Mechanics (Dynamics) | Review Questions 26 Solution Manual | Aki-Ola Elective Mathematics

Time to tackle some Review questions from the Chapter 26, Mechanics, in the Aki-Ola Further/Elective Mathematics book.

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Step-by-Step solution to all Review Exercises under Mechanics (Dynamics)
Simple and clear explanation of each step
Corresponding Youtube video to bolster understanding

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The Aki-Ola Series Further/Elective Mathematics was specifically written to cater for the needs of students preparing to sit for the West Africa Senior School Certificate Examination (WASSCE). The book is popular among WASSCE students, particularly Ghanaian students reading Advanced Mathematics at the Senior High School. Despite the author’s stepwise explanations of concepts, some students and teachers may find the Review Exercises challenging.

It is against this backdrop that I take the pain to present a step-by-step solution to the Review Exercises. This article provides solution to the Review Exercise 26 (Mechanics, Dynamics).

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Table of Contents

Review Exercise 26 ( Dynamics)

Question 1

A particle is projected vertically upwards with a speed of $25 ms^{-1}$ from a point on the ground. Take g=10 $ms^{-2}$ ].

Find

the position of the particle after $4s$ ;
the maximum height reached;
the time taken to reach the maximum height;
the time (s) when the particle is $30m$ above the ground.

Solution

The position of a particle is known if we know the distance it has traveled from its initial position. Hence, we calculate distance(s). Because throwing a particle up is against direction of gravity, we use $-g$ .

Variables known: initial velocity $(u)=25 ms^{-1}$ , $g=10 ms^{-2}, t=4s$

Variable to find: distance $(s)$

Suitable Equation: $s=ut-\frac{1}{2}gt^2 \par$

We begin solution:

$s=ut-\frac{1}{2}gt^2$

$s=(25)(4)-\frac{1}{2}(10)(4)^2$

$s=100-5 \times 16$

$s=100-80$

$s=20m$

At maximum height, $v=0$ , $u=25ms^{-1}$ .

NB: Height is a vertical distance.

$\Rightarrow$ We are finding the distance covered. That is, $s$ .

Variables known: $u=25 ms^{-1}, v=0, g=10 ms^{-2}$

Variable to find: distance $(s)$

Suitable Equation: $v^2=u^2-2gs$

We begin solution:

$v^2=u^2-2gs$

We make $s$ the subject

$s=\frac{u^2-v^2}{2g}$

$s=\frac{(0)^2-(25)^2}{2(10)}$

$s=\frac{625}{20}$

$s=31.25m$

We calculate the time taken to reach the maximum height.

Variables known: $u=25 ms^{-1},$ $v=0,$ $g=10 ms^{-2}.$

Variable to find: time $(t)$

Suitable Equation: $v=u-gt$

We begin solution:

$v=u-gt$

$t=\frac{u-v}{g}$

$t=\frac{25-0}{10}$

$t=2.5$

Thus, $t=2.5s$ .

Alternative Approach 1

Variables known: $u=25 ms^{-1},$ $v=0,$ $s=31.25m$

Variable to find: time $(t)$

Suitable Equation: $s=\frac{1}{2}(u+v)t$

We begin solution:

$31.25=\frac{1}{2}(25+0)t$

$31.25=12.5t$

$t=\frac{31.25}{12.5}$

$t=2.5$

Thus, $t=2.5s$ .

Alternative Approach 2

Variables known: $u=25 ms^{-1},$ $s=31.25m,$ $g=10 ms^{-2}$

Variable to find: time $(t)$

Suitable Equation: $s=ut-\frac{1}{2}gt^2$

We begin solution:

$s=ut-\frac{1}{2}gt^2$

$31.25=25t-\frac{1}{2}(10)t^2$

$31.25=25t-5t^2$

Re-writing in standard quadratic equation form,

$5t^2-25t+31.25=0$

Using the quadratic formula to solve for $t$ .

$t=\frac{-b \pm \sqrt{(-25)^2-4(5)(31.25)}}{2a}$

$t=\frac{5}{2}=2.5$ (repeated roots)

Thus, $t=2.5s$ .

Question 2

A trolley $P$ of mass 10 $kg$ moving with a velocity of 23 $m/s$ in the direction $040^\circ$ collides with another trolley $Q$ of mass $1.5 \times 10^3 kg$ moving with a velocity of 9 $m/s$ in the direction $330 ^\circ$ . Immediately after the collision, the trolley $P$ begins to move with a velocity of 8 $m/s$ in the direction $300^\circ$ . Assuming conservation of linear momentum, calculate the magnitude and direction (correct to the nearest degree) of the velocity of $Q$ immediately after collision.