WASSCE Further Maths: Vectors and Mechanics Practice Questions & Answers

Let the tensions be $T_1 = 10\text{ N}$T1​=10 N at $30^\circ$30∘ and $T_2$T2​ at $60^\circ$60∘ to the vertical.
Resolving horizontally for equilibrium: $T_1 \sin 30^\circ = T_2 \sin 60^\circ$T1​sin30∘=T2​sin60∘
$10(0.5) = T_2\left(\frac{\sqrt{3}}{2}\right) \implies 5 = T_2\frac{\sqrt{3}}{2} \implies T_2 = \frac{10}{\sqrt{3}}\text{ N}$10(0.5)=T2​(23​​)⟹5=T2​23​​⟹T2​=3​10​ N.
Resolving vertically: $W = T_1 \cos 30^\circ + T_2 \cos 60^\circ$W=T1​cos30∘+T2​cos60∘
$W = 10\left(\frac{\sqrt{3}}{2}\right) + \frac{10}{\sqrt{3}}(0.5) = 5\sqrt{3} + \frac{5}{\sqrt{3}} = 5\sqrt{3} + \frac{5\sqrt{3}}{3} = \frac{15\sqrt{3} + 5\sqrt{3}}{3} = \frac{20\sqrt{3}}{3}\text{ N}$W=10(23​​)+3​10​(0.5)=53​+3​5​=53​+353​​=3153​+53​​=3203​​ N.

The formula for the horizontal range is $R = \frac{u^2 \sin 2\theta}{g}$R=gu2sin2θ​.
Substitute $u = 20\text{ m/s}$u=20 m/s and $\theta = 30^\circ$θ=30∘:
$R = \frac{20^2 \sin(2 \times 30^\circ)}{10} = \frac{400 \sin 60^\circ}{10} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}\text{ m}$R=10202sin(2×30∘)​=10400sin60∘​=40×23​​=203​ m.

The magnitude of a vector $\mathbf{v} = x\mathbf{i} + y\mathbf{j}$v=xi+yj is $|\mathbf{v}| = \sqrt{x^2 + y^2}$∣v∣=x2+y2​.
Here, $|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$∣v∣=32+(−4)2​=9+16​=25​=5.

Velocity $v = \frac{ds}{dt} = 3t^2 - 12t + 9$v=dtds​=3t2−12t+9.
Set $v = 0$v=0 to find the times when the particle is at rest: $3(t^2 - 4t + 3) = 0 \implies (t-1)(t-3) = 0$3(t2−4t+3)=0⟹(t−1)(t−3)=0. So, $t = 1\text{ s}$t=1 s or $t = 3\text{ s}$t=3 s.
Acceleration $a = \frac{dv}{dt} = 6t - 12$a=dtdv​=6t−12.
At $t = 1$t=1, $a = 6(1) - 12 = -6\text{ m/s}^2$a=6(1)−12=−6 m/s2.
At $t = 3$t=3, $a = 6(3) - 12 = 6\text{ m/s}^2$a=6(3)−12=6 m/s2.
The possible values are $\pm 6\text{ m/s}^2$±6 m/s2.

The third force (equilibrant) must have a magnitude equal and opposite to the resultant of the two forces.
The magnitude of the resultant $R$R of two perpendicular forces is $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ N}$R=32+42​=9+16​=25​=5 N.
Thus, a force of $5\text{ N}$5 N is required to maintain equilibrium.

The time of flight is given by $T = \frac{2u \sin \theta}{g}$T=g2usinθ​.
Substitute the given values: $T = \frac{2 \times 20 \sin 30^\circ}{10} = 4 \times 0.5 = 2.0\text{ s}$T=102×20sin30∘​=4×0.5=2.0 s.

First, calculate $\mathbf{a} \times \mathbf{b}$a×b using the determinant method:
$\mathbf{a} \times \mathbf{b} = \mathbf{i}( (1)(2) - (-1)(-1) ) - \mathbf{j}( (2)(2) - (-1)(1) ) + \mathbf{k}( (2)(-1) - (1)(1) )$a×b=i((1)(2)−(−1)(−1))−j((2)(2)−(−1)(1))+k((2)(−1)−(1)(1))
$= \mathbf{i}(2 - 1) - \mathbf{j}(4 + 1) + \mathbf{k}(-2 - 1) = \mathbf{i} - 5\mathbf{j} - 3\mathbf{k}$=i(2−1)−j(4+1)+k(−2−1)=i−5j−3k.
The magnitude $|\mathbf{a} \times \mathbf{b}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$∣a×b∣=12+(−5)2+(−3)2​=1+25+9​=35​.

Using the first equation of motion: $v = u + at$v=u+at.
Since it starts from rest, $u = 0$u=0. $a = 2\text{ m/s}^2$a=2 m/s2 and $t = 5\text{ s}$t=5 s.
$v = 0 + (2 \times 5) = 10\text{ m/s}$v=0+(2×5)=10 m/s.

Using the equation $v^2 = u^2 + 2gs$v2=u2+2gs where $u = 0$u=0 (since it is dropped).
$v^2 = 0 + 2(10)(45) = 900$v2=0+2(10)(45)=900.
$v = \sqrt{900} = 30\text{ m/s}$v=900​=30 m/s.

The weight of the rod $W = mg = 5 \times 10 = 50\text{ N}$W=mg=5×10=50 N acts at its midpoint, which is $2\text{ m}$2 m from $A$A.
The support $C$C is $1\text{ m}$1 m from $B$B, making it $4 - 1 = 3\text{ m}$4−1=3 m from $A$A.
Taking moments about $A$A to eliminate the reaction at $A$A: $R_C \times 3 = 50 \times 2 \implies 3R_C = 100$RC​×3=50×2⟹3RC​=100.
$R_C = \frac{100}{3} = 33.33\text{ N}$RC​=3100​=33.33 N.