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Cannon Momentum problem

  • Thread starter Joe91090
  • Start date
  • #1
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Homework Statement



So I know that a cannon weighs 2220 kg and a cannon ball weighs 19.5 kg and that when you fire the cannon at 102 m/s the speed of the cannon is .895m/s in the opposite direction. part B of the problem asks how much faster would the ball travel is the cannon was mounted rigidly disregarding friction.

Homework Equations



mv1 = mv2

The Attempt at a Solution



I did 2220(x)= (19.5)(102) to find the speed of the cannon in part A so for part B i did
2220(0) = (19.5)(x) but im not sure if 19.5 m/s is correct.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
For Part B) all you have with your equation is 0.

Maybe consider the relative velocity of the cannonball to the barrel?
 
  • #3
54
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wouldnt the cannon barrel be 0 also since its a part of the rigidly mounted cannon ?
 
  • #4
209
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wouldnt the cannon barrel be 0 also since its a part of the rigidly mounted cannon ?
You write:
2220(0) = (19.5)(x)
So x=0 (x=v2? why the x?)

but im not sure if 19.5 m/s is correct.
How did you get 19.5m/s in the first place?

And, are you sure this is a (conservation of) momentum problem since it involves 'external' forces?
 
  • #5
54
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yes i wrote x = velocity and 19.5 kg and 2220 kg are the weights given in the problem
 
  • #6
209
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I edited my post so you might wanna look at it again.

Where did you write x=velocity ?
 
  • #7
54
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i could rewrite it (2220)(0)=(19.5)(v2)

Im not 100% sure this is the right equation but part B it says the total energy stays the same, to disregard friction and all other parameter retained the same as in part A
 
  • #8
209
1
i could rewrite it (2220)(0)=(19.5)(v2)

Im not 100% sure this is the right equation but part B it says the total energy stays the same, to disregard friction and all other parameter retained the same as in part A
Aha, you didn't mention that in your first post.
So then the kinetic energy of the cannon and cannon ball together in the first situation, equal the kinetic energy of the ball in the second situation.

And I'm pretty sure that your equation isn't correct, because v2 would be zero (v2=0, '2' being the cannon ball). That doesn't seem to make sense... I would get myself another cannon :)
 
  • #9
54
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so would i use 1/2mv^2=1/2mv^2 to solve ?

never mind because that would still make v2= 0 for the cannonball
 
  • #10
209
1
so would i use 1/2mv^2=1/2mv^2 to solve ?
If it's given in the original problem that the energy released in the first situation, is equal to the energy released in the second equation, yes.

But let's be a bit more specific here...

.5mvA1² + .5mvB1² = .5mvA2² + .5mvB2²

A is the cannon, B the cannon ball, 1 is the first situation, and 2 the second.
 
  • #11
54
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so i got 889.14 and 101439 for the kinetic energy of the cannon and cannon ball for part A and i add these together to get 102328 for the kinetic energy of the cannon ball in part B.

so i did 102328=1/2(19.5)v^2

i got 102.44 for v is this correct ????
 
  • #12
54
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does Va2= 0 ??


if it does then i still get 102.44
 
  • #13
209
1
so i got 889.14 and 101439 for the kinetic energy of the cannon and cannon ball for part A and i add these together to get 102328 for the kinetic energy of the cannon ball in part B.

so i did 102328=1/2(19.5)v^2

i got 102.44 for v is this correct ????
I get the same answer using this method. I'm not too sure whether this method is correct since you haven't been very clear in your problem statement...

does Va2= 0 ??
Yes
 
  • #14
54
0
i submitted .44 as an answer and it says im incorrect

Here is the original problem :


A cannon with a mass of 2220kg fires a 19.5 kg ball horizontally. the cannonball has a speed of 102m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to rool horizontally. What is the speed of the cannon immediately after it is fired ? I submitted .895 m/s and it said i was correct. Part B) The same explosive charge is used so the total energy of the cannnon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all the other parameters maintained the same ? answer in m/s.
 
  • #15
209
1
i submitted .44 as an answer and it says im incorrect

Here is the original problem :


A cannon with a mass of 2220kg fires a 19.5 kg ball horizontally. the cannonball has a speed of 102m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to rool horizontally. What is the speed of the cannon immediately after it is fired ? I submitted .895 m/s and it said i was correct. Part B) The same explosive charge is used so the total energy of the cannnon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all the other parameters maintained the same ? answer in m/s.
Hmm that's strange. Maybe someone else can help you out.
 
  • #16
3,003
2
The problem here is that your statement is not clearly worded.

So I know ... that when you fire the cannon at 102 m/s the speed of the cannon is .895m/s in the opposite direction.
How do you know this? Was this given? Was this part A of the problem? Because you mention part B but never mention what part A was.

Also, I am assuming you meant to say that "fire the cannonball at 102 m/s.....the speed of the cannon is .895m/s in the opposite direction..." else there is a contradiction here :smile:
 
  • #17
56
0
1. calculate the total amount of energy for situation 1

2. all energy will become kinetic energy for the canonball in situation 2

3. ballspeed in situation 2 minus ballspeed in situation 1 = your answer
 
  • #18
209
1
1. calculate the total amount of energy for situation 1

2. all energy will become kinetic energy for the canonball in situation 2

3. ballspeed in situation 2 minus ballspeed in situation 1 = your answer
That's what we've done =)
 

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