleration or the coefficient of friction? You could give the coefficient of friction and have the first part be to work out the friction force. Or, just give both the coefficient and the force, to explain how you would know it.

Christian Lawson-Perfect on Acceleration and weight: apply $F=ma$ and $W=mg$ 9 years, 5 months ago

Saved a checkpoint:

I've capitalised "Newtons" and changed "it's mass" to "its mass".

Christian Lawson-Perfect commented on Particle in vertical equilibrium, accelerating horizontally 9 years, 5 months ago

Oh, and again you need to say in the prompt what precision you want, when you set a precision restriction in the marking.