(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)
(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}).