Linear Impulse and Momentum Question

[whateverhello]

For example, consider a bullet (moving horizontally) that impacts a block (at rest) on an incline.

From what I know,
momentum1 + impulse = momentum2

I have the solution to the problem, and it says (for the impact),
(mass of bullet)(speed of bullet) + 0 = (mass of bullet+block)(speed of bullet+block)



I don't understand why the impulse is 0 though. According to the video above, which poses a similar problem, he says since we're talking x-direction, no need to worry about the gravity, and normal forces. But in this problem, since it's on an incline, shouldn't the x-component of the gravity be considered? (Note, the positive x-direction is taken to be along the ramp, so gravity considered would be mgsinθ. Is it because the impact time is so minuscule that FΔt = 0?

 

[A.T.]

block (at rest) ...I don't understand why the impulse is 0

Because the block is at rest.

 

[whateverhello]

Because the block is at rest.

Can someone elaborate on what he means, thanks. Judging from his reply, I must be missing something very basic here. This impulse stuff in general is giving me problems.

 

[nasu]

When we study collisions, in the first approximation it is assumed that the system of colliding objects is isolated, i.e. the resultant of the external forces is zero. In tehse conditions we can use conservation of momentum.
The reason this assumption works pretty well in many real situations is that the collision time is very short so that the impulse of external forces is very small when compared with the impulse of the large collision forces.

You don't have to assume zero impulse. But then you will need to know the duration of collision to count it in.

 

[whateverhello]

When we study collisions, in the first approximation it is assumed that the system of colliding objects is isolated, i.e. the resultant of the external forces is zero. In tehse conditions we can use conservation of momentum.
The reason this assumption works pretty well in many real situations is that the collision time is very short so that the impulse of external forces is very small when compared with the impulse of the large collision forces.

You don't have to assume zero impulse. But then you will need to know the duration of collision to count it in.

Ahhhhh, ok. So this is where whether a force is impulsive or non-impulsive comes into play right? Follow-up question, so I know weight and spring forces are non-impulsive, but what about normal forces? For example if the block were pushed up against a wall so that it won't move no matter what, how would the impulse equation look like?

 

[A.T.]

Can someone elaborate on what he means, thanks.

The 0 in your solution is the initial momentum of the block at rest.

 

[whateverhello]

The 0 in your solution is the initial momentum of the block at rest.

Uhhh no? It's the impulse of the external forces.

 

[A.T.]

Uhhh no? It's the impulse of the external forces.

It could be either, but if that's what the solution says.

 

[person_random_normal]

I don't understand why the impulse is 0 though.

Hey man,
the impulse is not zero !
two impulses of equal magnitudes are acting in opposite directions ,on the bullet and another on the block, so net impulse acting on the system is zero !
[BUT MIND THAT - I'm only talking about the situation for the direction along the incline , for direction perpendicular to it the normal force exerted on the bullet by the incline will destroy the corresponding component of the velocity of that bullet ! ]

I'm assuming that the incline is fixed