Momentum/impulse question

  • Thread starter dnt
  • Start date
In summary, the ball collides with a wall and the wall exerts a negative force of -155i N on the ball. The ball has a vertical velocity of -11i m/s after the collision and the wall has a negative impact on the ball's total change in momentum. The ball was in contact with the wall for .00619 seconds.
  • #1
dnt
238
0

Homework Statement



A .30 kg ball with initial velocity of 21i - 1.3j colldes with a wall. The wall exerts a force of -155i N on the ball and its final horizontal component of velocity is -11i m/s.

(a) Find the vertical velocity component.

(b) Find the total change in momentum.

(c) How long was the ball in contact with the wall?

Homework Equations



change in momentum = impulse = Ft

The Attempt at a Solution



I think my main problem is I am not sure if this is a conservatoin of momentum problem or not - is momentum conserved after the collision with the wall? I am having a hard time relating the 155 N force applied from the wall to the ball.

I know that if I can find the vertical component of velocity after hitting the wall, I can easily find the change in momentum (part b) by:

m (vf - vi) where vf and vi are the overall resultant velocity vectors.

and i know that if i get that i can set it equal to Ft to find the time in contact with the wall (part c) but i cannot figure out part a. can someone please help. thanks.
 
Physics news on Phys.org
  • #2
Strictly speaking momentum is conserved, but that's not very useful to solve the problem, because the change of momentum of the ball equals the change of momentum of the wall plus the earth!

You said (correctly) change of momentum = impulse = Ft. Apply that equation to the ball.

Force and momentum are vector quantities. What does the question say about the direction of the force on the ball?
 
  • #3
change in momuntum = m(change in v)

which is (.30 kg)(vf - 21.04 m/s) = (-155N)t

{i got the 21.04 m/s by using pythagorean theorem to 21i - 1.3j and found the resultant vector...correct?}

anyways, now i have two variables. i know if i get vf, then i can solve for t easily. but i cannot get vf! still stuck...

Force and momentum are vector quantities. What does the question say about the direction of the force on the ball?

since it is negative the force on the ball is in the OPPOSITE direction compared to when the ball struck the wall.
 
  • #4
doing some more thinking about it...am i getting closer:

m(vf-vi)=Ft but perhaps i should consider it in the horizontal only...

(.03)(-11 - 21) = (.155)(t)

t = .00619 seconds?

is that right?

but i still cannot figure out the final vertical velocity (part a). can i use one dimensional kinematics to find it (using vertical components only)?

vf = vi + at

vf = (1.3j) + (9.81)(.00619 sec)

vf = 1.3608j m/s

any help is much appreciated. I am really trying to get this.
 
  • #5
Your decimal points are sliding around so the answer is not right, but that is essentially it. Now what direction have you assumed the force is acting in? What does that say about vertical change in velocity?
 
  • #6
i assumed the direction was acting in the horizontal direction so it should not affect vertical velocity. however, gravity should make it increase. hence my confusion.
 
  • #7
It would. The usual line is that the collision time is short enough that the correction from gravity is not important. In this case you could make a point that it is (it's a SLOW collision). Not really the point of the exercise though I don't think...
 
  • #8
are you implying that vertical velocity isn't going to change because of such a short time period?
 
Last edited:
  • #9
No. I'm saying it will change (and in this case by a fair amount compared with vf). I am saying I don't think that correction is the point of the exercise. As for the final horizontal velocity - that was given to you as part of the problem statement. How could you 'figure it out'?
 
  • #10
ok tried to fix it:

HORIZONTAL: (.03)(-11 - 21) = (-155)(t)

t = .0619 seconds

VERTICAL:
vf = vi + at

vf = (1.3j) + (9.81)(.0619 sec)

vf = 1.907j m/s

so i solve for time BEFORE solving for final vertical velocity? is this correct?
 
  • #11
Dick said:
No. I'm saying it will change (and in this case by a fair amount compared with vf). I am saying I don't think that correction is the point of the exercise. As for the final horizontal velocity - that was given to you as part of the problem statement. How could you 'figure it out'?

i know that's why i edited it out - i overthought the problem and i confused myself! i then realized what i typed and took it out. sorry.
 
  • #12
Looks fine. Make it clear that vf has been corrected for gravitational acceleration during the collision. That may not be part of the expected answer.
 
  • #13
you mean for the 1.907 m/s part? how do i make it clear? isn't showing my work sufficient?

btw thanks for the help!
 
  • #14
Sure. Clearly showing work is sufficient. You're welcome - but I didn't help that much.
 

1. What is momentum and how is it related to impulse?

Momentum is a measure of an object's motion, calculated by multiplying its mass and velocity. Impulse, on the other hand, is the change in an object's momentum over time. They are directly related, as an object's change in momentum is equal to the impulse applied to it.

2. How is momentum conserved in a closed system?

In a closed system, the total momentum remains constant. This means that the momentum of all objects within the system before and after any interaction or collision will be equal. This is known as the law of conservation of momentum.

3. Can the impulse applied to an object change its direction of motion?

Yes, the impulse applied to an object can change its direction of motion. This is because impulse is a vector quantity, meaning it has both magnitude and direction. A change in direction of the impulse will result in a change in direction of the object's momentum.

4. How does mass affect an object's momentum and impulse?

Mass is directly proportional to an object's momentum and impulse. This means that as an object's mass increases, so does its momentum and the amount of impulse needed to change its momentum. This is because it takes more force to change the motion of a heavier object.

5. Can an object have momentum without having any mass?

No, an object cannot have momentum without having any mass. Momentum is calculated by multiplying an object's mass by its velocity, so an object with zero mass would also have zero momentum. Additionally, objects without mass, such as photons, do not have velocity in the traditional sense and therefore do not have momentum.

Similar threads

Replies
1
Views
487
  • Introductory Physics Homework Help
Replies
19
Views
950
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
879
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
454
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
3
Views
3K
Back
Top