What is the force imparted by the bullet on the block

Click For Summary

Homework Help Overview

The problem involves a bullet with a mass of 0.045 kg traveling at 560 m/s colliding with a stationary block of mass 0.5 kg. The collision duration is given as 0.35 seconds, and the question seeks to determine the force imparted by the bullet on the block.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of force using the formula F=(m(vf-vi))/t, with initial attempts yielding a negative force value. There is uncertainty regarding the interpretation of this value and its implications based on Newton's 3rd law.

Discussion Status

Participants are exploring various interpretations of the problem, including the implications of the block's mass and its motion post-collision. Some have suggested using conservation of momentum to find the final velocity of the block and bullet combination, leading to further calculations of force.

Contextual Notes

There is a noted ambiguity in the problem statement regarding whether the block remains stationary after the collision, and some participants question the phrasing of the question itself, suggesting it could be clearer about the block's ability to move.

hamza2095
Messages
28
Reaction score
1

Homework Statement


A bullet with a mass of 0.045 kg is shot at 560m/s fwd, and it gets dislodged in a stationary block that is 0.5kg.
The collision lasts 0.35 seconds. What is the force imparted by the bullet on the block?

Homework Equations


F=(m(vf-vi))/t

The Attempt at a Solution


F=(0.045kg(0-560m/s))/0.35s
F=-72N [FWD]

It doesn't make sense that the force the bullet has on the block is backwards 72N[BWD], so I'm guessing that's the force of the block on the bullet, therefore using Newton's 3rd law which if F Action= -F reaction i get the force of the bullet on the block is 72N [FWD]

Is this right, shouldn't the initial calculation be the final answer for the question? Help is appreciated
 
Physics news on Phys.org
hamza2095 said:

Homework Statement


A bullet with a mass of 0.045 kg is shot at 560m/s fwd, and it gets dislodged in a stationary block that is 0.5kg.
The collision lasts 0.35 seconds. What is the force imparted by the bullet on the block?

Homework Equations


F=(m(vf-vi))/t

The Attempt at a Solution


F=(0.045kg(0-560m/s))/0.35s
F=-72N [FWD]

It doesn't make sense that the force the bullet has on the block is backwards 72N[BWD], so I'm guessing that's the force of the block on the bullet, therefore using Newton's 3rd law which if F Action= -F reaction i get the force of the bullet on the block is 72N [FWD]

Is this right, shouldn't the initial calculation be the final answer for the question? Help is appreciated
Are you sure the "stationary block" remains stationary after the collision?

To me, the fact that the problem statement mentions the 0.5 kg mass of the block is a hint that the bullet+block combination has a non-zero velocity after the collision.
 
collinsmark said:
Are you sure the "stationary block" remains stationary after the collision?

To me, the fact that the problem statement mentions the 0.5 kg mass of the block is a hint that the bullet+block combination has a non-zero velocity after the collision.
I get 46.2m/s (fwd) for the velocity of the block and bullet using conservation of momentum, and with that i still get -66N fwd. But I think you use the velocity of the bullet relative to the block, which is still 0 because it's inside of it.
 
hamza2095 said:
I get 46.2m/s (fwd) for the velocity of the block and bullet using conservation of momentum, and with that i still get -66N fwd.

That sounds about right. :smile: The force applied to the bullet.

But I think you use the velocity of the bullet relative to the block, which is still 0 because it's inside of it.
You'll need to ensure that your frame of reference is the same before and after the collision.

You can't use a stationary frame before the collision, and then switch to a moving frame after, when calculating velocities and momentum values.
 
Last edited:
As an aside, the wording of the question could be better. It should make clear that the block is free to move, and it should not ask for "the force imparted". You can only calculate an average force, and forces are not imparted. Imparting implies transfer of some state, such as energy or momentum. Forces are applied or exerted.
 
collinsmark said:
You'll need to ensure that your frame of reference is the same before and after the collision.

You can't use a stationary frame before the collision, and then switch to a moving frame after, when calculating velocities and momentum values.
Okay I see what your saying, so using conservation of momentum i get that both the block and bullet are going 46.2m/s FWD. If i use that final velocity to find force I get -66 N [FWD], which is still a negative.
 
hamza2095 said:
Okay I see what your saying, so using conservation of momentum i get that both the block and bullet are going 46.2m/s FWD. If i use that final velocity to find force I get -66 N [FWD], which is still a negative.
You calculated the average force that slowed the bullet. That is therefore the force exerted on the bullet, but the question asks for the force the bullet exerted on the block.
 
haruspex said:
As an aside, the wording of the question could be better. It should make clear that the block is free to move, and it should not ask for "the force imparted". You can only calculate an average force, and forces are not imparted. Imparting implies transfer of some state, such as energy or momentum. Forces are applied or exerted.
So according to Newton's 3rd law the average force the bullet exerted on the block should be 66N [FWD], correct?
 
hamza2095 said:
So according to Newton's 3rd law the average force the bullet exerted on the block should be 66N [FWD], correct?
Yes.
 
  • Like
Likes   Reactions: hamza2095

Similar threads

  • · Replies 9 ·
Replies
9
Views
7K
Replies
14
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
3K
Replies
1
Views
2K
Replies
25
Views
3K
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K