One end of a light elastic string is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.4 kg. The string has natural length 0.6 m and modulus of elasticity 24 N. The particle is released from rest at O. Find the two possible values of the distance OP for which the particle has speed 1.5 m s⁻¹.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1] EPE = 24 (x -0.6)²/(2 × 0.6)
2. [M1] 0.4 x 1.5² /2 = 0.4gx – 24 (x−0.6)²/(2 × 0.6)
3. [M1] [20 x² - 28x + 7.65 = 0 or equivalent]
4. [A1] OP = 1.0279 m, 0.372 m (reject)
5. [M1] 0.4 x 1.5²/2 = 0.4gx
6. [A1] OP = 0.1125 m
7. [B1] (Alternative method) EPE = 24 x² /(2 × 0.6)
8. [M1] (Alternative method) 0.4 × 1.5² /2 = 0.4g(x + 0.6) − 24 x² /(2 × 0.6)
9. [M1] (Alternative method) [20 x² - 4x – 1.95 = 0 or equivalent ]
10. [A1] (Alternative method) [ x = 0.42787, – 0.22787 .reject] OP = 0.6 + 0.42787 = 1.0279
11. [M1] (Alternative method) 0.4 x 1.5² /2 = 0.4g( x + 0.6) [x = -0.4875]
12. [A1] (Alternative method) OP = 0.6-0.4875 = 0.1125