(a) A ball P moving with velocity (2u ms^{-1}), collides with a similar ball Q, of different mass, which is at rest. After collision, Q moves with (u ms^{-1}) and P with velocity (frac{1}{2} u ms^{-2}), in the opposite direction. Find the ratio of the masses of P and Q.
(b) Two forces of magnitudes 3 N and 7 N have a resultant of magnitude 5 N. Calculate, correct to one decimal place, the angle between the two forces.
(c) (AB = begin{pmatrix} -4 \ 6 end{pmatrix}) and (CB = begin{pmatrix} 2 \ -3 end{pmatrix}) are two vectors in the XY- plane. If V is the midpoint of AB, find CV.
Explanation
(2u m/s) (0 m/s) : (frac{1}{2} u m/s) (u m/s)
P Q P Q
Before collision : After collision
Momentum before collision = (M_{P} (2u) kg m/s)
Momentum after collision = (-frac{1}{2}u (M_{P}) + M_{Q} (u))
(2u (M_{P}) + frac{1}{2} u (M_{P}) = M_{Q} (u))
(frac{5}{2} M_{P} = M_{Q})
(frac{M_{P}}{M_{Q}} = frac{2}{5})
The ratio = 2:5
(b)
From cosine rule, (5^{2} = 3^{2} + 7^{2} – 2 times 3 times 7 cos (180 – alpha))
(25 = 9 + 49 – 42(-cos alpha))
(25 = 58 + 42cos alpha)
(cos alpha = frac{-33}{42} = -0.7857)
(alpha = cos^{-1} (-0.7857) = 141.79°)
(c)
(overrightarrow{VB} = frac{1}{2}overrightarrow{AB} = frac{1}{2} begin{pmatrix} -4 \ 6 end{pmatrix} = begin{pmatrix} -2 \ 3 end{pmatrix})
(overrightarrow{CV} + overrightarrow{VB} = overrightarrow{CB})
(overrightarrow{CV} = overrightarrow{CB} – overrightarrow{VB})
= (begin{pmatrix} 2 \ -3 end{pmatrix} – begin{pmatrix} -2 \ 3 end{pmatrix})
= (begin{pmatrix} 2 + 2 \ -3 – 3 end{pmatrix})
= (begin{pmatrix} 4 \ -6 end{pmatrix})