Lecture 4 - Solving Kinematics problems
In this lecture we will discuss problem solving approaches for kinematics.
If you need a pdf version of these notes you can get it here
Video of lecture
For some reason I haven't yet figured out only the sound was recorded from my camera, not the video, so here you have only the screen capture and audio.
Starting Point
- Draw a diagram
- Identify variables and relevant equations
- Identify known and unknown quantities
- Come up with a strategy to get the unknown quantities from the known
- Often it is best to solve problems using algebra as far as possible
Free Fall - Problem 2.61
A stone falling past a window: A free fall problem.
Variables: $(y,t) $
Equations: Define $y$ ↓ and $g=9.81\,\mathrm{ms}^{-2}$ ↓
$v= v_{0}+gt$
$y= y_{0}+v_{0}t+\frac{1}{2}gt^2$
$v^{2}=v_{0}^2+2g(x-x_{0})$
$g=9.81\,\mathrm{ms}^{-2}$
Knowns: Over a certain time interval we know the distance traveled.
Information required: Initial displacement when $v=0$.
2.61 Approach 1
Break the problem in to two motions. Using the equation $y= y_{0}+v_{0}t+\frac{1}{2}gt^2$ we can find out what the velocity was the beginning of the motion for which we know the distance traveled and time taken.
$(y-y_{0})-\frac{1}{2}gt^2=v_{0}t$
$v_{0}=\frac{2.2\mathrm{m}-0.5\times9.8\mathrm{ms}^{-2}\times(0.33)^2\mathrm{s}^{2}}{0.33\mathrm{s}}=5.05\mathrm{ms}^{-1}$
Now we take this velocity and make it the final velocity of a motion that took place immediately before this one. In this motion we know $v$ and $a$ and want to find the distance that the object fell before the motion we just looked at took place.
$v^2=v_{0}^2+2g(y-y_{0})$
Remembering we made our previous $v_{0}$ in to our $v$ for this motion we can say
$y-y_{0}=\frac{v^2-v_{0}^2}{2g}$
And the height the object starts above the windows is the same as the distance it falls through
$h=\frac{(5.05\mathrm{ms}^{-1})^2}{2\times9.81\mathrm{ms}^{-1}}=1.3\mathrm{m}$
2.61 Approach 2
A different approach using calculus.
Starting from time $t=0$ the displacement at any time can be expressed as
$\int_{0}^{y}\,dy=\int_{0}^{t}gt\,dt$
Let us define the time at which the stone passes the top of the window as $t_{1}$ and the time it passes the bottom of the window as $t_{2}$.
$y_{1}=\frac{1}{2}gt_{1}^{2}$
$y_{2}-y_{1}=\frac{1}{2}gt_{2}^{2}-\frac{1}{2}gt_{1}^{2}=\frac{g}{2}(t_{2}^{2}-t_{1}^{2})=\frac{g}{2}(t_{2}+t_{1})(t_{2}-t_{1})$
$\frac{y_{2}-y_{1}}{t_{2}-t_{1}}\frac{2}{g}=\frac{2.2\mathrm{m}}{0.33\mathrm{s}}\frac{2}{9.81\mathrm{ms^{-2}}}=1.36\mathrm{s}={t_{2}+t_{1}}$
As $t_{2}=t_{1}+0.33\mathrm{s}$
$t_{1}=\frac{1.36\mathrm{s}-0.33\mathrm{s}}{2}=0.515\mathrm{s}$
$y_{1}=\frac{1}{2}\times{9.81\mathrm{ms^{-2}}\times(0.515\mathrm{s^2})}=1.3\mathrm{m}$
Relative Velocity
Dealing with relative velocity is a particularly important application of addition and subtraction of vectors.
We can adopt a notation which can be helpful.
$\vec{v}_{AB}$: velocity of A relative to B
If we want to know the velocity of A relative to C if we know the velocity of A relative to B and the velocity of B relative to C
$\vec{v}_{AC}=\vec{v}_{AB}+\vec{v}_{BC}$
Whereas if we want to find the velocity of B relative to C
$\vec{v}_{BC}=\vec{v}_{AC}-\vec{v}_{AB}$
Problem 3.70
Relative to the water, the boat has a known velocity $v_{BW}$ (including direction $\theta_{BW}$).
The direction of the boat's velocity relative to the land can be deduced as $\tan\theta_{BL}=\frac{120\mathrm{m}}{280\mathrm{m}}$.
Problem 3.70 solution
 | Use the sine rule. $\frac{v_{WL}}{sin(\theta_{BW}-\theta_{BL})}=\frac{v_{BW}}{sin(90^{o}+\theta_{BL})}=\frac{v_{BL}}{\sin(90^{o}-\theta_{BW})}$ |
 | or use components. $\frac{v_{BW}\sin\theta_{BW}-v_{WL}}{v_{BW}\cos\theta_{BW}}=\tan\theta_{BL}$. |
(Recall that $\tan\theta_{BL}=\frac{120\mathrm{m}}{280\mathrm{m}}$).
