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Help Clarifying Answers-Linear Acceleration & Conservation of Momentum

  1. Apr 24, 2014 #1
    Hey all,

    I'm brushing up on some physics in my spare time, and I was wondering if someone knowledgeable here could help me clarify the reasons behind a couple answers...much thanks in advance! =)

    Question 1) The greatest linear acceleration of the center of mass of a baseball bat will be produced by pushing with a force F at...(picture attached)

    A) position 1​
    B) position 2​
    C) position 3​
    D) all the same​
    E) not enough information​

    Relevant equation: Force = mass[itex]_{center of mass}[/itex] * acceleration[itex]_{center of mass}[/itex]

    Attempt at solution: The solution tells me that it's "D) all the same," and that using F = ma, you can see that linear acceleration is not related to the position where the force is applied. However, this doesn't seem intuitive to me.

    If I have a bat sitting on a table, and I push at position 1 or 3, I would have expected there to be some but less linear acceleration (and more rotational acceleration), so that position 2 would have the greatest linear acceleration. I remember reading that rotational acceleration is calculated independently of linear acceleration.

    Am I overanalyzing and treating this as more than a statics problem? That as soon as it starts rotating if pushed in position 1 or 3, that the scenario is no longer valid because, realistically force F (if it continues in the same direction) is no longer pushing at the bat perpendicular to its motion? So in that case, I should think of this force more like a quick, impulsive force than a force over a prolonged period of time?

    Question 2) Suppose you are on a cart, initially at rest on a track with very little friction. You throw balls at a partition that is rigidly mounted on the cart. If the balls bounce straight back as shown in the figure, is the cart put in motion? (image attached)

    A) it moves to the right​
    B) it moves to the left​
    C) it remains in place​
    D) not enough information​

    Relevant equation:

    Conservation of momentum: 0 = m_person+cart* v_person+cart + m_ball * v_ball

    Attempt at solution: Taking the person, cart and balls as a system, there are no external forces in the direction of motion and we can say that momentum is conserved. The solution tells me that the answer is "A) it moves to the left." Again, if I use the equation to answer, I arrive at that conclusion. Or if I imagine putting a curtain covering the contraption and the only thing I see is a ball bouncing out, I can visualize that that must make the cart move to the left. HOWEVER...

    If I analyze it from the following approach, I get stuck (probably overanalyzing it again). Let's say the person starts out with the ball in his hand, primed to throw it (he's like a gun). He throws it, and the ball goes to the left, so the cart + person goes to the right in response (like recoil). Then the ball hits the cart partition and bounces off elastically. How do you justify that after that action the cart + person move to the left (and doesn't just stop the cart)? If anything, doesn't the ball hit the wall with a force F (that it left the person's hand in), and cancel it out?

    In a variation of this problem, I was told that if the ball is inelastic and smashes/sticks to the wall instead of bouncing off, that in this case the cart would stop moving to the right once the ball hits the wall.


    Again, any clarification would be great and super appreciated! =)

    Attached Files:

  2. jcsd
  3. Apr 24, 2014 #2


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    Hello riddle_me_phys,

    Welcome to Physics Forums! :smile:

    D) is the correct answer.

    It's an important concept to realize when studying "dynamics" the, the motion part of classical mechanics.

    The linear acceleration of an object, caused by an external force is, is independent of where on the object that force is applied.

    If it helps, try this. Either assume the bat is on a frictionless table, or assume that you and the bat are floating motionless in free space. You apply an external force (we won't concern ourselves with what happens with you, just the bat) by pressing on the bat, using your finger, with a constant force F, for a very short time, Δt.

    Suppose you start by pressing on the bat at its center of mass, position 2. During this time that you apply the force, the bat and your finger move a small distance, Δd2 (you may assume that Δd2 is much, much smaller than the length of the bat, if you wish). You'll notice that although the bat accelerates, it does not start rotating.

    Now start over. This time you'll use your finger to apply the force at position 1, all else the same. You apply a force at position 1 with a magnitude of F for the same amount of time, Δt. doing so, you'll find that the bat is easier to push by applying the same force, such that the distance that your finger travels, Δd1 is greater than the distance your finger moved when applying the force at position 2. In other words, Δd1 > Δd2. [Edit: by that I mean Δd1 is the distance your finger moved when applying the force at position 1. It's interesting to note that even in this experiment, the distance that the bat's center of mass is the same as it was in the first experiment, Δd2.] Taking out your measuring equipment, you'll also notice the following:

    • The final speed of the bat's center of mass is the same in both experiments, independent of where on the bat the force was applied. The direction of the bat's center of mass is also identical in both experiments. The direction that the center of mass moves is in the same direction as the force, independent of where on the bat the force was applied.
    • In the second experiment, when the force was applied at position 1, the bat also rotates at the end of the experiment. This rotation didn't happen when the force was applied at position 2.

    These results are quite interesting. As a matter of fact, in both cases, the linear momentum of the bat is, mv, and is equal to FΔt.

