Write a realistic word problem for which this is the correct equation

In summary: It is true that we don't have the complete exact original statement. As it stands, one can think of a variety of physical situations that could fit the answer.
  • #1
ChetBarkley
10
0
Homework Statement
(0.10kg)(40m/s)-(0.10kg)(-30m/s)=1/2(1400N)(delta t)
Relevant Equations
Impulse = change in momentum = Force * delta t
So far for the word problem I have: A 100 g particle, traveling at 40 m/s, collides inelastically with another 100g particle traveling towards it at 30 m/s.

Now from the equation provided we need the question to ask us to find delta t, and that's simple enough but I'm not sure what that 1/2 is doing front of the right hand side. Does it mean that we only want half the impulse, or that the particles collide elastically and we want the impulse of just one of them?
 
Last edited:
Physics news on Phys.org
  • #2
ChetBarkley said:
Homework Statement:: (0.10kg)(40m/s)-(0.10)(-30m/s)=1/2(1400N)(delta t)
Relevant Equations:: Impulse = change in momentum = Force * delta t

So far for the word problem I have: A 100 g particle, traveling at 40 m/s, collides inelastically with another 100g particle traveling towards it at 30 m/s.
Why two particles? Why not one particle being hit with something? Or, colliding with something.
ChetBarkley said:
I'm not sure what that 1/2 is doing front of the right hand side.
Good question.
 
  • #3
PeroK said:
Why two particles? Why not one particle being hit with something? Or, colliding with something.

Good question.
Well if it were one particle hitting something, like a wall, then what would the 0.100kg and the -30m/s represent?
 
  • #4
ChetBarkley said:
Homework Statement:: (0.10kg)(40m/s)-(0.10)(-30m/s)=1/2(1400N)(delta t)
Shouldn't you have (0.10kg)(40m/s)-(0.10kg)(-30m/s)=1/2(1400N)(delta t)?

Also, with this correction, you cannot have two 0.10 kg particles. The right hand side is an impulse which means that the left hand side must be the inelastic momentum change of a single 0.10 kg particle. I don't understand the "1/2" but if it belongs there, I suspect it is the tricky part to this question.
 
  • Like
Likes ChetBarkley
  • #5
ChetBarkley said:
Well if it were one particle hitting something, like a wall, then what would the 0.100kg and the -30m/s represent?
Mass of particle and initial velocity?
 
  • Like
Likes ChetBarkley
  • #6
kuruman said:
I don't understand the "1/2" but if it belongs there, I suspect it is the tricky part to this question.
I can now see how to get the 1/2 factor in. The impulse equation is ##J=m\Delta v=\bar F \Delta t##. One has to cook up a force function ##F(t)## to describe a contact force parametrized by ##F_0## such that $$\bar F\equiv \frac{\int_0^{T} F(t)~dt}{\int_0^{T} dt}=\frac{F_0}{2}.$$
 
  • Like
Likes sysprog
  • #7
kuruman said:
I can now see how to get the 1/2 factor in. The impulse equation is ##J=m\Delta v=\bar F \Delta t##. One has to cook up a force function ##F(t)## to describe a contact force parametrized by ##F_0## such that $$\bar F\equiv \frac{\int_0^{T} F(t)~dt}{\int_0^{T} dt}=\frac{F_0}{2}.$$
I think that this is insightful and helpful, but it seems to me that it doesn't directly address the question ##-## the title asks for help in formulating a word problem to which the equation specified (albeit probably partly incorrectly, due to the omission of kg that you postulated) is the correct answer; it doesn't ask for a more elucidative equation ##-## maybe a bat hitting a pitched ball would work, or if we want to set one side stationary, a golf club hitting a golf ball, or a bowling ball hitting a pin ##-## I also think that we haven't yet been given the complete exact original problem statement.
 
  • #8
sysprog said:
I think that this is insightful and helpful, but it seems to me that it doesn't directly address the question ##-## the title asks for help in formulating a word problem to which the equation specified (albeit probably partly incorrectly, due to the omission of kg that you postulated) is the correct answer; it doesn't ask for a more elucidative equation ##-## maybe a bat hitting a pitched ball would work, or if we want to set one side stationary, a golf club hitting a golf ball, or a bowling ball hitting a pin ##-## I also think that we haven't yet been given the complete exact original problem statement.
It is true that we don't have the complete exact original statement. As it stands, one can think of a variety of physical situations that could fit the answer. Maybe the complete statement will narrow all these to one choice.
 
  • Like
Likes sysprog

Related to Write a realistic word problem for which this is the correct equation

1. What is a realistic word problem that can be solved using a specific equation?

A bakery sells cakes for $20 each and pies for $15 each. If the bakery made a total of $420 in one day by selling 20 cakes and 12 pies, what is the cost of each cake and pie?

2. How do I know which equation to use for a given word problem?

To determine the correct equation, you should carefully read the word problem and identify the known values and the unknown value that needs to be solved for. Then, you can use your knowledge of mathematical equations to select the appropriate one.

3. Can I use the same equation for different word problems?

Yes, some equations can be used for multiple word problems as long as the known and unknown values are consistent. However, it is important to carefully read each word problem to ensure that the equation is applicable.

4. Is it necessary to show my work when solving a word problem using an equation?

Yes, it is important to show your work when solving a word problem using an equation. This not only helps you to keep track of your steps and ensure accuracy, but it also allows others to understand your thought process and verify your solution.

5. How can I check if my solution is correct for a word problem?

You can check your solution by plugging it back into the original equation and solving for the unknown value. If the result is equal to the known value given in the word problem, then your solution is correct.

Similar threads

Replies
1
Views
844
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top