Problem Solving Strategy for Motion in Two Dimensions

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Motion in Two Dimensions is a very important topic in Physics. In this post, I am going to demonstrate my strategies for solving this kind of problems. First, we revise the formula we have learnt :

  • v = u + at
  • v^2-u^2=2as
  • s=ut+\frac{1}{2}at^2

These formulas are useful. However, it may not be the best choice to solve problems using only formulas. In fact, equations would be tedious if you do not know how to apply it well.

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Problem Solving Strategy for Motion in Two Dimensions

Solve Things Symbolically


You might noticed that most of the physics problem I’ve posted are solved symbolically. This is because solving problems symbolically can have many advantages, especially in physics. From my experience, they are :

  1. Faster. We spend less time picking up our calculator.
  2. More Accurate. Rounding up the answer throughout the calculations may lead to a major error at the end. By solving symbolically, we will only need to perform the round-up once when we are finished with the final substitution.
  3. Less Mistake. It’s very easy to overlook 6 as a 0, especially when you are nervous in your exam. By solving symbolically, we are not likely to make mistakes for substitutions and arithmetic operations.
  4. Stronger Understanding of Problem. Does acceleration matters in this problem? If the time increased, will it travel further? Since symbolic answers normally looks nice, all these questions can be answered from it! Other than that, we can determine other special cases in the problem.
  5. Reusable. When you calculated g as 9.8 instead of the problem requirement g=10, you don’t have to solve the problem all over again but substitute the correct value.

Although it’s better solving problems symbolically, it really depends on the type of problem. For example, you won’t want to solve a system of 3 equations with only symbols. Now, let’s look at a typical problem regarding heat transfer, and solve it symbolically.

Untitled drawing - Edited

Two metals A and B with length 50\text{ cm} and 70\text{ cm} is wielded together. If we constantly heat up the end of metal A and B to 473\text{ K} and 323\text{ K}, what is the temperature of the wielding section of the rod?

( k is the thermal conductivity of metal, with unit Wm^{-1}k^{-1} )

Solution :

The wielding section will reaches thermal equilibrium when the heat transferred from metal A and B is the same. Hence,

\begin{aligned} \displaystyle Q_A &= Q_B \\ \\ k_AA\frac{T_A-T}{l_A} &= k_BA\frac{T-T_B}{l_B} \\ \\ T_Ak_Al_B - Tk_Al_B &= Tk_Bl_A - T_Bk_Bl_A \\ \\ T(k_Al_B+k_Bl_A) &= T_Ak_Al_B+T_Bk_Bl_A \\ \\ T\bigg(\frac{k_A}{l_A}+\frac{k_B}{l_B}\bigg) &= T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B} \\ \\ T &= \frac{T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B}}{\frac{k_A}{l_A}+\frac{k_B}{l_B}}\end{aligned}

With that formula in hand, we only have to substitute the values into their responding variable :

\begin{aligned} \displaystyle T &= \frac{T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B}}{\frac{k_A}{l_A}+\frac{k_B}{l_B}} \\ &= \frac{473\times\frac{200}{50}+323\times\frac{100}{70}}{\frac{200}{50}+\frac{100}{70}} \\ &= 433.5\text{ K}\end{aligned}

Since this problem often appears as a sub-problem in the UEC exam, it’s worth to memorize it. Next, let’s solve a similar problem, but UEC-leveled.

Untitled drawing - Edited

What is the value of T_1 and T_2?

Consider the heat transfer between metal P and Q,

\begin{aligned} \displaystyle T_1 &= \frac{473\frac{200}{0.5}+T_2\frac{100}{0.5}}{\frac{200}{0.5}+\frac{100}{0.5}} \\ 600T_1 &= 189200 + 200T_2 \\ 3T_1 - T_2 &= 926 \end{aligned}

Consider the heat transfer between Q and R,

\begin{aligned} \displaystyle T_2 &= \frac{T_1\frac{100}{0.5}+323\frac{400}{0.5}}{\frac{100}{0.5}+\frac{400}{0.5}} \\ 1000T_2 &= 200T_1 + 258400 \\ 5T_2 - T_1 &= 1292\end{aligned}

Solve the two system of equation :

\displaystyle \left\{  \begin{array}{l l}  3T_1-T_2 &= 866\\  5T_2-T_1 &= 1292  \end{array} \right.

T_1 = 430.14\text{ K} \\ T_2 = 344.43\text{ K}

It is true that the only way to be good at problem solving is through solving problems. However, there’s no point doing the same thing over and over. Fully understand the problem is a much better approach.

Solve Things Symbolically

Pulley Challenge

A block of mass 2M and two blocks of mass M are connected to a pulley as shown. While the system is static, the lower block of mass M is flicked up with a velocity v . Assume that the string connecting two blocks of mass M is long enough such that the blocks will not collide. What is the time taken for the rope between the 2 equal blocks be taut again?

Screenshot 2015-01-27 at 20.42.46

From a Problem

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Pulley Challenge

Projectile Strikes a Wall

A particle P is thrown horizontally from a tower of height H. After some time, it hits a wall and falls on the ground with an angle \theta with the horizontal. Given that the distance between the tower and the wall is D and the distance between the particle and the wall is d, what is the height of the tower H?

Screenshot 2015-01-25 at 18.33.13

Problem adapted from

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Projectile Strikes a Wall

Holding a cone

With two fingers, you hold an ice cream cone motionless upside down. The mass of the cone is m, and the coefficient of static friction between your fingers and the cone is \mu. When viewed from the side, the angle of the tip is 2\theta.

  1. What is the minimum normal force you must apply with each finger in order to hold up the cone?
  2. In terms of \theta, what is the minimum value of \mu that allows you to hold up the cone?

Assume that you can supply as large a normal force as needed.

Screenshot 2015-01-18 at 13.37.02

Problem Source : Introduction to Classical Mechanics

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Holding a cone

Block under an overhang

A block of mass M is positioned underneath an overhang that makes an angle \beta with the horizontal. You apply a horizontal force of Mg on the block. Assume that the friction force between the block and the overhang is large enough to keep the block at rest. What are the normal and friction forces (call them N and F_f ) that the overhang exerts on the block? If the coefficient of static friction is \mu, for what range of angles \beta does the block in fact remain at rest?

Screenshot 2015-01-18 at 12.53.22

Problem Source : Introduction to Classical Mechanics

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Block under an overhang