Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Latest interest

Having a good foundation of mathematics can help in other subjects that apply mathematical knowledge in practical situations. I'm referring to Physics, my latest interest!

For students of Physics, learn your mathematics well because "Mathematics is the queen of the sciences." in the words of Carl Friedrich Gauss.

If you are struggling with Physics currently, here is a helpful link that houses topical summaries: S-cool.

Keen learners may be interested to download an e-book (with explanations and worked examples) that I managed to scour from the Internet.

Some more e-books (in 3 parts) here: grade 10, grade 11, grade 12

Be sure to visit www.scruuphysics.com to access loads of learning resources compiled by experienced Physics educators.

Over time, I will write some Physics-related articles (in the form of solved problems) to address some learning issues. Do look out for them :)

Problem 1

Solution:

Consider the upward direction as positive.

Time taken to strike the ground = time taken to travel up to the maximum height + time taken to travel down from the maximum height.

First, we find the time taken to travel up to the maximum height.

Initially, v_i=v. At the maximum height, v_f=0.

v_f = v_i - gt, so 0 = v - gt \Rightarrow t = \frac{v}{g}.

We then use this time to find the displacement of the ball, s = \frac{1}{2}(v_i + v_f)t
\Rightarrow s = \frac{1}{2}(v + 0)\frac{v}{g} = \frac{v^2}{2g}

Next, we find the time taken to travel down from the maximum height, noting that the total displacement is -\left(\frac{v^2}{2g} + h \right).

Initially, v_i = 0. Now -\left(\frac{v^2}{2g} + h \right) = v_it + \frac{1}{2}(-g)t^2
\Rightarrow \frac{v^2}{2g} + h = \frac{1}{2}gt^2
\Rightarrow t = \sqrt[]{\frac{v^2}{g^2} + \frac{2h}{g}}

\Rightarrow t = \frac{v}{g}\:\sqrt[]{1 + \frac{2gh}{v^2}}

Adding both times, we reach our desired result.

Key formulae that we applied in this problem:

1. v = u + at

2. s = \frac{1}{2}(u + v)t

3. s = ut + \frac{1}{2}at^2

Problem 2

Solution:

Resolve the forces along the incline and consider the upward movement along the incline as positive.

(a) Since acceleration is zero, Fcos 30^{\circ} = 80 + mgsin 30^{\circ}
\Rightarrow F = \frac{80 + (20)(9.81)(sin 30^{\circ})}{cos 30^{\circ}} = 0.21\text{kN}

(b) Since acceleration is 0.75\:\text{m s}^{-2}, Fcos 30^{\circ} - (80 + mgsin 30^{\circ}) = ma
\Rightarrow F = \frac{(20)(0.75) + 80 + (20)(9.81)(sin 30^{\circ})}{cos 30^{\circ}} = 0.22\text{kN}

Problem 3

Solution:

(KE + PE) before impact = (KE + PE) after impact

\frac{1}{2}(0.38)v^2 + 0 = 0 + (0.38)(9.81)(0.24)

\Rightarrow v = 2.2\:\text{m/s}

If we have taken the mass of the bullet into consideration, LHS expression would have been (0.38 + 5.2 \times 10^{-3})(9.81)(0.24).



Solution:

By the conservation of momentum, (5.2 \times 10^{-3})V = 2.2 \times 0.38

\Rightarrow V = 160\:\text{m/s} (deduced)



Solution:

Loss of KE = \frac{1}{2}(5.2 \times 10^{-3})(160)^2 - (0.38)(9.81)(0.24)
                 = 66\:\text{J}

Key formulae that we applied in this problem:

1. Momentum = mv

2. KE = \frac{1}{2}mv^2

3. PE = mgh