# Conservation of Momentum and Thrown Objects

• Jessehk
In summary, the conversation discusses a physics problem involving the law of conservation of momentum and thrown objects. The problem involves a 50 kg girl in a stationary canoe holding two 10 kg cannon balls, and she throws them at 5 m/s. The final velocity of the canoe is calculated using the law of conservation of momentum, taking into account the frames of reference of the canoe and the boat. The solution is found to be approximately -0.87 m/s. The conversation ends with the person expressing gratitude for the help and sharing their solution.
Jessehk
[SOLVED] Conservation of Momentum and Thrown Objects

As part of our physics class, we've been given lots of problems typically solved by taking into account the law of conservation of momentum. I've had little-to-no trouble solving them, but one problem in particular is driving me nuts.

## Homework Statement

There is a 50 kg girl in a 50 kg stationary canoe holding 2 10 kg cannon balls. She throws a cannon ball at 5 m/s, and then throws the second at 5 m/s w.r.t the boat. What is the canoe's final velocity if the (incorrect) assumption is made that their is no friction between the canoe and the water? Answer: 0.87 m/s

## Homework Equations

$$\vec{p} = \vec{p \prime}$$

## The Attempt at a Solution

Ok, so I first calculate the speed of the canoe w.r.t the ground using the law.
$$0 = \vec{p_c} + \vec{p_b}$$

$$0 = 110 v_c + 10(5)$$

$$v_c = -5/11 m/s \approx -0.454545454... {m/s}$$

Next, my idea was that the fact the second ball is being thrown at 5 m/s w.r.t. the canoe while the canoe was moving was important. So I used the L. of C. of M. again but in the frame of reference of the canoe.

$$0 = 100v_c + 5(10)$$

$$v_c = -1/2$$

So I figured I would just add the speed in the frame of reference of the boat to the speed it was traveling before. That gives me
$$v_c \approx -0.95454545... m/s$$

Which is clearly not the right answer. Am I on the wrong track completely, or is there something simple I've missed? Any help would be appreciated. :)

When she throws the balls at 5 m/s with respect to the canoe, they mean with respect to the canoe after the ball is thrown. If the canoe's velocity (after the throw) is Vc (negative, of course), the ball's velocity is Vc + 5.

Thanks Doc Al. If I'm not mistaken, that leads me to this answer:

$$\vec{p} = (-5/11)(110) = \vec{p \prime} = 100v_c + 10(-5/11 + 5)$$

$$-50 = 100v_c + 500/11$$

$$v_c = -0.9545$$

Which is the same as my previous attempt. I think that makes sense, as I've just switched frames of reference. What am I doing wrong?

Looks to me like you're adding 5 m/s to the speed of the canoe before the throw, not after like I suggested.

I'd do it in two steps, having both throws be 5 m/s with respect to the canoe after the the throw. (The idea of changing frames is a good one; use that for the second throw.)

When you say after the throw, do you mean after the first ball is thrown (before which, the canoe is stationary), or after the second ball is thrown by the girl on the moving canoe?

After each throw. (That's how I would do it.)

Find the speed after the first throw (with the ball being 5 m/s w.r.t. the canoe after the throw). Then repeat (after switching frames) for the second throw.

Doc Al! Thank you so much for that help. It was exactly what I needed: it gave strong hints, and forced me to think.

In case anyone is interested:
$$0 = 110 v_c + 10(v_c + 5)$$

$$v_c \approx -0.417 m/s$$

In the frame of reference of the boat...
$$0 = 100v_c + 10(v_c + 5)$$

$$v_c = -5/12 m/s$$

Get back to the ground frame...
$$v_c = -5/12 - 5/11 \approx -0.87 m/s$$

Yay!

EDIT: it could arguably be solved more simply in the ground frame, but I find it easier to visualize in the canoe frame.

Last edited:

## 1. What is conservation of momentum?

Conservation of momentum is a fundamental law of physics that states that in a closed system, the total momentum is constant. This means that the total momentum before an event or interaction is equal to the total momentum after the event or interaction.

## 2. How does conservation of momentum apply to thrown objects?

When an object is thrown, it has an initial momentum that is determined by its mass and velocity. As the object moves through the air, it experiences various forces such as air resistance and gravity. However, according to the law of conservation of momentum, the total momentum of the object and any external forces acting on it will remain constant.

## 3. Can conservation of momentum be violated?

No, conservation of momentum is a universal law and cannot be violated. In any closed system, the total momentum will always remain constant. However, it may appear that momentum is not conserved if there are external forces acting on the system that are not taken into consideration.

## 4. How does the mass and velocity of a thrown object impact its momentum?

The momentum of an object is directly proportional to its mass and velocity. This means that a larger mass or a higher velocity will result in a greater momentum. This is why a heavier object thrown with the same velocity as a lighter object will have a higher momentum.

## 5. Can conservation of momentum be applied to all types of thrown objects?

Yes, conservation of momentum can be applied to all types of thrown objects, including projectiles, balls, and even liquids. As long as the object is in a closed system and is not experiencing any external forces, its total momentum will remain constant.

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