A canoe on a river
$\vec{v}_{CE}=\vec{v}_{CR}+\vec{v}_{RE}$
$\vec{v}_{CR}=\vec{v}_{CE}-\vec{v}_{RE}$
Problem 3.78
$\vec{v}_{RE}=\vec{v}_{RT}+\vec{v}_{TE}$
Projectile Motion
| $\Large a_{x}=0$ | $\Large a_{y}=-g$ |
| $\Large v_{x}=v_{x0} $ | $\Large v_{y}=v_{y0}-gt$ |
| $\Large x=x_0+v_{x0} t$ | $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$ |
| $\Large a_{x}=0$ | $\Large a_{y}=-g$ |
| $\Large v_{x}=v_{0}\cos\theta$ | $\Large v_{y}=v_{0}\sin\theta-gt$ |
| $\Large x=x_0+v_{0}\cos\theta t$ | $\Large y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |
Problem 3.88
Knowns: x,y at a given t
Useful equations:
| $v_{x}=v_{0}\cos\theta$ | $v_{y}=v_{0}\sin\theta-gt$ |
| $x=x_0+v_{0}\cos\theta t$ | $y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |
Set $x_{0}=0$ and $y_{0}=0$
| $x=v_{0}\cos\theta t$ | $y=v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |
Need an equation only in term of $v$ or $\theta$
$v_{0}=\frac{x}{\cos\theta t}$
$y=x\frac{\sin\theta t}{\cos\theta t}-\frac{1}{2}gt^{2}=x\tan\theta-\frac{1}{2}gt^{2}$
Solve for $\theta$ and then $v_{0}$
Alternatively, substitute in numbers first, which gives two equations containing $v$ and $\theta$ which can be solved simultaneously. (Much less elegant
Problem 3.56
Take a look a the approach used to find the motion path in Lecture 3. Instead of solving for y=0 solve for y=h. Note that the problem only wants you to find the solution for h>0. Also note that Mastering Physics requires that you type trig functions with brackets, e.g $\sin(\theta)$ not $\sin\theta$.
You want to solve the following equation for x
$h=x\frac{\sin\theta_{0}}{\cos\theta_{0}}-\frac{g}{(2v_{0}^{2}\cos^2\theta_{0})}x^2$
Rearranged in to standard quadratic form this is:
$x^2-\frac{2v_{0}^{2}}{g}\sin\theta_{0}\cos\theta_{0} x+h\frac{(2v_{0}^{2}\cos^2\theta_{0})}{g}=0$
I find the best way to move forward is to complete the square (though you can use the quadratic formula also)
$x^2-\frac{2v_{0}^{2}}{g}\sin\theta_{0}\cos\theta_{0} x+\frac{v_{0}^4}{g^2}\sin^2\theta_{0}\cos^2\theta_{0}-\frac{v_{0}^4}{g^2}\sin^2\theta_{0}\cos^2\theta_{0} +h\frac{(2v_{0}^{2}\cos^2\theta_{0})}{g}=0$
$x^2-\frac{2v_{0}^{2}}{g}\sin\theta_{0}\cos\theta_{0} x+\frac{v_{0}^4}{g^2}\sin^2\theta_{0}\cos^2\theta_{0}=\frac{v_{0}^4}{g^2}\sin^2\theta_{0}\cos^2\theta_{0} -h\frac{(2v_{0}^{2}\cos^2\theta_{0})}{g}$
$(x-\frac{v_{0}^2}{g}\sin\theta_{0}\cos\theta_{0})^2=\frac{v_{0}^2\cos^2\theta_{0}}{g^2}(v_{0}^2\sin^2\theta_{0}-2gh)$
$x=\frac{v_{0}^2}{g}\sin\theta_{0}\cos\theta_{0}\pm\frac{v_{0}\cos\theta_{0}}{g}\sqrt{v_{0}^2\sin^2\theta_{0}-2gh}$
Only one of the solutions for x is positive if $h>0$ and this is the one you're looking for.
$x=\frac{v_{0}\cos\theta_{0}}{g}(v_{0}\sin{\theta_{0}}+\sqrt{v_{0}^2\sin^2\theta_{0}-2gh})$
Monkey and Hunter
At the moment the hunter pulls the trigger the monkey can make a decision, should he hang on or drop from the tree in the hope the bullet will miss him?
Monkey and Hunter solution
In case my monkey shooting fails Physclips has a video you can step through frame by frame and a pretty detailed explanation here
Height of bullet with time:
$y=x\tan\theta-\frac{1}{2}gt^{2}$
Hunter's expectation:
$y=x\tan\theta$
At monkey's $x$ position, $x_{m}$ the bullet will be $-\frac{1}{2}gt^{2}$ below where monkey started…exactly the same distance the monkey will be below his starting point if he lets go.
This is a simple demonstration of the fact that gravity accelerate all object's equally independently of their horizontal velocity.
