(a) The position vectors of the points X and Y are \(x = (-2i +
(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) the angle between x and y ; (iv) the unit vector in the direction of \((x + y)\).
(b) A bullet of mass 0.084kg is fired horizontally into a 20 kg block of wood at rest on a smooth floor. If they both move at a velocity of \(0.24 ms^{-1}\) after impact; Calculate, correct to two decimal places, the initial velocity of the bullet.
Explanation
(a) \(x = (-2i + 5j) ; y = (i - 7j)\)
(i) \((3x + 2y) = 3(-2i + 5j) + 2(i - 7j)\)
= \((-6i + 15j) + (2i - 14j)\)
= \(-4i + j\)
(ii) \((y - 2x) = (i - 7j) - 2(-2i + 5j)\)
\((i - 7j) - (-4i + 10j) = (5i - 17j)\)
\(|y - 2x| = |-3i - 17j| = \sqrt{(5^{2}) + (-17 ^{2})}\)
= \(\sqrt{25 + 289} = \sqrt{314}\)
(iii) Let the angle between x and y be \(\theta\).
\(x . y = |x||y| \cos \theta\)
\((-2i + 5j) . (i - 7j) = |-2i + 5j||i - 7j| \cos \theta\)
\(-2 - 35 = (\sqrt{(-2)^{2} + (5)^{2}})(\sqrt{(1)^{2} + (-7)^{2}}) \cos \theta\)
\(-37 = (\sqrt{29})(\sqrt{50}) \cos \theta \implies -37 = (\sqrt{1450})\cos \theta\)
\(\cos \theta = \frac{-37}{\sqrt{1450}} = -0.9717\)
\(\theta = \cos^{-1} (-0.9717) = 166.34°\)
(iv) \(x + y = (-2i + 5j) + (i - 7j) = -i - 2j\)
\(|x + y| = \sqrt{(-1)^{2} + (-2)^{2}} = \sqrt{5}\)
Unit vector = \(\frac{-i - 2j}{\sqrt{5}}\)
(b).jpg)
\(m_{1} = 0.084kg ; m_{2} = 20 kg ; m = 20.084 kg\)
\(u_{1} = ? ; u_{2} = 0 m/s ; v = 0.24 m/s\)
Momentum before collision = \(0.084 u_{1} kg m/s\)
Momentum after collision = \(20.084 \times 0.24 = 4.82016 kg m/s\)
\(\therefore 0.084 u_{1} = 4.82016 \)
\(u_{1} = \frac{4.82016}{0.084} = 57.3829 m/s\)
The initial velocity of bullet is 57.38 ms\(^{-1}\).