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Discussion in 'International A And AS Level' started by hellodjfos;s'ff, May 16, 2018.
the answer was btw, 1200=60P +80/2(5.5^2-4^2)
btw for the last question i put all my answers in 2 sig figures becuase i got exact values but everyone said that i was supposed to put them in 3 sig figures? will my answers be considered wrong?
not at all, it is fine
is did it in a different way, i first found acceleration using v^2-u^2=2ax and found the a and then used the a in the equation work/distance-P=ma
Itll be accepted no worries
Alright tell me the questions 1, 2,5. 1 being the resistance P question, 2 being the train and power question, 5 being the coefficient of friction question.
1)A runner of mass 80kg increases his speed from 4ms^-1 to 5.5ms^1 against a restistive force P over a distance of 60m. Total work done is 1200J. Find the value of P
2)A train of mass 240000kg travels up a slope inclined at 4 degrees to the horizontal. The resistance to the train's motion is 18000N. At the instant when its speed is 15ms^-1, its deceleration is 0.2ms^-2. Find the power of the train's engine
5)A particle of mass 20kg is on rough plane inclined at 60 degrees to the horizontal. A force of magnitude P which is parallel to the line of greatest slope of the inclined plane is applied to to the particle to "maintain equilibrium" the greatest value of P is twice the smallest value of P. Find the coefficient of friction between the particle and the plane
Edit: My memory serves me well I see
that's vry specific and accurate. the answer was 10.5N right.
U sure the angle in Q2 is correct brow?
yes you can try to solve it now
dude for what price did you buy it, i might need p3 as well.
Ok can someone help me out on this:
My friend showed me his working for question 1. He put gain in kinetic+work done by runner=1200J(he claims that the paper said the work done by runner and achieving increase in speed was 1200J). He then found the driving force of the runner. Put that driving force into F=ma to find P. It looks right to me but he is getting 1N for P. What's he doing wrong, I can't spot it
1200 not only included the work done in increasing the velocity but also against the resistance.
The total work done by the runner is the gain in K.E and the work done against resistance to motion , which is the product of the constant resistive force and the distance travelled. So in this case,
Ok, he misread the question and I don't remember the full question either
wasnt the angle 41.4 ? cuz im pretty sure it was cos inverse 3/4 which is 41.4
Yeas it was
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