More complex problem involving momentum

In summary, in this problem, a 0.55 kg block of ice initially sliding at 3.5 m/s is kicked by a person causing it to slide at an angle of 28 degrees from its original direction. The person's foot was in contact with the block for 0.0025 seconds and the average force applied was 409.4 N. Using the equations for momentum, a unit vector in the direction of the block's momentum after the kick, the x-component of the block's momentum after the kick, the magnitude of the block's momentum after the kick, and the z-component of the block's momentum after the kick were all solved for. However, there were difficulties in finding the z-component of the unit vector
  • #1
PolyFX
31
0

Homework Statement


A 0.55 kg block of ice is sliding by you on a very slippery floor at 3.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.0025 seconds. The block eventually slides at an angle of 28 degrees from its original direction (labelled theta in the diagram). The overhead view shown in the diagram is approximately to scale. The arrow represents the average force your toe applies briefly to the block of ice.

What is the unit vector in the direction of the block's momentum after the kick?

What is the x-component of the block's momentum after the kick

What is the magnitude of the block's momentum after the kick?

find the z-component of the block's momentum after the kick

What was the magnitude of the average force you applied to the block?

Homework Equations


momentum = (mass)(velocity)
change in momentum = Fnet(time)

The Attempt at a Solution



I honestly have no clue what do do for this particular question. For example, what do I use the given angle (28 degrees) for?

This is how I tried the question:

First multiply the mass of the the block by the speed (0.55 x 3.5)
This will yield the momentum of the block

I don't know where to proceed beyond this point.

I need to find the momentum of the block after the kick. How do I do this? I only know the duration for which the contact occured? All help will be greatly appreciated.
Thank you.
 
Physics news on Phys.org
  • #2
Say the block initially is moving along the x-direction. You correctly calculated the initial momentum to be 0.55x3.5 kg m/s. That is the initial x-component of the momentum. Now can you find the x-component of the momentum after the kick? Knowing this and the 28o angle, can you find the y-component of the momentum after the kick?
 
  • #3
kuruman said:
Say the block initially is moving along the x-direction. You correctly calculated the initial momentum to be 0.55x3.5 kg m/s. That is the initial x-component of the momentum. Now can you find the x-component of the momentum after the kick? Knowing this and the 28o angle, can you find the y-component of the momentum after the kick?

Hi Kuruman,

I'm having trouble finding the momentum of the block after the kick. I have the initial momentum and I also have the time. Don't I also need Fnet? Or do we assume that Fnet has a x component of 0?

If so then, Final Momentum = Initial Momentum +(Fnet)(t)
So Final Momentum = Initial Momentum?

Actually there was an example in the book with a puck sliding on ice and in that example they also stated that there was no change in momentum. However, I don't know if the same applies to this case.

Thank You
 
  • #4
PolyFX said:
Hi Kuruman,

I'm having trouble finding the momentum of the block after the kick. I have the initial momentum and I also have the time. Don't I also need Fnet? Or do we assume that Fnet has a x component of 0?

If so then, Final Momentum = Initial Momentum +(Fnet)(t)
So Final Momentum = Initial Momentum?

Actually there was an example in the book with a puck sliding on ice and in that example they also stated that there was no change in momentum. However, I don't know if the same applies to this case.

Thank You

Note that momentum is linear and there was no net force in the direction of the block's initial motion, so that component of momentum could not have been affected. It is important that they said the force was applied perpendicularly to the motion, so that is the only momentum component that has changed. Ie., if you label the original non-zero momentum component the x-component, this component may be labelled the y- or the z-component, something perpendicular to the x-component.
Now that you know you have the x-component of the final momentum, you can use the angle and trigonometry to find the missing z-component, as they label it in your question.
 
  • #5
slider142 said:
Note that momentum is linear and there was no net force in the direction of the block's initial motion, so that component of momentum could not have been affected. It is important that they said the force was applied perpendicularly to the motion, so that is the only momentum component that has changed. Ie., if you label the original non-zero momentum component the x-component, this component may be labelled the y- or the z-component, something perpendicular to the x-component.
Now that you know you have the x-component of the final momentum, you can use the angle and trigonometry to find the missing z-component, as they label it in your question.


Hi,

Well I got most parts of this question correct but I was not so lucky on others.

So far,

I have found out that the x component of the blocks momentum after the kick is 1.925
I have found out the magnitude of the momentum after the kick is 2.180
I have also found out that the magnitude of the force applied on the block was 409.4 N

However, I cannot get the z component of the unit vector in the direction of the blocks momentum after the kick nor can I get z component of the blocks momentum after the kick.

Here's how I approached the question

Since the x component of the momentum does not change and is constantly at 1.925 I drew a right angled triangle with the x component as the base. I had the angle of the direction of the block which was 28 degrees. Therefore I used the equation

Tan28 = 1.925/Z (where z is the unknown we are trying to find)

Therefore Z = approximately 1.0235

I then used the pythagorean theorem to find out the hypotenuse which was in this case the magnitude of the blocks momentum (2.180).

Thus, I got the momentum of the block after the collision to be (1.925,0,1.0235)
I then divided this vector by 2.180(to get the unit vector) and ended up with (0.883,0,0.4694). both my x and y components are correct but my z component is wrong.

To find the average force applied on the block I divided 1.0235 (z component) by the time interval and got 409.4 N as the answer.

I'm a bit baffled because if the z component that I calculated was incorrect then how did my answer for the average force on the block turn out to be correct (I used the incorrect z value in the calculations)?

Furthermore, if the magnitude and the x component that I calculated were correct. Shouldn't I be able to rearrange the pythagorean theorem to solve for Z^2? This will still yield my initial answer of 1.0235.

I guess there's always the possibility that somehow I just got lucky during the calculation for the average force?

-Thank You
 
  • #6
I don't see where the question asks for the z-component of the unit momentum vector.
 

1. What is momentum and how is it defined?

Momentum is a physical quantity that measures the motion of an object. It is defined as the product of an object's mass and its velocity.

2. How is momentum conserved in a closed system?

In a closed system, the total momentum before an event or interaction must be equal to the total momentum after the event or interaction. This is known as the law of conservation of momentum.

3. How does momentum affect collisions?

Momentum plays a crucial role in collisions. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, only momentum is conserved as some kinetic energy is lost to other forms of energy.

4. Can momentum be negative?

Yes, momentum can be negative. The direction of an object's momentum is determined by its velocity. If an object is moving in the opposite direction of its positive chosen direction, its momentum will be negative.

5. How is momentum related to force?

Momentum and force are related through Newton's second law of motion, which states that the force applied to an object is equal to the change in momentum over time. This relationship is expressed as F = ma, where F is force, m is mass, and a is acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
5K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top