Newton's Third Law of a bullet

  • #1
Yesterday, one of my cousins asked me about Newton's third law. In this discussion he asked me this question: Why does not a bullet bounce back after striking the surface of glass as it does in the case of steel, considering that in both the cases the force with which it strikes the glass or steel is equal to the force applied by the glass or steel to the bullet?
I gave him a long explaination but he was not convinced (even I wasnt convinced :biggrin: ). At night when I went to the bed I thought about my explaination and found that it was greatly flawed. Can anyone please explain this to me?

Secondly, How can I find the force with which a bullet of mass m kg strikes a surface of an object with the constant velocity of v m/s? I know the kinetic energy at that instant will be (1/2)mv^2. Dont know where to go from here.

Thanks alot in advance. :smile:
Abd. S.
 
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Answers and Replies

  • #2
Tide
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Your example is a little complicated due to the shattering the glass and deformation of the bullet. However, the bullet does definitely bounce back! If you were travelling along at the same velocity as the bullet is initially travelling then you would observe the bullet head off in the reverse direction when it strikes the glass (YOU keep travelling in the forward direction). The change in momentum of the bullet is equal and opposite the change in momentum of the glass.

As to the Third Law, force is the rate of change of momentum. The change in momentum of the two objects are equal and opposite. Obviously, they interact with each other for precisely the same interval of time. Therefore, they exert equal and opposite forces on each other.
 
  • #3
PerennialII
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So essentially the point is in the inability of the 3rd law & Newtonian mech to include any information beyond rigid objects (unless we implement them when it gets "a bit more" complicated).
 
  • #4
First of all thanks for your replies.
So what do you guys think I should say to my cousin? I just have to say something to him. He is stuck on one point: Newtons third law is not true. I gave him other examples but he says that prove it in this case and he will believe me. :mad:
IRT Tide:
I dont think it will move in reverse direction. It will just slow down a bit. The difference in our velocity will make the difference between us larger and larger but I really dont think that it'll move in reverse direction. :confused:

Any thoughts on me second question? I'll be much pleased if someone can offer any kind of help.
Regards
Abd. S.
 
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  • #5
PerennialII
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With respect to your second q, the concept of an impulse is a good thing to have around when solving these sorts of problems :

http://www.glenbrook.k12.il.us/gbssci/phys/Class/momentum/u4l1b.html#impulse [Broken]

IRT Tide:
I dont think it will move in reverse direction. It will just slow down a bit. The difference in our velocity will make the difference between us larger and larger but I really dont think that it'll move in reverse direction.

Don't understand it either or see the point of it.

The answer I'd give to your cousin is that if the bullet obeys Newton's laws such that it sees the wall as being rigid, the bullet does not have any idea whether the wall is made of glass or steel. It will ricochet because the wall, by definition, is rigid (=impenetrable). In reality this is not a case, but in order to account for the real material behavior occurring as a result of the impact, you would have to account for 1) deformations of both components and 2) fracture and damage to both components [steel would undergo deformation, glass would shatter to million pieces]. They are occurrences which obey Newton's laws, but are not included (=implemented) in the simplistic way the problem here is treated (=would require an analysis of completely different order of complexity).
 
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  • #6
Doc Al
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same force -- different objects

DeathKnight said:
Yesterday, one of my cousins asked me about Newton's third law. In this discussion he asked me this question: Why does not a bullet bounce back after striking the surface of glass as it does in the case of steel, considering that in both the cases the force with which it strikes the glass or steel is equal to the force applied by the glass or steel to the bullet?
I gave him a long explaination but he was not convinced (even I wasnt convinced :biggrin: ). At night when I went to the bed I thought about my explaination and found that it was greatly flawed. Can anyone please explain this to me?
Newton's 3rd law merely says that whatever force the bullet exerts on the glass will be exactly equal and opposite to the force the glass exerts on the bullet. But the same force acting on two different objects will not have the same effect! A force F acting on a piece of glass shatters it, but that same force F acting on a speeding bullet just slows it down a bit and deforms it.

As Tide explains, the change in momentum of bullet and glass will be equal and opposite. But the bullet has a large initial momentum and thus just slows a bit. (What exactly happens to the bullet depends on the details of the collision.)

Secondly, How can I find the force with which a bullet of mass m kg strikes a surface of an object with the constant velocity of v m/s? I know the kinetic energy at that instant will be (1/2)mv^2. Dont know where to go from here.
The force that a bullet exerts on something that it strikes depends on too many variables for it to be calculated. For example, you can imagine that the force would be much less if the bullet shot through a pile of fluffy pillows than it would if it were shot into a steel wall.
 
