16a. A ball P moving with velocity 2 u m/s, collides with a similar ball Q, of different mass, which is at rest. After the collision, Q moves with u m/s and...

16a.Let the mass of ball P be \( m \) and the mass of ball Q be \( M \).  

Conservation of linear momentum (before and after collision):  
\(m \times 2u + M \times 0 = m \times \left(-\frac{1}{2}u\right) + M \times u\) 
2mu = -\(\frac{1}{2}\)mu + Mu  
Divide through by \( u \) (assuming \( u \neq 0 \)):  
2m = -\(\frac{1}{2}\)m + M
 M = 2m + \(\frac{1}{2}\)m = \(\frac{5}{2}\)m
Thus, the ratio of the mass of P to the mass of Q is  
m: M = 2: 5

bi.  Let the angle between the two forces be \( \theta \).  

By the parallelogram law (or cosine rule for vector addition):  
\(R^2 = 3^2 + 7^2 + 2 \times 3 \times 7 \times \cos\theta\)
\(5^2 = 9 + 49 + 42\cos\theta\)
\(25 = 58 + 42\cos\theta\)
\(42\cos\theta\) = 25 - 58 = -33
\(\cos\theta = -\frac{33}{42} = -\frac{11}{14} \approx -0.7857\)
\(\theta = \cos^{-1}(-0.7857) = 141.8^\circ\)

c.  Given: \(\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}, \quad
\overrightarrow{CB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)

V is the midpoint of AB, so its position vector satisfies  
\(\vec{V} = \frac{\vec{A} + \vec{B}}{2}.\)
Vector \(\overrightarrow{CV}\) is  
\(\overrightarrow{CV} = \vec{V} - \vec{C} = \overrightarrow{CB} - \frac{1}{2}\overrightarrow{AB}.\)
Substitute the given vectors:  
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\) 
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)