# Force when blocks collide

1. Aug 29, 2015

### Ocata

Suppose a 10 kg block is set on a frictionless surface of infinite length. I then apply a constant force in the form of a constant push such that the block accelerates at a constant rate of 1 m/s^2.

The force I apply will be F=ma= 10kg*1m/s^2= 10N.

Suppose I apply this force for 10 seconds. Then at=v = (1m/s^2)(10s)= 10m/s.

I am applying 10N of force for the duration of the 10 second push. So the block is experiencing a net force of 10N.

I stop pushing the block at exactly 10 seconds so that the block is traveling at a constant velocity of 10m/s.

Now, suppose there is a second block of 100kg on the frictionless surface set at rest directly in the path of the block in motion.

When the moving block hits the stationary block, what forces occur?

All I know I that the force I applied to the block was 10N and the net force on the block after 10 seconds was 0N.

How does force fit into the scenario upon impact and after impact?

2. Aug 29, 2015

### Ocata

What I am trying to ask is can I say something like, "block A hit block B with x amount of force?"

If my hand pushing a block causes a force, it is not my hand that "contains" a force, but rather my hand that "causes a force when it causes an acceleration of the block.

So I would say the block in motion at a constant velocity does not in itself contain a force, but a force occurs when the two blocks interact in some way that causes at least on of their velocities to change.

So maybe I want to ask, how can I calculate the force that the 10kg block in motion will have on the 100kg stationary block?

3. Aug 29, 2015

### Staff: Mentor

Equal and Opposite forces act.

The result is that the second block gains whatever momentum the first block loses.

4. Aug 29, 2015

### Staff: Mentor

It is possible to calculate the average force exerted---but you'll need additional information, such as the length of time that the bodies are in contact.

5. Aug 29, 2015

### CWatters

As Nascent Oqygen said.. You need to know more about the type of collision that occurs.

Do you really need to know the force or would it be sufficient to know how fast the two blocks end up going and in what directions? Sometimes that's easier to work out.

6. Aug 29, 2015

### Ocata

So there is a time element in there somewhere that needs to be known first?

For instance, the only reason I know how fast the 10kg block is going is because I specified for how long I applied a force.

So if I say: A 10kg block travelling at a velocity of 10m/s strikes a 100kg block and maintains contact for 2 seconds. Then the 10kg block applied a force onto the 100kg for some amount of time. I'm not sure how long the force of one block would be applied to the other block but I would imagine the force would be applied for however much time the 100kg block is in a state of acceleration. Ah! perhaps that's the key! Maybe I need to know how much acceleration occurs in the 100kg block due to the collision from the 10kg block at 10m/s.

If I know the acceleration, then I will know the force because I already know the mass.

Hmm...the chicken and the egg... If I can find the acceleration of the 100kg block during impact, then I can calculate force via f=ma. But if I can find the duration of the strike and final velocity of the 100kg block, I can find the force via f=m(v/t). But how do I find either or any of these?

Maybe its just impossible?

Maybe I should employ the conservation of momentum?

mv + mv = mv + mv

10(10) + 100(0) = 10(v) + 100(v)
100 = 1000v
1/10(m/s) = v

Hmm, I don't think .10 m/s tells me much because how do I know how long it took to reach that speed?

I feel like I'm chasing my tail here lol

7. Aug 29, 2015

### Ocata

Hi CWatters,

Please note: I took a while to write the previous post because I had to take some phone calls in between so the previous posted was posted before I read your previous.

To answer your question, I would like to know both actually. Velocity and force.

"Force of impact" seems to be a common phrase. In most instances where collisions/impacts/strikes etc occur, the first aspect that seems to be communicated is force of impact. One one thing hits on other thing, it seems more natural to wonder about the force of the impact.

If a car is being tested for its crash safety, do they car about the final speed or the change of speed over time? Hence the acceleration of the person inside the vehicle?

When they measure Mike Tyson's punch on the discovery channel, they didnt measure the velocity of the bag after impact. They measured his force on the punching bag..etc.

So it seems interesting to me to know about the force as well as the final velocity.

Thanks

Last edited: Aug 29, 2015
8. Aug 29, 2015

### Ocata

Oh okay, so i have to set a few parameters first. I was beginning to think along these lines.

So I'll restate the problem with a little faster final velocity and a contact duration of 2 seconds.

A 10kg block is pushed with a force of 100N for 100 seconds. F/m = a = 100/10 = 10 m/s^2
at = v = 10(100) = 1000 m/s

mv + mv = mv + mv

10(1000) + 100(0) = 10(v) + 100(v)
10000 = 1000v
10(m/s) = v

Ft = mv
F = mv/t
F = (100kg)(10m/s)/(2s) = 500N

So the Force of the 10kg block on the 100kg block is 500N?

