M2 June 2018 Q2
2.
The points \(A\), \(B\) and \(C\) lie on a smooth horizontal plane. A small ball of mass 0.2 kg is moving along the line \(AB\) with speed 4 m s\(^{-1}\). When the ball is at \(B\), the ball is given an impulse. Immediately after the impulse is given, the ball moves along the line \(BC\) with speed 7 m s\(^{-1}\). The line \(BC\) makes an angle of 35\(^\circ\) with the line \(AB\), as shown in Figure 1.
| Scheme | Marks |
|---|---|
| Using column vectors or \(\mathbf{i}\) and \(\mathbf{j}\): \(\begin{pmatrix}I\cos\theta\\I\sin\theta\end{pmatrix} = 0.2\begin{pmatrix}7\cos 35\\7\sin 35\end{pmatrix} - 0.2\begin{pmatrix}4\\0\end{pmatrix}\) | M1 |
| \(\left(= \begin{pmatrix}0.347\\0.803\end{pmatrix}\right)\) | A1 |
| \(|I| = \sqrt{0.347^2 + 0.803^2}\) | DM1 |
| \(|I| = 0.875\) | A1 |
| (4) |
Notes
M1 Must be subtracting. Need both components. Could consider components separately
A1 Correct unsimplified equation. Accept +/-
DM1 Use Pythagoras to find magnitude. Dependent on the previous M1
A1 0.87 or better
a alt
Alternative using vector triangle | M1 |
| Cosine rule: \(|I|^2 = 1.4^2 + 0.8^2 - 2 \times 1.4 \times 0.8 \times \cos 35^\circ\) | DM1 A1 |
| \(|I| = 0.875\) (N s) (0.87 or better) | A1 |
M1 Allow with velocities rather than impulse/momentum
DM1 A1 Dependent on the previous M1
| Scheme | Marks |
|---|---|
| \(\tan\theta = \dfrac{0.803}{0.347}\ \ \ \ \left(\text{or } \cos\theta = \dfrac{0.347}{0.875}\right)\) | M1 A1ft |
| \(\theta = 66.6^\circ\ \ (67)\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1 Trig ratio of a relevant angle (using velocities or impulse/momentum)
A1ft Correct expression for correct \(\theta\). Ft on values from (a). Do not ISW
A1 Or better from correct work.
b alt
| Sine rule: \(\ \dfrac{\sin\theta}{1.4} = \dfrac{\sin 35}{|I|}\) | M1 A1ft |
| \(\theta = 66.6^\circ\ \ (67)\) | A1 |
A1 Or better