Solving Question: Speed of Ball with Rigid Cannon

  • Thread starter Thread starter srhly
  • Start date Start date
  • Tags Tags
    Stuck
Click For Summary
A cannon with a mass of 1960 kg fires a 24.5 kg ball at a speed of 120 m/s. When the cannon is free to roll, its recoil speed is calculated to be approximately 1.4907 m/s. If the cannon is mounted rigidly, all kinetic energy is transferred to the cannonball, resulting in a new speed of about 120.75 m/s. The discussion emphasizes the conservation of energy and momentum in determining the ball's speed under different conditions. The calculations suggest a slight adjustment in the recoil speed may be necessary for accurate results.
srhly
Messages
17
Reaction score
0
I had a two-part question but I've figured out the first part. A cannon with mass of 1960 kg fires a 24.5 kg ball horizontally. The cannonball has a speed of 120 m/s after it has left the barrel. For the first part I found that if the cannon was free to roll horizontally, its speed immediately after it was fired was 1.4907 m/s. The part I'm stuck on says imagine the same charge is used so total energy of cannon plus cannonball system is the same. Disregarding friction how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s.
 
Physics news on Phys.org
What if you thought of momentum instead of energy? It is also conserved.
 
srhly said:
I had a two-part question but I've figured out the first part. A cannon with mass of 1960 kg fires a 24.5 kg ball horizontally. The cannonball has a speed of 120 m/s after it has left the barrel. For the first part I found that if the cannon was free to roll horizontally, its speed immediately after it was fired was 1.4907 m/s. The part I'm stuck on says imagine the same charge is used so total energy of cannon plus cannonball system is the same. Disregarding friction how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s.

It usually is adviceable to write down the exact question verbatim. Sometime, what you interpret as being asked is different than what it is actually asking. If I have understood your question here correctly, this is what it is asking for:

1. Since you already found the first part, you know the KE of the cannonball and the cannon separately.

2. If the cannon is fixed, then ALL the KE is now transferred to the cannonball. So the cannonball now will have the KE of what it had before PLUS the KE from the non-moving cannon. So this is its new KE for the cannonball. From here, find the new speed.

Zz.
 
srhly said:
I had a two-part question but I've figured out the first part. A cannon with mass of 1960 kg fires a 24.5 kg ball horizontally. The cannonball has a speed of 120 m/s after it has left the barrel. For the first part I found that if the cannon was free to roll horizontally, its speed immediately after it was fired was 1.4907 m/s. The part I'm stuck on says imagine the same charge is used so total energy of cannon plus cannonball system is the same. Disregarding friction how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s.
For item #1, did you calculate the cannon's recoil speed correctly?? The recoil speed should be (1.5 m/sec). Although the difference is small, it may indicate a problem somewhere. (Perhaps in copying the problem??) When using (1.5 m/sec) for item #1, the item #2 cannonball speed (with the cannon fixed) should be approx (120.75 m/sec).


~~
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
1K
Replies
22
Views
4K
  • · Replies 7 ·
Replies
7
Views
13K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K