  • #7
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DeathKnight said:
Why does not a bullet bounce back after striking the surface of glass as it does in the case of steel, considering that in both the cases the force with which it strikes the glass or steel is equal to the force applied by the glass or steel to the bullet?
DeathKnight, the total impulse that the bullet applies on the glass is smaller than the total impulse that the bullet applies on steel. It is also the case for the average force.
 
  • #8
krab
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DeathKnight said:
Secondly, How can I find the force with which a bullet of mass m kg strikes a surface of an object with the constant velocity of v m/s? I know the kinetic energy at that instant will be (1/2)mv^2. Dont know where to go from here.
F=ma, so you cannot find the force unless you know the bullet acceleration (it will be negative, or deceleration in this case). The velocity does not help you unless you know the difference in velocity before and after collision, AND either the time increment over which the collision occurred, or the distance the bullet traveled while in collision.
 
  • #9
Tide
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DeathKnight said:
I dont think it will move in reverse direction. It will just slow down a bit.

It will definitely move in the reverse direction! I specified that you were moving in a frame of reference that has the same velocity as the initial velocity of the bullet. YOUR velocity doesn't change when the bullet impacts the glass. Obviously, to a "stationary" observer, the bullet simply slows down a bit but that was not the situation I desribed.
 
  • #10
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Comment: the yield strength of one kind of glass (probably typical, but I can't say for sure) is about 3.4 ksi (3400 psi/23 MPa). The yield strength of mild steel is around 40 ksi, and is over 100 ksi for many alloy steels. Glass breaks a lot more easily than steel...;)
 
  • #11
PerennialII
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It will definitely move in the reverse direction! I specified that you were moving in a frame of reference that has the same velocity as the initial velocity of the bullet. YOUR velocity doesn't change when the bullet impacts the glass. Obviously, to a "stationary" observer, the bullet simply slows down a bit but that was not the situation I desribed.

Yeah, sounds very reasonable, but I see it as a somewhat unnecessary construct for explaining the problem at hand.
 
  • #12
Tide
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PerennialII said:
Yeah, sounds very reasonable, but I see it as a somewhat unnecessary construct for explaining the problem at hand.

I think it cuts to the heart of the matter. YMMV.
 
  • #13
PerennialII
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I think it cuts to the heart of the matter. YMMV.

I can see it as an analytical explanation about the event. Taking a look of the thread overall I think the matter was explained, although the Newtonian aspects and what actually transpires in the collision may be somewhat difficult to parse.
 
  • #14
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DeathKnight said:
Yesterday, one of my cousins asked me about Newton's third law. In this discussion he asked me this question: Why does not a bullet bounce back after striking the surface of glass as it does in the case of steel, considering that in both the cases the force with which it strikes the glass or steel is equal to the force applied by the glass or steel to the bullet?.

The bullet exerts a force on the glass or steel wall and there is an opposing force of the wall on the bullet. This opposing force is the result of forces between the atoms that make up the wall. If these forces are not strong enough the wall will not stay in one piece and bits of the wall will move, which will slow the bullet down.

In a glass wall the forces between the atoms are less strong than in a steel wall. In the case of a glass wall the result of the impact will be a slowed down bullet and flying pieces of glass. In the case of the steel wall the result of the impact will be a reverse in the motion direction of the bullet. However, the total momentum never changes: The slowed down bullet plus the moving pieces of glass have the same momentum that the bullet had before its impact with the wall. Similarly, the bullet that has changed direction after collision with the steel wall has the same momentum that it had before its impact with the wall

The difference between the path of a bullet hitting a steel wall and the path of a bullet hitting a glass wall is caused by the differences between the forces holding together a steel wall and the forces holding together a glass wall.
 
  • #15
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Ok, this question is fairly simple...lets see if I can make any sense when I try to explain it. Ok, so the bullet and the glass is only one example, don't get tripped up on this because it's a complicated example. Try this one,

try kicking a football with your bare foot...what happens? depends on how hard you kick it...if you kick it really hard it will most likely hurt your foot. Why? because at the moment of contact between the ball and your foot Newtons 3rd law applies...the ball applies as much force to your foot as you apply to the ball...if this were not the case then why is your foot red, or possibly bleeding if you kick it hard enough? Newtons 3rd law does not rely on what happens after the collision...it only applies during the instant of the collision. Everything that happens after the collision are based on different principles. what are the masses of the two objects? Is it a perfectly elastic collision? Is it a perfectly inelastic collision? What what the initial velocity of the ball and your foot...if the football were flung hard enough at your foot it would be your foot that moves backward, but your foot would still apply equal force at the moment of the collision

So the glass does apply the same ammount of force to the bullet at the instant of the collision...then due to forces such as initial velocity, mass, and the elastic/inelastic properties of the two objects the bullet is able to penetrate the glass. Your cousin might want to try looking into the laws of momentum and collisions, it seems to me that those are the areas he is confused about. Hope this helped.
 

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