Mark Scheme for January 2001 M1 exam

Hey, does anyone have the mark scheme for M1? If you do would you be able to post it on this thread or PM it to me. Thanks very much.

It's the last 2 questions I'm stuck on, I'll post them here incase anyone can help.

6) A parachutist drops from a helicopter H and falls vertically from rest towards the ground. Her parachute opens 2 seconds after she leaves H and her speed then reduces to 4ms(-1). For the first 2 seconds her motion is modelled as that of a particle falling freely under gravity. For the next 5seconds the model is motin with constant deceleration, so that her speed is 4ms(-1) at the end of this period. For the rest of the time before she reaches the ground, the model is motion with constant speed of 4ms(-1).
(a) Sketch a speed-time graph to illustrate her motion from H to the ground.
(b) Find her speed when the parachute opens.

A safety rule states that the helicopter must be high enough to allow the parachute to open and for the speed of a parachutist to reduce to 4ms(-1) before reaching the ground. Using the assumptions made in the above model,

(c) find the minimum height of H for which the woman can make a drop without breaking this safety rule.

Given that H is 125m above ground when the woman starts her drop.
(d) find the total time taken for her to reach the ground.
(e) state one way in which the model could be refined to make it more realistic.

7) A sledge of mass 78kg is pulled up a slope by means of a rope. The slope is modelled as a rough plane inclined at an angle ∂ to the horizontal, where tan ∂ = 5/12. The rope is modelled as light and inextensible and is in a line of greatest slope of the plane. The coefficient of friction between the sledge and the slope is 0.25. Given that the sledge is accelerating up the slope with acceleration 0.5 ms(-2),
(a) find the tension in the rope.
The rope suddenly breaks, Subsequently the sledge comes to instantaneous rest and then starts sliding down the slope.
(b) Find the acceleration of the sledge down the slope after it has come to instantaneous rest.

Reply 1

mizfissy815

Question - 6

a)

b) v= u+at
v=0+(9.81)*(2)
v= 19.62 ms^-1

c)Area travelled in the 7 seconds- Area of triangle + area of trapezium
Area of triangle = 0.5bh
Distance = 0.5 x 2 x 19.62 = 19.62
Distance= 19.62 m

Area of trapzium= 0.5(a+b)h
Distance= 0.5(4+19.62) x 5
Distance = 59.05 m

Total distance travelled in first 7 seconds= 19.62 + 59.05 = 78.67 m

d) Area under the rectangle= 125 - 78.67= 46.33m
A= l x w................. (v x t)
46.33= 4 x w
w= 11.58s

Total time= 7 + 11.58 =18.58s

e) Consider air resistance.

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Q- 7

a) T- mgsinθ -µR = ma
T- 78gsin22.6 - 0.25(78gcos22.6) =78 (0.5)
T - 294.055 -176.6 = 39
T= 39 + 294.055 + 176.6
T= 509.7N

b) mgsinθ - µR =ma
78gsin22.6 - 0.25(78gcos22.6) = 78a
78gsin22.6 - 176.6 = 78a
a= 1.51 ms^-2

H.JPG8.8KB

Reply 2

KgeoK

Thanks so much for that!