    But there is something different about each experiment. In the second case, the bat rotates. Where did the extra energy come from? Why it came from W = FΔd. In the second experiment, you applied the force over a greater distance, thus you gave more energy to the bat.

    Yes, that is also correct. The cart/person system will ultimately move to the left as the ball continues on to the right.

    Correct again. If the ball stuck to the wall, the final velocity of the system would be zero. The same would be true if the person caught the ball, instead of letting it fly off to the right.

    An important part of this problem is analyzing the ball's momentum during its bounce off the wall. Momentum is a vector that has both magnitude and direction. When the ball bounces off the wall (assume a perfectly elastic collision), it starts with speed v and bounces off with speed somewhere close to v, but in the opposite direction. The total change in the ball's momentum is roughly 2mv for the bounce (it's slightly less than that). You could get more detailed than that if you knew the masses of the ball, the person and the cart. doing so, you'd find that the ball's final horizontal speed times the ball's mass is equal the mass of the cart+person, times the final speed of the cart/person.
    Last edited: Apr 24, 2014
  4. Apr 24, 2014 #3
    For the first problem, keep in mind that force is just a vector equation.

    $$\sum\vec{F} = m\vec{a_{G}}$$

    And this yields you ##n## independent equations for any vector space belonging to ##R^n##

    This means if you're working in 2 dimensions, you have 2 force equations. It's pretty easy to see that there's really no way to change the magnitude of your acceleration vector for your center of mass, because your acceleration vector is always just going to be a scalar multiple of the sum of your forces.

    Now, moment, on the other hand, is a different ball game. For rigid bodies you can sum the moments around any point you like, and you'll be able to do more with that. Also, you can calculate the relative acceleration of one point as compared to any other point on a rigid body by using the equations for relative motion.

    By summing the moments around a certain point, you can solve for ##\alpha## with

    $$\sum M_{G} = I\alpha = Fd_{G}$$

    And with relative acceleration of a rigid body for some point P around center of mass G,

    $$\vec{a_{P}} = \vec{a_{G}} + \alpha\hat{k}\times \vec{R_{GP}} - \omega^2\vec{R_{GP}}$$

    So this kind of shows how the linear acceleration does change direction, but never really changes magnitude. The real game changer ends up being the moment.
  5. Apr 25, 2014 #4

    rude man

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    Regarding question 2, keep in mind that the center of mass of an isolated system cannot move. So since there is zero friction, as the balls bounce off the left wall and are then permitted to fly away to the right, the cart will move to the left just enough so that the center of mass of the cart plus flying balls will always remain at the same position.

    If the balls were caught by the thrower after they hit the left wall the cart would not move appreciably since the c.m. of the cart hasn't moved.
  6. Apr 25, 2014 #5
    Thanks collinsmark, Rellek, and rude man for the clarifications!!

    Collinsmark, I very much appreciate you taking the time to write out that explanation. Regarding the linear acceleration problem, I accept that Δd1 will be greater than Δd2, but how were you able to conclude that the center of mass will have the same final displacement/speed in all the scenarios, independent of where the force is applied? I do understand that the bat starts at rest, so for all 3 positions, if the center of mass has the same final speed over the same time period, then the acceleration of the center of mass will be the same regardless.

    I think I understand the cart problem now in terms of where I was getting stuck. With the elastic collision, there is a 3rd "event" that I wasn't thinking of, where the ball goes from zero to v away from the wall, which causes a force to the left on the cart and person. This is what the inelastic collision doesn't have which is why it causes the cart to stay stopped.

    Rude man, couldn't the center of mass of an isolated system move? In the same system, if the person just picked up the balls and walked off the cart as far away as he could go, wouldn't that move the center of mass...oooh...now we'd be taking into account the earth and ground. I see! =)
  7. Apr 25, 2014 #6

    rude man

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    Yeah, if there was no friction he couldn't walk away! He'd slip & maybe slide a ways, but the c.g. of the cart + himself would not move. And if there is friction, as you say, you have now involved the whole Earth! The c.m. of the Earth still hasn't moved.
  8. Apr 25, 2014 #7


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    I don't believe I can be successful at deriving the results without first either invoking Newton's laws of motion, or alternately invoking conservation of momentum (but conservation of momentum is derived from Newton's laws of motion, so that would lead to a circular argument.) Similarly, I could conceivably use Lagrangian or Hamiltonian mechanics, but these too are based on conservation laws that ultimately are based on Newton's laws of motion. (Newton's third law of motion, "for every action there is an equal and opposite reaction," plays an important role in conservation of momentum.)

    So in the end it comes down to where do
    [tex] \sum_i \vec F_i = m \vec a [/tex]
    [tex] \sum_i \vec \tau_i = I \vec \alpha [/tex]
    come from?

    They come from Isaac Newton's mind. But they've been shown to hold true, time and time again for speeds less than relativistic speeds (where velocities involved are << c, the speed of light).

    I don't believe there is a way to derive these laws without running into circular logic problems.
    Last edited: Apr 25, 2014
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