M3 June 2013 (R) Q3
3. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 2 m and modulus of elasticity 20 N. The other end of the spring is attached to a fixed point \(A\). The particle \(P\) is held at rest at the point \(B\), which is 1 m vertically below \(A\), and then released.
(a) Find the acceleration of \(P\) immediately after it is released from rest. (4)
The particle comes to instantaneous rest for the first time at the point \(C\).
(b) Find the distance \(BC\). (6)
| Scheme | Marks |
|---|---|
| Weight + thrust = mass x accn. | M1 |
| \(0.5 \times g + \dfrac{20 \times 1}{2} = 0.5a\) | B1(thrust) A1ft |
| \(a = g + 20 = 29.8 \approx 30\) (m s\(^{-2}\)) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Change in GPE \(= mg(x + 1)\) | B1 |
| EPE at B \(= \dfrac{20 \times 1^2}{2 \times 2}\) or EPE at C \(= \dfrac{20 \times x^2}{2 \times 2}\) | B1 |
| Conservation of energy: \(\dfrac{20 \times 1^2}{2 \times 2} + mgh = \dfrac{20 \times x^2}{2 \times 2}\) \(h = x + 1\) | M1A1 |
| \(5 + 0.5g(x + 1) = 5x^2\) | |
| \(5x^2 - 0.5gx - (5 + .5g) = 0\) | |
| \(x = \dfrac{0.5g + \sqrt{(0.5g)^2 + 20(5 + 0.5g)}}{10} = 1.98\) | M1dep |
| Distance \(BC = 1 + 1.98 = 2.98\) (m) | A1 |
| (6) | |
| (10 marks) |