A small ball B is connected to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The ball is projected with speed 1 m s¯¹ vertically downwards from a position 0.4 m vertically below A, and reaches its greatest speed at the point 0.7 m below A.
- (i)Show that the mass of B is 0.9 kg.[2]
- (ii)Calculate the greatest speed of B.[4]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.
- [M1] mg = 12(0.7-0.4) / 0.4 Use T = λx / L
- [AG] m = 0.9 kg A1
- [B1] EPE = 12(0.7 – 0.4)2 / (2 × 0.4) Correct EPE term
- [M1] 0.9v² / 2 = 0.9g(0.7 – 0.4) + 0.9 × 12 / 2 – 12(0.7 – 0.4)² / (2 × 0.4) Attempts a 4 term energy equation
- [A1] Correct equation
- [A1] v = 2 m s¯¹
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