# Problem Solving Strategy for Motion in Two Dimensions

Image Credit : wired.com

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.

Before introducing my strategy, I’ll share another useful but not commonly used formula

$\displaystyle y = \frac{\sec^2{\theta}}{2v^2}g\text{ }x^2 + x\tan \theta$

This formula is used to calculate the displacement of the particle at the $y$-axis for given displacement at the $x$-axis. The derivation of the formula is actually pretty simple.

The partial displacement at the $y$-axis and $x$-axis is respectively

\begin{aligned} \displaystyle y &= u_y^2t+\frac{1}{2}gt^2 \\ x &= u_xt \end{aligned}

Substitute $t$ to the first equation

\begin{aligned} \displaystyle y &= \frac{u_y}{u_x}x+\frac{1}{2}g\bigg ( \frac{x}{v \cos \theta} \bigg )^2 \\ &= x\tan\theta+\frac{1}{2v^2\cos^2\theta}g\text{ }x^2 \\ &=\frac{\sec^2\theta}{2v^2}g\text{ }x^2+x\tan\theta\end{aligned}

Example Problem 0.1 :

A canon ball is being launched at speed $10\text{ ms}^{-1}$ and $60^\circ$ in respect to the horizontal. If the target is at height $1\text{ m}$, what is the horizontal distance should the canon be placed?

$g = -10\text{ ms}^{-2}$

Solution 0.1 :

Apply the formula above, we have

\begin{aligned} \displaystyle y &= \frac{\sec^2\theta}{2v^2}g\text{ }x^2+x\tan\theta \\ 1 &= \frac{\sec^2 60^\circ}{2(10)^2}g\text{ }x^2+x\tan 60^\circ \\ 1 &= \frac{1}{5}x^2+\sqrt{3}x \end{aligned}

Solve the quadratic equation we get $x = 0.62\text{ m}$ or $x = 8.04\text{ m}$.

Strategy 1 : Graphic ( 图解法 )

Graph is a useful tool that is commonly ignored by most students. Solving problems using graphics can help you visualize the problem. The best graph, in my opinion, is the $v-t$ graph, because it can show us two things:

• Displacement – area under the function
• Acceleration – slope of the function

Example Problem 1.1 :

Consider the free fall of a particle at position $A$ with height $H$. Given that $AB = BC$ and the time for the particle to travel from $B$ to $C$ is $2\text{ s}$.

1. What is the time traveled from $A$ to $B$, in seconds?
2. What is the height $H$, in meters?

Solution 1.1

Practice Problem 1.2 :

A particle at rest starts moving with acceleration $a_1$. Until it reaches the speed $v$, it immediately decelerates with $a_2$ until it stops. The total time consumed is $t$. Prove that:

1. The displacement of the particle is $\frac{1}{2}vt$.

1. 质点的位移为 $\frac{1}{2}vt$.

Strategy 2 : Relative Motion ( 相对运动 )

This strategy is often used to solve problems that has more than two moving objects involved. The formula of relative motion is :

$v_a' = v_a - v_b$

Once you got the idea, the formula is not necessary to be memorized.

Example Problem 2.1 :

A boat can travel at a speed of $3\text{ ms}^{-1}$ on still water. A boatman wants to cross a river whilst covering the shortest possible distance. In what direction should he row with respect to the bank if the speed of the water is

1. $2\text{ ms}^{-1}$
2. $4\text{ ms}^{-1}$

Assume that the speed of the water is the same everywhere.

Solution 2.1

Practice Problem 2.2 :

There are two particles at points $A(0,0)$ and $B(40,0)$ respectively. Particle $A$ moves upwards at speed $3\text{ ms}^{-1}$ while particle $B$ moves towards left at speed $4\text{ ms}^{-1}$. What is the minimum distance can they get between them?

Practice Problem 2.2 :

A sandstorm $100\text{ m}$ wide is heading towards Adam at speed $2\text{ ms}^{-1}$ from distance $1000\text{ m}$ away. Since Adam is infamous for his laziness, he wants to cross the sandstorm running as slow as possible. What is the minimum speed should Adam run without being hit by the sandstorm?

Practice Problem 2.3 :

Three small snails are each at a vertex of an equilateral triangle of side $60\text{ cm}$. The first sets out towards the second, the second towards the third and the third towards the first, with a uniform speed of $5\text{ cm min}^{-1}$ . During their motion each of them always heads towards its respective target snail. How much time has elapsed, and what distance do the snails cover, before they meet?

Practice Problem 2.4 :

A small ball falling vertically with constant velocity $v$ elastically strikes a massive inclined cart moving with a horizontal velocity $v$. What is the rebound velocity of the ball? ( The velocity of the cart doesn’t change after the collision )

Practice Problem 2.5 :

Two ships $A$ and $B$ are originally at a distance $d$ from each other depart at the same time from a straight coastline. Ship $A$ moves along a straight line perpendicular to the shore while ship $B$ constantly heads for ship $A$, having at the each moment same as the latter. After a sufficiently great interval of time the second ship will be following the first one at a certain distance $x$. Find $x$.

Strategy 3 : Energy Conservation ( 功能原理 ) :

The Energy Conservation is useful especially when dealing with projectile motion. Energy can be changed from one form to another without losing its total internal energy. In this case, the internal energy of a particle is consists of its kinetic energy and potential energy.

\begin{aligned} \displaystyle E &= E_k + E_p \\ \\ &= \frac{1}{2}mv^2 + mgh \end{aligned}

Practice Problem 3.1

A ball with mass $1\text{ kg}$ is thrown upwards with speed $30\text{ m/s}$. It is given that when the ball reaches its highest point, the work done by surrounding air on the ball is $100\text{ J}$. What is the highest point can the ball reaches?

Practice Problem 3.2 :

Christopher accidentally fell inside a cylindrical well of depth 15 meters and radius 8 meters. Jumping from the center of the bottom of the well, what is the minimum launch speed that Christopher needs to achieve, in order to jump out of the well?

A ball slides along a friction-less ramp of length $L$. As it reaches the bottom, it hits the board and slides up along the slope again. If the speed of the ball after the collision is $\frac{4}{5}$ times the speed of the ball just before the collision, find the total distance traveled by the ball when it comes to a complete stop.