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Homework Help: Working equation for change in time (Impulse)

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    This http://www.alcaweb.org/cgi-bin/WebObjects/ALCA.woa/wa/resource?id=jQ3f23g01p" [Broken] takes one to the assignment. Students had to determine specific equations based on list and in turn derive a working equation to calculate the total time it took for a bowling ball to come to a stop AFTER it hit the ground.

    2. Relevant equations

    Started with

    1) mgh = 1/2mv2
    2) F(ground on ball) = mg + ma
    3) Ft = mv

    3. The attempt at a solution

    Working equation derived from above three equations simplified down to

    t = 0/g

    Assumptions - "F" in both equations 2 & 3 are same force. All "m" values divide out. How does "v" in equation relate to "v" in equation 3?

    ... is there a way to salvage this one? As a teacher am curious on where it when off track.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 10, 2010 #2


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    Hi tsacket, welcome to PF. The problem with this is that you have two unknowns, the decelerating force, F and the deceleration time t. One can figure out the product F*t, but it is not possible to determine how much of the product is "F" and how much is "t", at least not without additional information or assumptions.

    I am not sure how you got t = 0/g or whether you actually meant to write that. The v in equation 1 is the speed of the ball just before it hits the ground. The v in equation 3 is essentially the first v. More correctly, the impulse equation is

    Ft = m Δv = m(vfinal - vinitial) = m(0 - vinitial)= - mv

    The minus sign indicates that the force on the ball is opposite to the initial velocity v.
  4. Apr 10, 2010 #3
    Thanks kuruman for you replay .... this helps me flush this out in my mind and try to salvage this.

    1) Concerning "F" in impulse equation, am I way off at proposing that the equation 2 in original post address this unknown. Would "F" = mg + ma (mg is force applied by ground to bowling ball weight + additional force (ma) applied by ground to decelerate bowling ball?

    2) Concerning t=0/g

    We combined three equations to initially start with this....

    FΔt = mΔv
    with assumption that F = F(ground on ball) then
    (mg +ma) Δt = mΔv or m(g+a) Δt = mΔv or (g+a)Δt = Δv where masses divide out leaving

    (g + Δv/Δt) Δt = vf - vi

    vf = 0 leaving

    (g + Δv/Δt) Δt = vf

    to get at vf (initial velocity at point of contact to ground) can we use mgh= 1/2mv2 or v = (2gh)1/2 when substituted we get ...

    (g + Δv/Δt) Δt = (2gh)1/2

    Question remaining is does Δv also equal to (2gh)1/2? If so than it all reduces down to t = 0/g ... it seems.

    Am I blowing smoke here? :-) I had two groups of students come to same conclusion and I seem to be an an impasse as well.

    Thanks...in advance
  5. Apr 10, 2010 #4


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    It seems that you are using the symbol "F" in two different contexts. If you say that

    FΔt = mΔv, then F = mΔv/Δt or F = ma. By Newton's Second Law, mass times acceleration is the net force, i.e. the sum of all the forces, FNet. Therefore the symbol "F" must stand for "net force". Now the net force is the sum of all the forces is,

    FNet = Fground on ball - FGravity

    If you make the assertion that F = Fground on ball, then basically you are saying that the force of gravity is negligible compared with the force exerted by the ground, in which case no wonder you end up with g = 0. It is OK to make this assertion however, because the ball stops very fast in all realistic situations which means that the acceleration provided by the ground is many times the force of gravity. Then you need to write the (approximate but valid) equation

    Fground on ballΔt = mΔv

    However, this does not solve the original problem that I mentioned, namely you cannot separate F from Δt in the product without additional information or assumptions.

    Yes, you can say that the change in velocity is Δv = 0 - (2gh)1/2.
  6. Apr 11, 2010 #5
    Yes Kuruman that is what I was shooting for... I total missed mark on my second formula. The net force equation was my problem. Blew that one.. amateur hour. :redface:

    Given this correction FNet would equal 0; hence,
    Fground on ball = FGravity = mg so...

    mgΔt = mv
    mgΔt = m (0 - (2gh)1/2)
    gΔt = (2gh)1/2
    Δt = (2gh)1/2 / g (final working equation?)

    if correct we would only need to measure the height from our third floor window to ground to determine the approximate Δt for bowling ball to come to a complete stop AFTER hitting the ground.? Dimensionally, units reduce down to seconds so that at least works.
  7. Apr 11, 2010 #6


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    No, no and no. If FNet equals zero, the acceleration will also be zero according to Newton's Second Law. The ball accelerates throughout its motion, but its acceleration changes when it hits the ground. There are two parts to the motion.

    Part I - The ball is in free fall before it hits the ground
    In this case the net force is the force of gravity mg directed down. The acceleration is g and directed also down. The initial velocity is zero and the final velocity is (2gh)1/2 directed down. This final velocity becomes the initial velocity for the next part. The total time the ball is in the air is given by the kinematic equation, tf = (2h/g)1/2.

    You can use mgΔt = mv, but in this case Δt is not the stopping time, it is the time it takes for the ball to change its velocity from zero to -(2gh)1/2, i.e. the time of flight. The change in velocity is Δv = -(2gh)1/2 - 0.

    Part II - The ball hits the ground and is stopped by it in time interval Δt
    In this case there are two forces acting on the ball, gravity, which is still mg directed down, and the force exerted by the ground, Fground. This latter force is directed up and is greater than mg, so that the sum of the two forces acting on the ball, i.e. the net force, is directed up. Here is the logic

    1. The speed of the ball is decreasing as soon as it hits the ground.
    2. This means that the acceleration must be opposite to the velocity.
    3. Since the ball's velocity is initially down, the acceleration must be up.
    4. If the acceleration is up, according to Newton's Second Law, the net force must also be up.
    5. The net force is the sum of an "up" force exerted by the ground and a "down" force that is gravity. Since their sum is up, the force exerted by the ground has a larger magnitude than the force of gravity.

    In this case you can write
    (Fground - mg) Δt = mΔv
    where Δt is the stopping time and the change in velocity is Δv = 0 - (-(2gh)1/2)

    I don't think you can get around the problem of separating Fground from Δt simply by manipulating the equations.
  8. Apr 11, 2010 #7

    Agreed ... there is no way to measure/calculate for Fground or Δt. Sure would have been interesting activity if we could.

    Thanks for helping me work through it. If okay will print this page for my students. With your help it has been instructive. Don't mind showing my students I don't know everything, but still have a love for it.
  9. Jun 7, 2010 #8
    Can't we estimate it, though? In a physics course I am taking, we had to prove why or why not skydiving on the moon from 4 km would kill you. The only way to prove this, short of real experimentation, is to figure out if the net force of the impact exceeds current demonstrated levels of force, and at which resulted in death of a 50th percentile male.

    If Δt is the time it takes for an object to change from one velocity to another, wouldn't we just need to know the length of the object? (Diameter, in this case)

    Δt = l/Δv

    This is the only logical way I could come to a sound reasoning against jumping to one's death in an environment with no drag resistance.

    Example, if the skydiver, 1.78 m long, is traveling feet-first at an instantaneous velocity of 113.14 m/s right before impact, and the moon is orbiting at an average velocity of 4.627 m/s, a difference of 108.51 m/s, wouldn't the time it takes him to change velocities, from his own to the moon's, be

    Δt = 1.78 m/108.51 m/s = .016 seconds?

    If the object moves after the force of the impact, before coming to a stop, couldn't we use that distance, too, in the same equation?
    Last edited: Jun 7, 2010
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