Trajectory of an exploding shell

In summary, the conversation discusses the problem of finding the horizontal speed of a fragment from a shell fired at a 59 degree angle. By using conservation of linear momentum at the point of explosion and taking into account the masses of the two fragments, the horizontal speed can be determined. The hint suggests considering the direction and speed of the velocity at the top of the trajectory.
  • #1
Bizznaatch
2
0
A 36-kg shell is fired from a gun with a muzzle velocity 185 m/s at 59 degrees above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment?

Hint: Use conservation of linear momentum at the point of explosion.

My gut approach to this problem was that I could use the angle to find the vertical and horizontal components of velocity. If we assume no friction, the horizontal component of acceleration should remain constant. It is apparently the wrong answer. I don't really understand how to use conservation of linear momentum here? Any help is much appreciated! Thanks!
 
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  • #2
Perhaps if you posted your working we could help you along. Further, try to use the hint regarding conservation of momentum rather than velocities. An important point here is that the shell splits into to pieces and therefore you have to take into account the masses of the pieces, not just the velocities.
 
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  • #3
Welcome to PF!

Bizznaatch said:
My gut approach to this problem was that I could use the angle to find the vertical and horizontal components of velocity. If we assume no friction, the horizontal component of acceleration should remain constant. It is apparently the wrong answer. I don't really understand how to use conservation of linear momentum here? Any help is much appreciated! Thanks!

Hi Bizznaatch! Welcome to PF! :smile:

(btw, you mean the horizontal component of velocity should remain constant - the horizontal acceleration, of course, is zero)

Hint: What is the direction of the velocity at the top of the trajectory?

So what is its speed there? :smile:
 

FAQ: Trajectory of an exploding shell

1. What factors affect the trajectory of an exploding shell?

The trajectory of an exploding shell is affected by several factors, including the initial velocity of the shell, the angle at which it is fired, air resistance, and the force of gravity. Wind speed and direction can also play a role in altering the trajectory.

2. How does air resistance impact the trajectory of an exploding shell?

Air resistance, also referred to as drag, can significantly alter the trajectory of an exploding shell. As the shell moves through the air, it experiences resistance which slows it down and changes its path. This can cause the shell to fall short of its intended target.

3. Can the trajectory of an exploding shell be predicted accurately?

While there are mathematical models and calculations that can be used to predict the trajectory of an exploding shell, it is not always possible to accurately predict its exact path. Factors such as wind and air resistance can cause unexpected changes in trajectory.

4. How does the angle of firing affect the trajectory of an exploding shell?

The angle at which an exploding shell is fired plays a significant role in its trajectory. If the shell is fired at a higher angle, it will travel further but with a steeper drop. A lower angle will result in a shorter distance but a flatter trajectory.

5. What is the maximum height achieved by an exploding shell?

The maximum height that an exploding shell can reach is dependent on its initial velocity and angle of firing. In most cases, the maximum height will be reached when the shell is fired at a 45-degree angle. However, air resistance and other factors can impact the maximum height achieved by the shell.

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