Finding the correct change of momentum in this system,

[HazyMan]

Homework Statement

A box of mass M rests on a flat surface. A bullet of mass m and velocity u moves in a straight line and collides with the box and it gets stuck in it, FORMING ONE NEW BODY. What's the force that the bullet applies to the box?

Relevant Equations

F=Δp/Δt

I tried out the $$F=Δp/Δt$$ equation and i came up with a change of momentum of the box-bullet fusion $$(M+m)V-0$$, but the textbook says otherwise (using a change of momentum of JUST the box itself, excluding the bullet). According to the textbook, the correct change is $$MV-0$$, without the added mass of the bullet being in the expression. Which of the two is really correct and why? I am supposing that the momentum the box gains is the momentum it gains considering the fact that it's fused with the bullet, while the book seems to ignore that the bullet is stuck in the box, and acts as if the box's gained momentum is JUST ITS OWN MASS times its velocity. I could really use a hand in this situation, thanks for reading!

 

[haruspex]

What's the force that the bullet applies to the box?

Impossible to say without knowing the time taken for the two velocities to become equal.

i came up with a change of momentum

Change in momentum of what, exactly? If you want the change in momentum of bullet+box then you need to take into account the bullet's original momentum.

without the added mass of the bullet being in the equation

What equation? So far you have only listed expessions.

 

[HazyMan]

Impossible to say without knowing the time taken for the two velocities to become equal.

Change in momentum of what, exactly? If you want the change in momentum of bullet+box then you need to take into account the bullet's original momentum.

What equation? So far you have only listed expessions.

Edits done. Hopefully things are more clear now.

 

[HazyMan]

Impossible to say without knowing the time taken for the two velocities to become equal.

Now about the time taken for the two velocities to become equal, what do you mean? One thing i did not include in the problem (Because i thought it was not required) was that the time taken for the bullet to lose its speed after the collision is 0.01s.

 

[jbriggs444]

Now about the time taken for the two velocities to become equal, what do you mean? One thing i did not include in the problem (Because i thought it was not required) was that the time taken for the bullet to lose its speed after the collision is 0.01s.

You have a starting velocity of the bullet.
You can calculate an ending velocity of the bullet.
You know the time taken for the bullet to accelerate from the one velocity to the other.
Can you calculate the bullet's average acceleration for the 0.01 s that it is accelerating?

 

[HazyMan]

You have a starting velocity of the bullet.
You can calculate an ending velocity of the bullet.
You know the time taken for the bullet to accelerate from the one velocity to the other.
Can you calculate the bullet's average acceleration for the 0.01 s that it is accelerating?

it should be equal to $$a=Δu/0.01$$. The given initial velocity in the book problem is 200m/s and i'd say the ending velocity is 0 since it collides onto the box, so the acceleration should be $$a=-200/0.01$$, but i think I'm missing something regarding the ending velocity..

 

[haruspex]

change of momentum of the box-bullet fusion

What is the initial momentum of the box-bullet system? It is not 0.

the textbook says otherwise (using a change of momentum of JUST the box itself
...
Which of the two is really correct

Change in momentum of the box-bullet system is different from that of the box alone, so you can have different expressions, both being correct (but yours for the box-bullet system is not).
So probably you mean which is useful in solving the problem?
Since the question asks for the force the bullet applies to the box you need to keep them separate. If you treat them as a combined system that would become an internal force and disappear from the equations.

One thing i did not include in the problem (Because i thought it was not required)

Without that, how were you going to use
Relevant Equations: F=Δp/Δt
?

 

[HazyMan]

What is the initial momentum of the box-bullet system? It is not 0.

That's what i thought later on. What still kept me scratching my head was that despite the fact that $$MV−0$$ makes more sense considering that the union has no initial momentum, the box itself had no initial velocity so multiplying the box's mass with the velocity of the union didn't quite exactly make much sense to me.

Without that, how were you going to use
Relevant Equations: F=Δp/Δt?

Well i was mainly focusing on the Δp aspect only, which was what had me confused while Δt didn't bother me.

Change in momentum of the box-bullet system is different from that of the box alone, so you can have different expressions, both being correct (but yours for the box-bullet system is not).
So probably you mean which is useful in solving the problem?
Since the question asks for the force the bullet applies to the box you need to keep them separate. If you treat them as a combined system that would become an internal force and disappear from the equations.

It just didn't add up to my head that the mass of the box itself is being multiplied by the velocity of the bullet-box system. I guess what makes it add up is that there is no way the box itself could have moved without the bullet's force applied onto it but still, the box by itself and the bullet-box system definitely differ in mass...

 

[haruspex]

the union has no initial momentum

Yes it does. The initial momentum of the box-bullet system is the sum of the momenta of its parts. The box part has no initial momentum but the bullet part does.

the mass of the box itself is being multiplied by the velocity of the bullet-box system

The mass of the box is multiplied by its own velocity. After coalescence, that just happens to be the same as that of the bullet and hence of the box-bullet system.

 

[HazyMan]

Yes it does. The initial momentum of the box-bullet system is the sum of the momenta of its parts. The box part has no initial momentum but the bullet part does.

The mass of the box is multiplied by its own velocity. After coalescence, that just happens to be the same as that of the bullet and hence of the box-bullet system.

Alright, i think i understand. Thanks for the help! That final sentence definitely put an end in my problem

 

[PeroK]

One thing i did not include in the problem (Because i thought it was not required) was that the time taken for the bullet to lose its speed after the collision is 0.01s.

This is why you call yourself "hazy man"?

 

[HazyMan]

This is why you call yourself "hazy man"?

...what?

 

[PeroK]

...what?

A bit hazy on the details!

 

[HazyMan]

A bit hazy on the details!

aaahhh! Well, no. The nickname HazyMan is related to an old favourite game of mine