How far below the aiming point does the messy food strike Chretien's image?

[mandi_ah]

HI!
I've looked everywhere for help for this question but i just can't seem to understand it... The question is below. It is from the SIN contest from 2001, question number 4. I wish they would have the answers posted up on the Waterloo website, but they only have the most recent contests :uhh:

Homework Statement

The Air farce "chicken cannon" uses compressed air of pressure 10 atmospheres (1 atm=10^5 N/m^2) to accelerate a 5 kg clump of messy food through a tube of diameter 10 cm which is 1.0 m long. The "cannon" is fired horizontally at a picture of Prime Minister Chretien that is 4.0 m away. How far in cm below the aiming point does the messy food strike Chretien's image? You may assume that the pressure in the barrel stays the same during firing.

a) 1. 64 cm
b) 12.8 cm
c) 9.37 cm
d) 5. 04 cm
e) 2. 50 cm

Homework Equations

Pressure= Force/Area
Force=mass x acceleration

The Attempt at a Solution

P = 1 x 10^6 N/m^2
A= lw
= (10 cm)(1 m)
= 1000 cm^2
= 10 m^2

F= P x A
= (1x10^6 N/m^2)(10m^2)
= 1 x 10^7 N

accleration= Force/mass
= (1 x 10^7 N)(5 kg)
= 2 x 10^7 m/s^2


i don't know.. am i on the right track?

-thanks :shy:

 

[tim_lou]

its easiest to use work-energy. but your method is definitely correct. However, acceleration = net force/ mass, you are neglecting gravity and the force exerted by the air above the firing thing.

 

[AngeloG]

It's a good idea again to draw a picture of what's happening with the forces acting at each point.

Before, During and After.

But you're on the right track to split this up into several things =).

Near the end it would be a good idea to use vectors given you know the time, position, velocity or other variables that might be useful [you decide and find this out for yourself] =).

 

[AlephZero]

Your "area" is wrong. You need the area where the pressure is applied to the food - the length of the barrel doesn't come into that.

Also be careful with the units. 1000 cm^2 is not 10 m^2.

Those 2 mistakes explain why you got such a high acceleration - 2 million times the acceleration due to gravity isn't likely to be the right answer!

 

[mandi_ah]

ya i just realized that my area was wrong, and it made a big difference

 

[yanisyanis11]

apparently the answer is e) 2.50cm
first of all we need to calculate the velocity at which the ball leaves the tube;
F = Pxd
F = 10^6x(0.0025)(3.14)
F = 7850N
a= (F/m)
a= (7850N/5kg)
a= 1570 m/s^2

(V2)^2= (v1)^2 +2ad d in this case in the length of the tube and v1 is zero
(V2)^2 =2(1570) 1m
V2 = 56.0m/s

now using this, we need to consider the situation horizontaly (where there is no forces acting on the object and vertically when there the force of gravity acting on the object.

therefore;let x represent the distance horizontally 4m

x= (V2)t t is the time taken for the food to hit the ground
4/(56.0m/s) = t
t= 0.07s

now using this, we need to consider the situation horizontally where;
y = (0.5)gt^2

y is How far in cm below the aiming point does the messy food strike Chretien's image

y= (0.5)g (0.07)^2
y= 0.02487m
y= 2.50cm

therefore the answer is e).