(a) The position vectors of the points X and Y are \(x = (-2i + 5j)\) and \(y = (i - 7j)\) respectively. Find : (i) (3x + 2y) ; (ii) \(|(y - 2x)|\) ; (iii) ...

Answer Details

(a) With \(\mathbf{x} = -2\mathbf{i} + 5\mathbf{j}\) and \(\mathbf{y} = \mathbf{i} - 7\mathbf{j}\):

(i) \(3\mathbf{x} + 2\mathbf{y} = (-6\mathbf{i} + 15\mathbf{j}) + (2\mathbf{i} - 14\mathbf{j}) = -4\mathbf{i} + \mathbf{j}\).

(ii) \(\mathbf{y} - 2\mathbf{x} = (\mathbf{i} - 7\mathbf{j}) - (-4\mathbf{i} + 10\mathbf{j}) = 5\mathbf{i} - 17\mathbf{j}\), so \(|\mathbf{y} - 2\mathbf{x}| = \sqrt{25 + 289} = \sqrt{314} \approx 17.7\).

(iii) \(\cos\theta = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}||\mathbf{y}|}\), where \(\mathbf{x}\cdot\mathbf{y} = (-2)(1) + (5)(-7) = -37\), \(|\mathbf{x}| = \sqrt{29}\), \(|\mathbf{y}| = \sqrt{50}\):

\[\cos\theta = \frac{-37}{\sqrt{29}\,\sqrt{50}} = \frac{-37}{38.08} = -0.9717 \Rightarrow \theta \approx 166.3^\circ.\]

(iv) \(\mathbf{x} + \mathbf{y} = -\mathbf{i} - 2\mathbf{j}\), \(|\mathbf{x} + \mathbf{y}| = \sqrt{5}\). Unit vector \(= \dfrac{-\mathbf{i} - 2\mathbf{j}}{\sqrt{5}} \approx -0.45\mathbf{i} - 0.89\mathbf{j}\).

(b) Conservation of momentum (block initially at rest). Let \(u\) be the bullet's initial speed:

\[0.084\,u = (0.084 + 20)(0.24) \Rightarrow u = \frac{20.084 \times 0.24}{0.084} = \frac{4.82016}{0.084} \approx 57.38\ \text{m s}^{-1}.\]