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Bullet Projectile Motion Experiment

This is a classic physics experiment which counter to our intuition.  We have  a situation where 1 ball is dropped from a point, and another ball is thrown horizontally from that same point.  The question is which ball will hit the ground first?

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(diagram from School for Champions site)

Looking at the diagram above you might argue that the ball that is dropped falls to the floor quicker as it has a shorter path.  Or, you might think that the ball thrown sideways would travel faster to the ground because of its initial horizontal velocity.  Both of these views are wrong however – as both balls will land at exactly the same time.  To understand why, let’s look at the 2 situations in turn.

The ball launched sideways

To show that both balls would hit the ground at the same time we need to split the motion into its x and y components.  We have

x = v t \cos \theta 
y = vt \sin \theta - \frac{1}{2} g t^2

Where the angle theta is the angle of launch, v is the initial velocity, g is the gravitational constant 9.8 m/s.  If we have a launch from the horizontal direction, then this angle is 0, which gives the simplified equations:

x = vt

y = 0.5gt2

if we relabel y as the vertical distance (d), then we have:

\ t =\ \sqrt {\frac{2d}{g}}

which is the time taken (ignoring air resistance etc) for an object launched horizontally to fall a distance d, where g is the gravitational constant 9.8 m/s.

So if we have a ball launched at a speed of 1 m/s from a height of 1m, it would hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

So we can use this value of t to see how far in the x direction it has travelled:

x = vt

x = 1(0.45)

x = 0.45m.

The ball dropped vertically

We still start with:

x = v t \cos \theta 
y = vt \sin \theta - \frac{1}{2} g t^2

But this time we have no initial velocity as so we simply get:

x = 0

y = 0.5gt2

or as before, if we relabel y as the vertical distance (d), then we have:

\ t =\ \sqrt {\frac{2d}{g}}

So with a ball dropped from a height of 1m, it would also hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

But this time the distance in the x direction will of course be 0.

Showing this graphically 

We can also show this graphically using the tracker software.  This allows you to track the motion of objects in videos.  So using the video above we can set the axis, and the height of the table

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and then the motion capture software actually plots the parabola of the ball’s motion.

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This first graph shows the change in the y direction with respect to time for the ball launched horizontally.  We have large steps because the video was in super slow motion, so there were frames of very little movement.  Nevertheless we can clearly see the general parabola, with equation:

y = -0.43x2 -1.2x + 107

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The second graph shows the change in y direction with respect to time for the ball dropped vertically down.  As before we have a clear parabola, with equation:

y = -0.43x2 -1.2x + 106

Which is a remarkably close fit.  So, there we go, we have shown that the vertical motion of our 2 objects are independent of their horizontal motion.

 

 

 

 

This is a classic puzzle which is discussed in some more detail by the excellent Wired article.  The puzzle is best represented by the picture below.  We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch.  However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired.  The question is therefore, at what angle should the hunter aim so that he still hits the monkey?

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(picture from the Wired article – originally from a UCLA physics textbook)

The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls.  You can see a video of this experiment at the top of the page.

You can use tracking software (such as the free software tracker ) to show this working graphically.  Tracker provides a video demo with the falling monkey experiment:

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As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.

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We can then generate a graph to represent this data.  The red line is the height of the bullet with respect to time.  The faint blue line (with yellow dots) is the height of the monkey with respect to time.  We can see clearly that the red line can be modeled as a quadratic.  The blue line should in theory also be a quadratic (see below):

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but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink.  Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.

We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t2+39t +0.1

So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t2+39t +0.1

which will be around 0.45 seconds after the gun is fired.

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(A more humane version, also from Wired – where we can throw the monkey a banana)

Newtonian Mathematics

The next question is can we prove this using some algebra?  Of course!  The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa).  So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls.  So they still collide.  Mathematically we have:

The vertical height of the bullet given by:

y = V0t – 0.5gt2

Where V0 is the initial vertical speed, t is the time, g is the gravitational force (9.8)

The vertical height of the monkey is given by:

y = h – 0.5gt2

where h is the initial vertical height of the monkey.

Therefore these will intersect when:

V0t – 0.5gt2 = h – 0.5gt2
V0t = h
V0/h = t

And for any given non-zero value of V0 we will have a t value – which represents the time of collision.

Well done – you have successfully shot the monkey!

If you like this you might also like:

Throwing cannonballs off the Leaning tower of Pisa – why weight doesn’t affect falling velocity

War Maths – how cannon operators used projectile motion to win wars

 

Galileo:  Throwing cannonballs off The Leaning Tower of Pisa 

This post is inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

Galileo Galilei was an Italian mathematician and astronomer who (reputedly) conducted experiments from the top of the Tower of Pisa.  He dropped various objects from in order to measure how long it took for them to reach the bottom, coming to the remarkable conclusion that the objects’ weight did not affect the speed at which it fell.  But does that really mean that a feather and a cannonball would fall at the same speed?  Well, yes – as long as they were dropped in a vacuum.  Let’s have a look at how we can prove that.

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Newton’s Laws:

For an object falling through the air we have:

psgV – pagV – FD = psVa

ps = The density of the falling object
pa = The density of the air it’s falling in
FD = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

To understand where this equation comes from we note that Newton second law (Force = mass x acceleration) is

F = ma

The LHS of our equation (psgV – pagV – FD) represents the forces acting on the object and the RHS (psVa) represents mass x acceleration.

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. pa will be many magnitudes smaller than than ps – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

psgV – FD ≈ psVa

Next, we can note that in a vacuum FD (the drag force) will be 0 – as there is no air resistance.  Therefore this can also be ignored to get:

psgV ≈ psVa

g ≈ a

 But we have a = dU/dt where U = velocity, therefore,

g ≈ a
g ≈ dU/dt
g dt ≈ dU

and integrating both sides will give:

gt ≈ U

 Therefore the velocity (U) of the falling object in a vacuum is only dependent on time and the gravitational force.  In other words it is independent of the object’s mass.  Amazing!

This might be difficult to believe – as it is quite unintuitive.  So if you’re not convinced you can watch the video below in which Brian Cox tests this out in the world’s largest vacuum chamber.

If you liked this post you might also like:

War maths – how modeling projectiles plays an essential part in waging wars.

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