Is this at least in the right ballpark as far as thinking about it?

9. Aug 29, 2015

### Ocata

Hi can someone please provide some guidance. Much appreciated.

10. Aug 29, 2015

### Staff: Mentor

That is not the way to write a conservation of momentum equation! It requires subscripts.

You are assuming that the pair stick together and have a common velocity?

11. Aug 29, 2015

### Ocata

My apologies, didnt have access to a desktop computer at the time i was writing the formula. I work outside and my only access on the weekends is through my phone. I thought it would be more convenient and that the idea of what goes where would make sense. I will fix the format.

12. Aug 29, 2015

### Ocata

$m_{1}v_{1a}+m_{2}v_{1b} = m_{1}v_{2a}+m_{2}v_{2b}$

Where $v_{2a} = v_{2b} = v_{f}$

$10kg(100m/s)+ 100kg(0) = 10kg(v_{f})+100kg(v_{f})$

v_{f}= 10m/s

Then F= ma = 100kg(10m/s)/(2s)

F= 500N

Yes, for a certain amount of time, I would assume they both would have to share a velocity in the same way if i was pushing a block with some amount of force, my hand and the block would be traveling at the same velocity during the duration of the applied force. Is this a correct assumption?

Last edited: Aug 29, 2015
13. Aug 29, 2015

### Staff: Mentor

In a perfectly inelastic collision, the two blocks would end up traveling at the same velocity after the collision. In any other type of collision, the two blocks will be traveling the same velocity for an instant as the first first block slows down and the second block speeds up.

14. Aug 29, 2015

### Staff: Mentor

Usually you will be given information relating directly or indirectly to K.E., such as assume a perfectly elastic collision, or that the first body stops dead, or it rebounds with a particular speed, etc.

15. Aug 29, 2015

### Ocata

Okay so for a perfectly inelastic collision, the duration of contact would be indefinitely, but surely an actual force (change in velocity) could only occur for a limited time. Not sure what time frame that would be or if that time frame varies depending on mass and velocity upon impact.

For a perfectly elastic collision, what time frame is considered an instant? Is it measurable? Is it like .00000001 seconds?

16. Aug 29, 2015

### Ocata

If we were to assume a perfectly elastic collision, given the scenario described above which was

A 10kg object traveling at 100m/s collides with a 100kg stationary object (such that all motion remains linear) and the collision is perfectly elastic. What is the force with which the 10kg object strikes (or applies) to the 100kg object?

Would the calculation change? Assuming my calculation above even makes sense to begin with..

17. Aug 30, 2015

### Staff: Mentor

10 + 100 = 110

18. Aug 30, 2015

### Staff: Mentor

The length of time the bodies interact is not related to how well energy is conserved. Two solid quartz balls might clash for 0.01 secs, while model train carriages of similar weight fitted with spring buffers might interact for over 1 sec, yet in each case the collision may be near perfectly elastic.

19. Aug 30, 2015

### Nokx

The calculation would change, because you would have to find the final velocities such that you not only conserve momentum but also kinetic energy (KE=mv^2/2).
Also, force varies over time in any collision. Using Ft=mv will give you an average force during the duration t of the collision the objects, thus you need to know how long the two objects were in contact. This will not give the maximum force, only the average during the period of collision.

20. Aug 30, 2015

### CWatters

Indeed it's very common for Newspaper to write "the force of the impact damaged/destroyed..." but that's not really good physics. It would be better to say the energy of the impact damaged/destroyed... You can have a very large force acting on an object that does no work on it. A high rise building for example exerts very large forces on the ground under it but you would hope it does no work on it.

Consider hammering a post into the ground. You could hammer it in with one large blow from a pile driver, a dozen small blows from a sledge hammer or a steady push using the weight of the bucket on a JCB. It might be reasonable to imagine that no matter how you do it the energy required to get it in the ground would be a constant that depends on things like the hardness of the ground, the diameter of the post and how pointed or blunt it is. However the peak force may well depend on the method used.

Sure you can measure the force one object applies on another but what do they actually mean by his punching force? For example when his glove first makes contact the force will be very low, then it will rise to some peak value as the glove and bag are compressed, then it will fall again as his arm reaches maximum extension. So what do they quote? Peak force? Average? The peak force will probably depend on the hardness of the gloves and punch bag. Measuring and comparing the force between boxers only makes any sense if the gloves and the bag are the same. You could probably register a higher peak force by using a concrete punch bag and concrete glove. The duration of the force would probably be shorter though.