Conservation of Momentum Problem?

Click For Summary

Homework Help Overview

The problem involves conservation of momentum in a scenario where a cat on a skateboard throws a book to change its velocity. The discussion centers on determining the total mass of the system and the mass of a second book needed to achieve a specific velocity after the first book is thrown.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply conservation of momentum to find the total mass and the mass of a second book needed for a specific velocity. Some participants question whether the interpretation of the problem is correct, particularly regarding the timing of when the second book is thrown.

Discussion Status

Participants are exploring different interpretations of the problem and discussing potential errors in calculations. Some guidance has been offered regarding rounding errors, but there is no explicit consensus on the correct approach or interpretation of the problem's requirements.

Contextual Notes

There is uncertainty regarding the timing of the book throw and whether the mass of the second book should be calculated based on the total mass after the first book is thrown or before. The original poster expresses confusion about the correctness of their method and the values used in calculations.

r_swayze
Messages
65
Reaction score
0
A cat stands on a skateboard that moves without friction along a level road at a constant velocity of 2.00 m/s. She is carrying a number of books. She wishes to stop, and does so by hurling a 1.20 kg book horizontally forward at a speed of 15.0 m/s with respect to the ground. (a) What is the total mass of the cat, the skateboard, and any remaining books? (b) What mass book must she now throw at 15.0 m/s with respect to the ground to move at −2.00 m/s?

I got the answer to part A, here is my work for part A:

m1vi1 + m2vi2 = m1vf1 + m2vf2

m1 = total mass of the cat, the skateboard, and any remaining books
vi1 = 2.0 m/s
vf1 = 0 m/s
m2 = 1.2 kg book
vi2 = 2.0 m/s
vf2 = 15.0 m/s

m1vi1 = -m2vi2 + m1vf1 + m2vf2

m1 = (-m2vi2 + m1vf1 + m2vf2) / vi1

m1 = (-1.2*2.0 + m1*0 + 1.2*15.0) / 2.0
m1 = 7.8 kg

Now for part B, I thought I had the correct answer but apparently it is wrong. Here is my work:

From part a
m1 + m2 = total mass of of the cat, the skateboard, and any remaining books
m1 + m2 = 7.8 kg + 1.2 kg
m1 + m2 = 9 kg = mt

mtvi = m1vf1 + m2vf2

9*2 = m1*-2.0 + m2*15.0
18 = -2m1 + 15m2

So m1 + m2 should equal to total mass = 9 kg

m1 + m2 = 9
m1 = 9 - m2

Plug into equation

18 = -2(9 - m2) + 15m2
18 = -18 + 2m2 + 15m2
36 = 17m2
m2 = 2.11 kg

Have I erred somewhere? If you plug in 2.11 kg and 6.88 kg, it comes out roughly correct. Should I be using 9 kg in part B?

Any help is much appreciated
 
Physics news on Phys.org
Your solution must be correct but a rounding error. m2=2.1176 kg, should be rounded to 2.12 kg.

ehild
 
ehild said:
Your solution must be correct but a rounding error. m2=2.1176 kg, should be rounded to 2.12 kg.

ehild

I tried 2.12 as well, it was still incorrect. Is my method correct?
 
I think, your solution is correct, but we may misunderstand the text. Maybe it asks the mass of book that the cat has to throw after it has stopped, to gain -2 m/s velocity. Try it...

But also the official solution can be wrong.

ehild
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
18K
  • · Replies 4 ·
Replies
4
Views
2K