Projectiles, circular motion and delayed objects Problem set B Answers

1. (a) (i) 61.4° above the horizontal (ii) 31.5° below the horizontal
2. (b) 1.5 s
3. (c) \(\mathbf{r} = \begin{pmatrix}8t\\14.7t − 4.9t^2\end{pmatrix}\); greatest height 11.0 m

32.4 m

1. (a) \(|\mathbf{r}| = 1.8\) for all \(t\), so a circle of radius 1.8 m
2. (b) \(\mathbf{v} = \begin{pmatrix}-9\sin 5t\\9\cos 5t\end{pmatrix}\) m s⁻¹
3. (c) \(\mathbf{v}\cdot\mathbf{r} = 0\)
4. (d) 9 m s⁻¹

1. (a) 0.9 m s⁻¹
2. (b) 5 m s⁻¹

1. (a) \(\mathbf{v} = \begin{pmatrix}-8\sin 2t\\8\cos 2t\end{pmatrix}\), \(\mathbf{a} = \begin{pmatrix}-16\cos 2t\\−16\sin 2t\end{pmatrix}\)
2. (b) \(k = −4\)
3. (c) Towards the centre of the circle (the origin)
4. (d) 16 m s⁻²

1. (a) \(\mathbf{r} = \begin{pmatrix}3\sin 0.7t\\−3\cos 0.7t\end{pmatrix}\); \(|\mathbf{r}| = 3\), so a circle of radius 3 m
2. (b) Initial position \(\begin{pmatrix}0\\−3\end{pmatrix}\) m, radius 3 m
3. (c) 8.98 s
4. (d) \(\dfrac{2\pi\times 3}{8.98} = 2.1\) m s⁻¹, the speed

1. (a) \(\mathbf{r} = \begin{pmatrix}4t − 1.5t^2\\t + \dfrac{2}{3}t^3\end{pmatrix}\) m
2. (b) 21.1 m

22.3 m

\(b = 14.7\)

1. (a) \(\mathbf{r} = \begin{pmatrix}5t\\16t − 4.9t^2\end{pmatrix}\)
2. (b) \(\mathbf{r} = \begin{pmatrix}5(t − 3)\\16(t − 3) − 4.9(t − 3)^2\end{pmatrix}\)

1. (a) \(b = 23.75\)
2. (b) 33.2 m

1. (a) \(d = \sqrt{72^2 + (164.1 − 29.4t)^2}\)
2. (b) 5.58 s

1. (a) \(|\mathbf{r}| = a\), constant: a circle of radius a
2. (b) \(|\mathbf{v}| = an\)
3. (c) \(T = \dfrac{2\pi}{n}\)
4. (d) \(|\mathbf{a}| = an^2 = \dfrac{v^2}{a}\)
5. (e) 3.2 m s⁻¹, 1.57 s, 12.8 m s⁻²

1. (a) \(\mathbf{v}\cdot\mathbf{r} = 0\)
2. (b) \(\mathbf{a} = -k^2\mathbf{r}\)