# Some simple momentum problems

1. Mar 31, 2006

### virtuoso_735

A 54.0 kg dancer leaps 0.33 m high. With what momentum does the dancer reach the ground?

Well, I tried the momentum formula: p=mv. I plugged in 54.0 kg for the mass, and the velocity I put in 9.8 but of course this is wrong. I don't know how to get the velocity for a falling object (very embarassing!). Anyways, does anyone know how to get the velocity of a falling object if they jump .33 m high like in the problem so I can get the momentum? Thanks.

Here is another problem I am having trouble with:

A space probe with a mass of 7.630 103 kg is traveling through space at 110 m/s. Mission control decides that a course correction of 30.0° is needed and instructs the probe to fire rockets perpendicular to its present direction of motion. If the gas expelled by the rockets has a speed of 3.200 km/s, what mass of gas should be released?

Well, I'm not even sure which equation to use and how to do it. I tried a few equations including

mv=mv+MV and then mv=MV, trying to solve for M, but it is wrong everytime. Does the 30 degrees have anything to do with it? Can anyone help? Thanks.

Last edited: Mar 31, 2006
2. Mar 31, 2006

### dav2008

You have to remember that 9.8 m/s2 is the acceleration that objects experience due to gravity near the earth.

With that in mind you can use the equation:
$$v_f^2 = v_0^2 + 2 a \Delta x$$
to determine the final velocity.

For the second one, the only equation you should keep in mind is that the momentum of a system is constant unless it's acted upon by an outside force.

In this case you should equate the momentum of the space probe before the rockets are fired to the momentum of the gas plus the momentum of the remaining mass of the spaceship.

You have to be careful and keep in mind the vector nature of the momentum.

Last edited: Mar 31, 2006
3. Apr 1, 2006

### virtuoso_735

Thanks for the help. The equation helped a lot.

Here's another problem I'm having trouble with

The velocity of a 640 kg auto is changed from 10.0 m/s to 44.0 m/s in 69.6 s by an external, constant force. What is the magnitude of the force?

Do I use the equation F(change in time)=m(change in velocity)?

4. Apr 1, 2006

### virtuoso_735

Nevermind...I got it. :)

5. Apr 1, 2006

### dav2008

Just keep in mind that in general terms the following is true: $$F=\frac {d\vec{p}}{dt}$$.