1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

HELP 3 physics problems that i am stuck on

  1. Sep 1, 2004 #1
    HELP!!! 3 physics problems that i am stuck on

    Help me out guys. I am in a college physics class and a week and a half into the class i am stuck.

    1.) A bullet is moving at 244m/s. It hits a tree and penetrates 8.34mm. (A)What is the average acceleration as it slows? (B) Assuming constant acceleration, how long does it take for the bullet to stop.

    I keep coming up with over 7mill m/s for part A... so obviously I am doing something wrong. If someone can just get me started on the right track I think I can figure the rest out.

    2.) A ball is thrown straight up at 36m/s How long will it take for the ball to return to it's starting position?

    I know it is one of the free fall equations, but i just cant figure out which... also, air resistance is not a factor.

    3.) An astronaut lands on a strange, new planet. She discovers she can jump a maximum of 27m @ 6.0m/s. What is the magnitude of acceleration of gravity on the new planet?

    Any help on any of these would be greatly appreciated. Thanks.

    Stacy in NC
     
  2. jcsd
  3. Sep 1, 2004 #2
    1.) A bullet is moving at 244m/s. It hits a tree and penetrates 8.34mm. (A)What is the average acceleration as it slows? (B) Assuming constant acceleration, how long does it take for the bullet to stop.

    I keep coming up with over 7mill m/s for part A... so obviously I am doing something wrong. If someone can just get me started on the right track I think I can figure the rest out.

    my note: use vf^2 = vi^2 + 2ad

    you shouldn't get 7mil m/s. Try again.

    2.) A ball is thrown straight up at 36m/s How long will it take for the ball to return to it's starting position?

    I know it is one of the free fall equations, but i just cant figure out which... also, air resistance is not a factor.

    my note: figure out the time it takes for an object to reach 36m/s with only gravitational acceleration. Then multiply it by 2 since the trajectory is symmetrical.

    use vf = vi + at

    3.) An astronaut lands on a strange, new planet. She discovers she can jump a maximum of 27m @ 6.0m/s. What is the magnitude of acceleration of gravity on the new planet?

    my note: again use vf^2 = vi^2 + 2ad

    Any help on any of these would be greatly appreciated. Thanks.

    Stacy in NC

    P.S - just do more studying and you should do fine...
     
  4. Sep 1, 2004 #3

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    1) The kinetic energy of the bullet is

    [tex]E = \frac{1}{2} m v_0^2[/tex]

    The tree has to dissipate all of this energy by applying a force to the bullet. The relationship between work, force, and distance is

    [tex]W = \mathbf F \cdot \mathbf d[/tex]

    In this case, you know how much work is done (E) and the distance it was done over (d), and you can solve for F. You're not so much interested in the force as you are in the acceleration, so you can use Newton's second law to replace F with ma:

    [tex]\frac{1}{2} m v_0^2 = Fd = mad[/tex]

    The m's cancel, and you're left with a simple expression for the acceleration. This acceleration is in the millions of m/s2. This makes sense, when you realize that a bullet initially travelling at 244 m/s would cover 8.34 mm in about 34 millionths of a second:

    http://www.google.com/search?hl=en&lr=&ie=UTF-8&q=8.34+mm+/+244+m/s&btnG=Search

    The acceleration has to be quite large.

    To find the time it takes the bullet to stop, just using the standard kinematic formula:

    [tex]s(t) = v_0 t + \frac{1}{2} a t^2[/tex]

    Where s(t) is the position of the bullet at time t. Plug in 8.34 mm for s(t), 244 m/s for [itex]v_0[/itex], and the acceleration you found above, then solve for time. The answer you get should be a few tens of microseconds.

    2) If you're going to memorize any equations, memorize this one:

    [tex]s(t) = v_0 t + 1/2 a t^2[/tex]

    That's all you need to solve this problem. The acceleration is just g, and you know v_0. If the ball starts at ground level, then s(t=0) is 0. Solve this equation for the other permissible value of t.

    3) You can derive the maximum height of a trajectory easily enough, but here's the formula:

    [tex]y_\textrm{max} = \frac{v_0^2}{2a}[/tex]

    Plug in the numbers and solve for a.

    - Warren
     
    Last edited: Sep 1, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: HELP 3 physics problems that i am stuck on
Loading...