(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.
\(2u m/s\) \(0 m/s\) : \(\frac{1}{2} u m/s\) \(u m/s\)
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, \(5^{2} = 3^{2} + 7^{2} - 2 \times 3 \times 7 \cos (180 - \alpha)\)
\(25 = 9 + 49 - 42(-\cos \alpha)\)
\(25 = 58 + 42\cos \alpha\)
\(\cos \alpha = \frac{-33}{42} = -0.7857\)
\(\alpha = \cos^{-1} (-0.7857) = 141.79°\)
(c) 
\(\overrightarrow{VB} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
\(\overrightarrow{CV} + \overrightarrow{VB} = \overrightarrow{CB}\)
\(\overrightarrow{CV} = \overrightarrow{CB} - \overrightarrow{VB}\)
= \(\begin{pmatrix} 2 \\ -3 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
= \(\begin{pmatrix} 2 + 2 \\ -3 - 3 \end{pmatrix}\)
= \(\begin{pmatrix} 4 \\ -6 \end{pmatrix}\)
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