# Confused on velocity

1. Jan 19, 2005

### Tabe

Here's my predicament...
A 30.0 kg shopping cart full of groceries sitting at the top of a 2.0 meter hill rolls until it hits a tree stump at the bottom of the hill. Upon impact, a .25 kg can of peaches flies out of the cart. What is the velocity of the can as it leaves the cart?
I tried solving the problem, but the answer that I got really doesn't make any sense. First I solved for the final velocity of the cart before it hit the tree stump, and I got 6.26 m/s. Then, when I solved for the velocity of the can, I got an answer of 750 m/s, which is very unrealistic. Can someone point me in the right direction, or point out where I went wrong?

2. Jan 19, 2005

### MathStudent

Does the question indicate the slope of the hill or the angle of inclination??
Does it offer any other information.

This is definitly a conservation of momentum problem

3. Jan 19, 2005

### Tabe

No, the problem is very vague. It doesn't specify anything about the hill, or give any more info on the can itself. If it is a conservation of momentum problem, wouldn't that mean that the can would be traveling really fast once it left the cart?

4. Jan 19, 2005

### Tabe

When I solved for the final velocity, I solved for potential gravitational energy, I multiplied mgh, and I got 588.6. Then I set that equal to kinetic energy, which is 1/2mv^2. I solved for velocity, and I got 6.26 m/s. From there, I took the velocity and used a kinematics equation to solve for the velocity of the can.

5. Jan 19, 2005

### Tabe

Ok, let me check what I did. I just wanted to make sure, because it really didn't sound very realistic to me. Great! I redid the problem, and everything came up equivalent to each other.
Thanks for your help. It's really gonna help me on my exam tomorrow! :)

6. Jan 19, 2005

### beth314159

An odd word problem...

Neglecting friction, etc., the cart and can are travelling together with a speed 6.26 m/s at the bottom of the hill. Now, if the cart stops dead, and the can slides off the cart, then, neglecting friction, or any sort of collision with the cart on the way off, the can is still going 6.26 m/s. The momentum of the cart goes to zero after it collides with the tree; the momentum of the can stays the same, and so does its velocity. (Where does the cart's momentum go? Well, the tree + earth system recoils a little!)

-Beth

7. Jan 19, 2005

### jdstokes

Use the conservation of energy equation to find the velocity of the cart at the bottom of the hill
\begin{align*} K_1 + U_1 & = K_2 + U_2 \\ mgh & = \frac{1}{2}mv^2 \\ v & = \sqrt{2gh} \\ & = 6.26\,{\rm m/s} \end{align*}

When the cart hits the tree, it delivers an impulse of

$J = (30.0\,{\rm kg})(6.26\,{\rm m/s})$

to the can. The impulse on the ball is equal to the its momentum change,

\begin{align*} J & = m(v_2 - v_1) \\ (30.0\,{\rm kg})(6.26\,{\rm m/s}) & = (0.25\,{\rm kg})(v_2 - (6.26\,{\rm m/s})) \end{align*}

so the final velocity of the can is 757 m/s.

8. Jan 19, 2005

### MathStudent

That's a very valid point. I didnt consider the earth and tree stump as part of the system in my original model, but now that you point that out it makes perfect sense. Thank you.