Force on Cannon After Firing: Solving the Problem

[dionysus]

1. A 160.0Kg cannon initially at rest shoots an 8.00Kg cannon ball at a speed of 240.0 m/s. The cannon moves backwards 0.40 meters. What is the force on the cannon after firing the cannon ball?

I've solved for the velocity of the cannon, getting 12 m/s, but don't know exactly where to go from there.


2. An 80kg astronaut is in space, at rest. He throws a 5kg tool at 6.0 m/s to another astronaut, with a mass of 65 kg, who catches it.
c. How much energy is used by the astronaut throwing the tool

Is it wrong to use 1/2(m)(v)^2? I used that to get an answer of 90 J, but I've been told the answer was 96 J.

Any help is appreciated, I know this is probably very basic to some of you.

 

[PhanthomJay]

1. A 160.0Kg cannon initially at rest shoots an 8.00Kg cannon ball at a speed of 240.0 m/s. The cannon moves backwards 0.40 meters. What is the force on the cannon after firing the cannon ball?

I've solved for the velocity of the cannon, getting 12 m/s, but don't know exactly where to go from there.


2. An 80kg astronaut is in space, at rest. He throws a 5kg tool at 6.0 m/s to another astronaut, with a mass of 65 kg, who catches it.
c. How much energy is used by the astronaut throwing the tool

Is it wrong to use 1/2(m)(v)^2? I used that to get an answer of 90 J, but I've been told the answer was 96 J.

Any help is appreciated, I know this is probably very basic to some of you.

You should post one question at a time. Very little about Physics is basic, in my opinion.

In your first question, assuming a horizontal mount, you have the correct velocity of the cannon moving backwards at 12m/s initially, but it comes to a stop in 0.45 meters. Something has got to stop it, like ground friction or the cannon support. Otherwise, it would move forever. When it stops, it has no more speed. It must therefore have decelerated from 12m/s to 0 in 0.45 meters. One way to solve this problem is to calculate that deceleration, then what would you have to do to solve for the stopping force on the cannon? Or are you familiar with work energy methods?

For the second question, have you considered that the astronaut moves backwards at a certain speed?

 

[dionysus]

Thanks, you were right about the astronaut moving backwards, that solved that problem.

As for the cannonball one, I think I'm supposed to assume its a frictionless surface/system, since I'm given no information about that. With that in mind:

Solving for time: (.4 m) / (12 m/s) = .0333 s
Momentum: (8 kg)(240 m/s) = 1920 kgm/s
Force: (1920 kgm/s) / (.0333 s) = 57600N

Does that seem right assuming no friction/deacceleration?
For momentum I just used mass times velocity, and for force I used impulse (it was at rest to begin with) divided by time.

 

[PhanthomJay]

Thanks, you were right about the astronaut moving backwards, that solved that problem.

As for the cannonball one, I think I'm supposed to assume its a frictionless surface/system, since I'm given no information about that. With that in mind:

Solving for time: (.4 m) / (12 m/s) = .0333 s
Momentum: (8 kg)(240 m/s) = 1920 kgm/s
Force: (1920 kgm/s) / (.0333 s) = 57600N

Does that seem right assuming no friction/deacceleration?
For momentum I just used mass times velocity, and for force I used impulse (it was at rest to begin with) divided by time.

That's a good way to do it, however, you have used time = distance/speed, whch is not correct. The correct formula is time = distance/average speed. The average speed of the cannon is not 12m/s. It can't get there instananeously from rest, nor can it decelerate to 0 instantaneously. So what is it's average speed? Then calculate the force as you have done, using the correct value for the average speed, which is an average force. Check your work using work energy methods, if you are familiar with that method.