A-LevelMathematicsDifferential equations9709/52 · May/June 2017 · Q48 Marks
A small object of mass 0.4 kg is released from rest at a point 8 m above the ground. The object descends vertically and when its downwards displacement from its initial position is x m the object has velocity v m s¯¹. While the object is moving, a force of magnitude 0.2v² N opposes the motion.
- (i)Show that v dv/dx = 10 – 0.5v².[2]
- (ii)Express v in terms of x.[4]
- (iii)Find the increase in the value of v during the final 4 m of the descent of the object.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [M1] Uses Newton's Second Law vertically
- [A1] AG vdv/dx = 10 - 0.5 v²
- [M1] Separates the variables and attempts to integrate
- [A1] – ln(10 – 0.5 v² ) = x (+ c )
- [M1] Attempts to find c using x = 0, v = 0
- [A1] v = √(20-20e¯ˣ)
- [M1] M1 if x values are substituted into their value for part (ii)
- [A1] Allow 0.04
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