Respuesta :
When approaching a momentum, you have to identify your initial condition and your final conditions. Next is to identify your target variable, in this case, it is the recoil velocity of the cannon.
Let's say our initial condition is when the cannon has the shell and has not fired yet. Well, since there is no velocity, there is no momentum:
(1250kg+8.1kg)(0m/s) = 0kgm/s
Next lets find the final condition, this is when the shell is just fired. So the shell is moving at a velocity and the cannon is moving at a velocity.
(1250kg)v + (8.1kg)(220m/s)
Using conservation of momentum, our initial condition must equal our final condition, thus we get:
0kgm/s = (1250kg)v +(8.1kg)(220m/s)
Solve for v, we get:
0 = 1250v + 1782
-1782 = 1250v
v = -1.4256 m/s
This makes sense because if the cannon is shooting the shell, we would expect the cannon to move back. And since the cannon is a lot heavier, the velocity would be smaller in magnitude, relative to the velocity of the shell.
So the answer is:
v = -1.4256 m/s
Hope this helps! :D
Let's say our initial condition is when the cannon has the shell and has not fired yet. Well, since there is no velocity, there is no momentum:
(1250kg+8.1kg)(0m/s) = 0kgm/s
Next lets find the final condition, this is when the shell is just fired. So the shell is moving at a velocity and the cannon is moving at a velocity.
(1250kg)v + (8.1kg)(220m/s)
Using conservation of momentum, our initial condition must equal our final condition, thus we get:
0kgm/s = (1250kg)v +(8.1kg)(220m/s)
Solve for v, we get:
0 = 1250v + 1782
-1782 = 1250v
v = -1.4256 m/s
This makes sense because if the cannon is shooting the shell, we would expect the cannon to move back. And since the cannon is a lot heavier, the velocity would be smaller in magnitude, relative to the velocity of the shell.
So the answer is:
v = -1.4256 m/s
Hope this helps! :D
