Impulse on ball involving angles

Click For Summary

Homework Help Overview

The problem involves a ball of mass 0.40 kg that is initially moving to the left and is then kicked at an angle of 45 degrees upwards to the right. Participants are tasked with finding the impulse and average force resulting from this interaction, given a specific collision time.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to treat momentum and velocity as vectors, suggesting the breakdown of these vectors into components to account for the angles involved. There is confusion regarding how to incorporate the angles into the calculations of momentum and impulse.

Discussion Status

Some participants have made progress in calculating the change in velocity components and are exploring how to find the total change in velocity using Pythagorean theorem. Guidance has been offered regarding the correct approach to vector addition and the calculation of impulse, though there is still some uncertainty about the average force calculation.

Contextual Notes

Participants mention a time constraint for submission and express concerns about the complexity of the problem relative to its marking weight. There is also a reference to personal circumstances affecting their ability to focus on the problem.

AdamNailor
Messages
22
Reaction score
0
1. A ball of mass 0.40kg is moving to the left at a velocity v0=-20 m/s. It is kicked and given a velocity, at 45 degrees upwards to the right, of vf30[/sub] m/s.

1) Find the impulse of the force

2) Find the average force

The collision time is 0.010s



2. I = (delta) P

I = F (delta)t



3.

1) before p = 0.4x-20= -8
after 30x0.4=12

Change in p = 20Ns Where do the angles some in?

2) F=(delta)P/t=2000

Which is obviously wrong. They didn't include the angles for fun, so there something wrong in my working. Do I have to choose a line to work out the momentum across? ie, linear momentum. A bit confused. Any help appreciated.
 
Last edited:
Physics news on Phys.org
Momentum and velocity are vectors and must be treated as such.

[tex]\Delta \vec{v} = \vec{v}_f - \vec{v}_i[/tex]

Hint: Break the vectors into components, then subtract them. That's where the angles come in.
 
Doc Al said:
Momentum and velocity are vectors and must be treated as such.

[tex]\Delta \vec{v} = \vec{v}_f - \vec{v}_i[/tex]

Hint: Break the vectors into components, then subtract them. That's where the angles come in.

OK, so, 30cos45=21.2

21.2-(-20)=41.2

correct?
 
AdamNailor said:
OK, so, 30cos45=21.2

21.2-(-20)=41.2

correct?
So far, so good. You've correctly found the change in the horizontal component of the velocity. What about the vertical component? (Then find the magnitude of the total change in velocity.)
 
Doc Al said:
So far, so good. You've correctly found the change in the horizontal component of the velocity. What about the vertical component? (Then find the magnitude of the total change in velocity.)

well, sin 45 and cos45 are the same, so the vertical change is just sin45=21.2

So the total change in velocity is the total of the two, 21.2+41.2=62.4

So no to find the change in momentum I times 62.4 by the mass to get 24.96 ? which is the impulse?
 
AdamNailor said:
well, sin 45 and cos45 are the same. so, the change is just sin45=21.2
OK, the change in the vertical component of velocity is 21.2 - 0 = 21.2.
So the total change in velocity is the total of the two, 21.2+41.2=62.4
No, you can't just add horizontal and vertical components of a vector. How do you find the magnitude of a vector from its components?
 
Doc Al said:
No, you can't just add horizontal and vertical components of a vector. How do you find the magnitude of a vector from its components?

Pythagorus. So [tex]\sqrt{41.2^2+21.2^2}=46.3[/tex]

so I = 46.3x0.4 = 18.5
 
AdamNailor said:
Pythagorus. So [tex]\sqrt{41.2^2+21.2^2}=46.3[/tex]

so I = 46.3x0.4 = 18.5
Good. (Impulse will have the same units as momentum, thus kg-m/s.)
 
And my working out tghe average force will be correct with this new value? ie,

F=18.5/0.01=1850N.

Thanks for you help btw. This has got to be handed in in 20 minutes! Worth 25 marks too. Silly really for how easy it is, once you've got started, I'm just too hungover to think properly and needed a nudge to get going. Not even sure there are 25 stages in the question to mark.
 
Last edited:
  • #10
AdamNailor said:
And my working out tghe average force will be correct with this new value? ie,

F=18.5/0.01=1850N.
Looks good.
 

Similar threads

Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
14
Views
3K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
10
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
9K