Quantifying Inertial Resistance

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1. Feb 27, 2016

NRa

Hi! I am having a little trouble with a question asked by a colleague.

There’s a ball B with a certain mass M, at rest. A small ball A of mass m is moving with speed v toward M.

If m=M, and the collision is perfectly elastic and the two objects perfectly rigid, than we know that A would come at rest and B will start to move with velocity V.

Now if M = 2m, than upon collision, A would rebound with a velocity V1 and B would start to move with a velocity V2.

These are one-dimensional collisions.

If I have to think about the forces that B exerts on A during the collision in the two cases above, than:

If the collisions are instantaneous, impulse imparted to A in second case is greater than in first. That means, when B has a greater mass, it offer greater inertial resistance. Is it right to say that the contact force from B in the second case is greater? Since normal force has its roots in electromagnetic forces, is it okay to say that now that the mass has increased, the normal force is greater?

How else would I quantify that inertial resistance? Does an increase in mass leads to an increase in rigidity?

2. Feb 27, 2016

drvrm

If you are considering elastic collisions the force exerted by first body on the second one is same /equal to the force exerted by 2nd body on first. that's why the change in momentum leads to its conservation as well as energy conservation . you write out the conditions -it may be possible that if one of them is having a large mass the velocity change goes to zero ..... or the total momentum of the first one also comes out to be zero( see momentum is a vector) ..try to think over it and write out equations with their signs correctly!

3. Feb 27, 2016

NRa

I know that the forces involved in each collision are equal and opposite, from Newton's 3rd Law.
Question is whether the force on ball A has the same magnitude in each case as well. My reasoning says that it's not. What i want is to improve upon that reasoning if it's right or otherwise, get a correct perspective.

4. Feb 27, 2016

Aniruddha@94

Since the impulse imparted to A in case 2 is greater than that in case 1 I think it's safe to say that the average force is greater in case 2 ( assuming of course that the collision time is the same). Also, the forces should be normal, but I'm not sure what you mean by quantifying inertial resistance. Increase in mass doesn't necessarily imply an increase in rigidity, shouldn't that be a property of the material used?

5. Feb 27, 2016

NRa

Exactly. That should be the property of the material and i am considering that both A and B have the same make in both cases. But where i am failing is how to quantify that inertial resistance, that increase in average force, in terms of the normal force. How do i explain this increment as a result of the increase in mass? One way to say that is perhaps, now there are more atoms in B in case 2. So greater electromagnetic forces results in stronger normal force...however, there is something not right with this point of view. It's too crude.

6. Feb 27, 2016

drvrm

Consider two particles, denoted by subscripts 1 and 2. Let m1 and m2 be the masses, u1 and u2 the velocities before collision, and v1 and v2 the velocities after collision.

The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision, and is expressed by the equation

Likewise, the conservation of the total kinetic energy is expressed by the equation

now you apply the conservation laws to the problem under consideration and putting up specific values see whether the forces operating come out to be different or same in both cases -then only one can come to 'proposal ' of change of materials rigidity.

7. Feb 27, 2016

drvrm

Is it right to say that -Whether Impulse is different in two cases of change in mass

Keeping in mind that the force as the rate of change of momentum, lets explore the concept of impulse, which is the product of a force and the time duration. We know that force is the rate of change of momentum.
If you multiply a rate of change by a duration of change, you will get an amount of change. The impulse is the product of the force and the time of action. It is the amount of momentum a certain force will produce over time.so can be measured by change of momentum of body 1 and body 2 in both the situation.

8. Feb 27, 2016

Staff: Mentor

If the objects are perfectly rigid then the force is infinite.

You have to get rid of this assumption if you want to investigate your question. The easiest thing is to consider them to be two springs each following Hookes law. Larger mass does not imply greater stiffness, but what is the consequence?

9. Feb 27, 2016

brotherStefan

You beat me to the punch. ;-)

10. Feb 28, 2016

NRa

Yes, yes i did have an inkling that this assumption mustn't be included to have the right picture of the situation.
So if we are to include the elastic set up for the two colliding objects now; thinking about spring-like forces, then how does adding more mass change the 'spring-coefficient', so to speak? Also, if in the second case, there's indeed a greater force, than the collision time should be smaller? Well, forget the time interval for now. The problem with the concept of impulse, especially in this one dimensional scenario which should give a straightforward answer, is that the instantaneous force operates in a covert manner during the collision, making the comparison of the magnitude in the two cases difficult.

Oh, and thank you to everyone who has responded. It's really encouraging.

11. Feb 28, 2016

drvrm

the forces are acting which leads to change in momentum of the bodies - so however 'covert' it is - their role can be estimated/calculated. actually that helps in knowing the forces- as in case of Rutherford alpha particle scattering the analysis led to knowledge about the forces(a new type).

12. Feb 29, 2016

NRa

I am not arguing that the change can't be calculated. I am asking how two compare the magnitudes of the forces in the two cases i have described in the original post. It's clear that the impulse in the second case is greater than the first. If i have to talk in terms of the force that resulted during those two unique collisions, how should i go about it... That's the question.

13. Feb 29, 2016

Aniruddha@94

Is it not? Isn't impulse the change in momentum after some event. If we look at only body A ( which has the same mass and same initial velocity in both the cases) isn't the change in its momentum greater in case 2 ?

14. Feb 29, 2016

NRa

If i consider two steel balls of the same mass; one at rest, other moving at some speed and colliding with the second one:

$m_A = 0.01 kg; V_A = 10 m/s; r_A = 6.68 mm$
$m_B = m_A = 0.01 kg; V_B = 0 m/s; r_B = r_A = 6.68 mm$

Kinetic energy travels at speed of sound through stainless st which is 5800 m/s. Before B starts to move, the distance the energy travels is $2 \times r_B = 0.01336 m$

Collision lasts till all the kinetic energy of A is transferred to B, so $\Delta t = \frac {2 \cdot r_B}{V_s} = 2.3 \mu s$
Change in momentum for A in case when mass of A = mass of B:

$| 0-m_A \cdot V_A | = 0.1 kgm/s$

$0.1 = F_{av} \cdot (2.3 \times 10^{-6}) \Longrightarrow F_{av}= 43413.2 N$

In second case,

$m_B = 2 \cdot m_A = 0.02 kg; V_B = 0 m/s; r_B = 8.42 mm$

This time, is it right to assume that all the kinetic energy of A is transferred to B in the same, kinetic, form?

What seems to be happening is that when A strikes B, it’s kinetic energy isn’t transferred to B in just the kinetic form; some of it is being stored as the potential energy and rest is used to put B into motion at a speed smaller than $V_A$ (actually, it’s 2/3 of $V_A$).

That stored potential energy of B then accelerates A into motion after it comes to rest during the collision. This must be the impulse on A that gives it a backward velocity.

But this doesn’t make any sense. In case 1, energy is travelling in purely kinetic form and as the mass increases, now some of the energy is stored in potential form as well??

15. Feb 29, 2016

drvrm

well i do not understand how you are devising new 'collision' set up ? if you move towards other ways of force being generated then you may not rightfully compare the two events as 'given in your original post'

16. Feb 29, 2016

Staff: Mentor

It doesn't. You can have a small mass with a spring coefficient like steel or a large mass with a spring coefficient like marshmallow.

Aack, no don't forget the time interval. In the second case you know that the impulse is greater. You must find out both the force and the time. The force is determined by using Hookes law and the time is determined from the kinematics.

17. Feb 29, 2016

Staff: Mentor

Specify the spring constants for each mass and calculate using Hookes law.