A2 October 2020 Q6
6.
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point \(O\). The particle is held with the string taut and \(OP\) horizontal. The particle is then projected vertically downwards with speed \(u\), where \(u^2 = \dfrac{9}{5}gl\). When \(OP\) has turned through an angle \(\alpha\) and the string is still taut, the speed of \(P\) is \(v\), as shown in Figure 5. At this instant the tension in the string is \(T\).
| Scheme | Marks | AO |
|---|---|---|
| Conservation of energy: | M1 | 3.1a |
| \(\dfrac{1}{2}mu^2 + mgl\sin\alpha = \dfrac{1}{2}mv^2 \qquad \left(v^2 = \dfrac{9gl}{5} + 2gl\sin\alpha\right)\) | A1 | 1.1b |
| Equation of motion: | M1 | 3.1a |
| \(T - mg\sin\alpha = \dfrac{mv^2}{l}\) | A1 | 1.1b |
| Complete strategy to find \(T\) in terms of \(\alpha\) | M1 | 2.1 |
| \(\begin{aligned}\Rightarrow T &= mg\sin\alpha + \dfrac{mv^2}{l} = mg\sin\alpha + \dfrac{9mg}{5} + 2mg\sin\alpha\\ &= 3mg\sin\alpha + \dfrac{9mg}{5}\ \ ^{*}\end{aligned}\) | A1* | 2.2a |
| (6) |
Notes
M1: Must include all terms. Condone sign errors and sin/cos confusion
A1: Correct unsimplified equation
M1: Must include all terms. Condone sign errors and sin/cos confusion
A1: Correct unsimplified equation
M1: Complete strategy to form an expression for \(T\) in terms of \(\alpha\) e.g. by using conservation of energy and the circular motion to form sufficient equations to obtain an equation in \(T\) only.
A1*: Obtain given answer from correct working
| Scheme | Marks | AO |
|---|---|---|
| String slack \(\Rightarrow T = 0 \Rightarrow \sin\alpha = -\dfrac{3}{5}\) | B1 | 3.1a |
| Use energy equation to find \(v\): | M1 | 1.1b |
| \(v^2 = \dfrac{9gl}{5} - \dfrac{3}{5}\times 2gl,\quad v = \sqrt{\dfrac{3gl}{5}}\) | A1 | 1.1b |
| (3) |
Notes
B1: Correct deduction
M1: Substitute value to find \(v^2\)
A1: Correct only
| Scheme | Marks | AO |
|---|---|---|
| Initial vertical component of speed \(= \dfrac{4}{5}\times\sqrt{\dfrac{3gl}{5}}\) | B1 | 1.1b |
| Use of suvat: \(0 = u^2 - 2gh = \dfrac{16}{25}\times\dfrac{3gl}{5} - 2gh\) | M1 | 3.1a |
| \(h = \dfrac{24l}{125}\) | A1 | 1.1b |
| Total height above \(O = \dfrac{3l}{5} + \dfrac{24l}{125} = \dfrac{99l}{125}\) | A1 | 2.2a |
| (4) | ||
| (13 marks) |
Notes
B1: Correct vertical component of velocity when string goes slack.
M1: Use of \(v^2 = u^2 + 2as\) or alternative complete method to find the additional height.
A1: Additional height correct
A1: Total height correct
Alternative (c)
| Scheme | Marks | AO |
|---|---|---|
| Initial horizontal component of speed \(= \dfrac{3}{5}\times\sqrt{\dfrac{3gl}{5}}\) | B1 | 1.1b |
| Conservation of energy: | M1 | 3.1a |
| \(mgh = \dfrac{1}{2}m\left(\dfrac{9}{5}\right)gl - \dfrac{1}{2}m\left(\dfrac{9}{25}\times\dfrac{3gl}{5}\right)\) | A1 | 1.1b |
| \(h = \dfrac{99l}{125}\) | A1 | 2.2a |
| (4) |
B1: Correct horizontal component of velocity when string goes slack.
M1: Use of conservation of energy or alternative complete method to find the height. All terms required. Condone sign errors.
A1: Correct unsimplified equation in \(h\) and \(l\)
A1: Correct answer