1
$\begingroup$

Two balls 'A' and 'B' are thrown vertically upwards with the same velocity. The mass of A is greater than that of B. We need to find which of the balls reaches a greater height (assuming the effect of air resistence is negligible).

I personally feel that A would travel further since its momentum is higher than that of B. But few people whom I discussed this question with had different opinion.

One said that they would both travel the same distance since $s=(ut) + (1/2)at^2$ The other said that B would travel further since the effect of gravity on it is less according the universal law of gravitation.

If someone could explain which of these is correct or maybe point out the flaw in either of them then it would help a lot. Thanks in advance.

$\endgroup$
0
$\begingroup$

If the objects have same initial velocity, they should travel the same distance. Momentum has no picture there.

But when they reach the peak height, they stop for a fraction of second (or lesser), that's where your mass comes into picture as their potential energies come into picture.

Momentum comes into picture when they come in contact with another object, or the ground. The heavier object would make a bigger (or deeper) impact on the ground (impact surface) than the lighter one. But as for travelling distance and velocity of travel, it should be the same as the equations are (clearly) mass independent.

$\endgroup$
5
  • $\begingroup$ What do you mean by "that's where your mass comes into picture as their potential energies come into picture."? Why would their potential energies "come into picture" only at that point? $\endgroup$
    – JiK
    Sep 8 '16 at 13:45
  • $\begingroup$ I just meant that the difference in mass would lead to different velocities with which the objects will fall as the potential energy converts into kinetic energy from that point. Before that point, as the problem was stated, the mass is not relevant to the problem. $\endgroup$
    – AtulBhatS
    Sep 8 '16 at 13:48
  • 1
    $\begingroup$ If drag is not taken into account, the difference in mass will not lead to different velocities when falling. The same equations apply: their acceleration is simply $g$. $\endgroup$
    – JiK
    Sep 8 '16 at 13:52
  • $\begingroup$ Very true. I guess it was a mistake on my part that I assumed the problem was not in an ideal environment and with drag. The mass would only affect the impact surface when the downward fall is broken/ended. $\endgroup$
    – AtulBhatS
    Sep 8 '16 at 13:57
  • $\begingroup$ I edited my question have a look $\endgroup$
    – HDatta
    Sep 8 '16 at 16:55
1
$\begingroup$

If the mass of an object is $m$ then its weight is $mg$ where $g$ is the gravitational field strength.

Assuming that no other forces act on the mass and apply Newton's second law

$F = ma \Rightarrow mg = ma \Rightarrow a = g$

where $a$ is the acceleration of the mass shows that the acceleration of a mass is independent of its mass.

So as your two masses had identical initial conditions and undergo the same acceleration then their motions must be the same.

$\endgroup$
10
  • $\begingroup$ But why isn't momentum affecting the motion in any way? $\endgroup$
    – HDatta
    Sep 8 '16 at 9:33
  • $\begingroup$ It is. You apply a force and the momentum changes. However both the force (weight) and momentum both depend on the mass $m$. So in effect the mass cancels out. A more massive object will suffer a greater change in momentum but is has a larger force, its weight, acting on it. $mg = \dfrac {\Delta m v}{\Delta t}$ Note that mass cancels. $\endgroup$
    – Farcher
    Sep 8 '16 at 9:41
  • $\begingroup$ Fine but is the momentum not changing the distance covered by both objects? $\endgroup$
    – HDatta
    Sep 8 '16 at 9:49
  • $\begingroup$ The key point is that the acceleration of the two masses is the same $=g = \dfrac {\Delta v}{\Delta t}$ so all your kinematic for constant acceleration apply and since the initial conditions (velocity and position) are the same the motion of the two masses will be the same. $\endgroup$
    – Farcher
    Sep 8 '16 at 9:56
  • $\begingroup$ Thank you very much, I need to explain this to the other people, so could you just explain it to me in a nutshell step by step. Thank you once again... $\endgroup$
    – HDatta
    Sep 8 '16 at 10:23
0
$\begingroup$

Effect of gravity does not depend on the mass of the balls. If you draw free body diagram each of these objects, you will see that masses are cancel each other. So, both of them has the same distance after they are thrown vertically. Greetings :)

$\endgroup$
7
  • $\begingroup$ But clearly the universal law of gravitation states that F = G(m1*m2){(d^2)^(-1)}. If earth is pulling the object then lets say m1 is earth's mass and m2 is the ball's mass. Therefore we can say that gravitational force on A is higher than that on B. So the acceleration would be higher on A as compared to that on B. $\endgroup$
    – HDatta
    Sep 8 '16 at 9:15
  • $\begingroup$ After you write the equation given above, you should equate this equation to the (mass of object*acceleration of the object). you will see that masses will be cancelled . $\endgroup$ Sep 8 '16 at 9:19
  • $\begingroup$ I see that the third is being proved wrong and the second one is being proved correct, but won't there be any effect on the motion due to the difference in momentum? $\endgroup$
    – HDatta
    Sep 8 '16 at 9:32
  • $\begingroup$ @Reptile1234 (a) What do you refer to by "the third" and "the second one"? (b) The motion of a ball is uniquely determined by knowing the initial position, the initial velocity, and the acceleration (which may vary during the time or not). In this case, all those are same for both balls. $\endgroup$
    – JiK
    Sep 8 '16 at 14:25
  • $\begingroup$ @CihanYesil after the masses get cancelled what r u concluding. $\endgroup$
    – HDatta
    Sep 8 '16 at 16:53
0
$\begingroup$

Your question stems from not knowing the difference between kinematics and dynamics. This link will help you What is the difference between "kinematics" and "dynamics"? to ask relevant questions

$\endgroup$
0
$\begingroup$

the thing you have missed to mention is effect of air resistance.If we include the effect of air resistance in this question then role of mass will come into play, but if you are not including air resistance then you will use s=ut+at^2/2, which is independent of mass so both bodies will traverse the same distance.Hope my message reach you.

$\endgroup$
1
  • $\begingroup$ Im sorry I hadn't mentioned few things properly please have a look at the edited question $\endgroup$
    – HDatta
    Sep 8 '16 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.