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 P with velocity \(\frac{1}{2}\) u m/s in the opposite direction. Find the ratio of the mass of P and Q.
b. Two forces of magnitude 3N and 7N have a resultant of magnitude 5N. Calculate, correct to one decimal place, the angle between the two forces.
c. AB\(^→\) \(\left| \begin{array}{cc} -4 \\ 6 \end{array} \right|\) and CB\(^→\) \(\left| \begin{array}{cc} 2 \\ 3 \end{array} \right|\) are two vectors in the XY plane. If V is the midpoint AB\(^→\). Find CV\(^→\)
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:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
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:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
