A-LevelMathematicsMechanics9709/52 · Oct/Nov 2019 · Q36 Marks
A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 12N. The other end of the string is attached to a fixed point O. The particle P is projected vertically downwards with speed 2 m s¯¹ from the point 0.5 m vertically below O. For an instant when the acceleration of P is 4 m s¯² downwards, find the extension of the string and the speed of P.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [M1] 0.5 x 4 = 0.5g - T
- [M1] T = 12e / 0.6
- [A1] e = 0.15
- [B1FT] EPE = 12 * 0.15² / (2 * 0.6) and distance fallen = 0.6 - 0.5 + 0.15
- [M1] 0.5v² / 2 = 0.5 * 2² / 2 + 0.5g(0.6-0.5+0.15) - 12 * 0.15² / (2 * 0.6)
- [A1] v = 2.85 m s⁻¹
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →