Momentum Problem: Is this correct?

[mailmas]

Homework Statement

There's a car, there's a cannon strapped on top of a car, the car travels with a velocity of 20 m/s and the cannon fires. The cannon fires a ball at 200 m/s. What is the velocity of the car? Velocities are with respect to the ground.

mass car = 1000
mass cannon = 100

Homework Equations

SUMATION p_initial = SUMMATION p_final

The Attempt at a Solution

100*20 + 1000*20 = 100*(-200+20) + 1000*V_car
22000 = 100* (-180) + 1000*V_car
40000 = 1000* V_car
V_car = 40 m/s

 

[TSny]

Is the direction that the ball is fired given?

Why write -200+20 for the final velocity of the ball? According to the problem statement, all velocities given are with respect to the ground.

 

[mailmas]

Is the direction that the ball is fired given?

Why write -200+20 for the final velocity of the ball? According to the problem statement, all velocities given are with respect to the ground.

The ball does go in the direction of the ground. I thought it would there would be momentum of equal magnitude but opposite direction acting on the cannon. That is where I got the -200 + 20 from for the velocity of the cannon. Is that correct?

 

[TSny]

The ball does go in the direction of the ground. I thought it would there would be momentum of equal magnitude but opposite direction acting on the cannon. That is where I got the -200 + 20 from for the velocity of the cannon. Is that correct?

I don't think so. You are given the mass of the cannon and the mass of the car. Are you given the mass of the ball?

 

[mailmas]

I don't think so. You are given the mass of the cannon and the mass of the car. Are you given the mass of the ball?

Oh shoot, I read the problem again and I am given the mass of the ball, it's 10kg.
So it would be:
10*20 + 100*20 + 1000*20 = 10*200 + 100*Vcannon + 1000*Vcar
22200 = 2000 + 100*Vcannon + 1000*Vcar
20200 = 100*Vcannon + 1000*Vcar

Would the Vcannon be equal to:
100*Vcannon = Massball*Vball => Vcannon = -20m/s? I'm confused now. Since it's moving at +20m/s would it go to 0?

 

[TSny]

Oh shoot, I read the problem again and I am given the mass of the ball, it's 10kg.
So it would be:
10*20 + 100*20 + 1000*20 = 10*200 + 100*Vcannon + 1000*Vcar
22200 = 2000 + 100*Vcannon + 1000*Vcar
20200 = 100*Vcannon + 1000*Vcar

OK, that looks right.

Would the Vcannon be equal to:
100*Vcannon = Massball*Vball => Vcannon = -20m/s? I'm confused now. Since it's moving at +20m/s would it go to 0?

Here, we will need to interpret what it means for the cannon to be "strapped to the top of the car". Does it mean that the cannon is not allowed to move relative to the car?

 

[mailmas]

OK, that looks right.

Here, we will need to interpret what it means for the cannon to be "strapped to the top of the car". Does it mean that the cannon is not allowed to move relative to the car?

Yes I believe that is right. So would it instead be something along the lines: Msystem*Vsystem = -2000 => 1100*(20-v) = -2000 => 20-v = -1.81 => V= 18.2 m/s

 

[TSny]

Yes I believe that is right. So would it instead be something along the lines: Msystem*Vsystem = -2000 => 1100*(20-v) = -2000 => 20-v = -1.81 => V= 18.2 m/s

I'm not following this. The equation Msystem*Vsystem = -2000 says that the total momentum of the system is negative. Why is that?

But, go back to where you had

20200 = 100*Vcannon + 1000*Vcar

If the cannon is not allowed to move relative to the car, what can you say about Vcannon and Vcar?

 

[mailmas]

I'm not following this. The equation Msystem*Vsystem = -2000 says that the total momentum of the system is negative. Why is that?

But, go back to where you had

20200 = 100*Vcannon + 1000*Vcar

If the cannon is not allowed to move relative to the car, what can you say about Vcannon and Vcar?

They're the same.

 

[TSny]

They're the same.

Yes.

 

[mailmas]

Yes.

So we could get the velocity of the system(car) be 20200/1100 = 18.36 m/s

 

[TSny]

So we could get the velocity of the system(car) be 20200/1100 = 18.36 m/s

Yes, I think that's right. This assumes that the cannon fires horizontally in the same direction that the car is traveling.

 

[mailmas]

Yes, I think that's right. This assumes that the cannon fires horizontally in the same direction that the car is traveling.

Thanks!