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Is my interpretation good? simple Kinetic E problem

  • #1

Homework Statement


A cat is chasing a mouse in a straight line. The cat's mass, mc, = 5.0kg. The mouse's mass, mm, is 0.035kg.

Problem 1. If the kinetic energy of the cat is 100 times the kinetic energy of the mouse can the cat catch the mouse?

Problem 2. If the cat and the mouse have the same speed, what is the kinetic energy of the cat?

Homework Equations


Conservation of energy. Kinetic energy = (1/2)mv2

The Attempt at a Solution


Solution for Problem 1.

Ekc = (100)Ekm
(1/2)mcvc2 = (100)(1/2)mmvm2
mcvc2 = (100)mmvm2
vc2 = [(100)mmvm2] / mc
vc2 = [ (100)(0.035kg)vm2 ] / 5.0kg
vc2 = (7/10) vm2
vc = (7/10)(1/2) vm
vc ≅ 0.83 vm

Therefore, the cat cannot catch the mouse. The cat's speed is 83% of the mouse's speed

{Is the result correct? Is my therefore statement correct? Is this the proper interpretation of the result?}

Solution for Problem 2. - If the speed of the cat is the speed of the mouse what is the cat's kinetic energy?

vc = vm
[(2Ekc)/mc](1/2) = [(2Ekm)/mm](1/2)
Ekc/mc = Ekm/mm
Ekc = mc Ekm / mm
Ekc = {mc (1/2mmvm2)} / mm
Ekc = (1/2) mcvm2
Ekc = 2.5kg vm2
Ekc = How can I find this value? All of the known problem info. is stated.

The correct answer according the answer page of the text is "140 times."

What approach should I take to solving this problem.

Many thanks!
 

Answers and Replies

  • #2
34,043
9,883
Therefore, the cat cannot catch the mouse. The cat's speed is 83% of the mouse's speed

{Is the result correct? Is my therefore statement correct? Is this the proper interpretation of the result?}
Correct.

You cannot calculate the kinetic energy of the cat in Joules, but you can express it in terms of the kinetic energy of the mouse. ##\frac{E_c}{E_m}## is something you can find.


By the way: You can use the result of the second part to answer the first one, there is no need to calculate the relative speed (but it is a valid solution).
 
  • #3
Thank you so very much!!!

Ekc = mc Ekm / mm
Ekc = [ mc / mm ] * Ekm
Ekc ≅ 142.86 * Ekm

So, the kinetic energy of the cat is roughly 140 times the kinetic energy of the mouse.

I didn't think it was possible to find the value in the Joules. It's nice to have some confirmation.

I'm not sure what you mean when you state, the result of the second part can be used to answer the first part. I believe part 1 and part 2 to be two distinct situations. In part 1, the problem states that the kinetic energy of the cat is 100 times the kinetic energy of the mouse. In part 2, we prove that this is not true. That is, in part 2 we prove that the cat's kinetic energy is 140 times the kinetic energy of the mouse if their speeds are equivalent.
 
  • #4
34,043
9,883
The cat needs 140 times the kinetic energy of the mouse to be as fast as the mouse. What do you expect for its speed if it just has 100 times the kinetic energy?
 
  • #5
The cat needs 140 times the kinetic energy of the mouse to be as fast as the mouse. What do you expect for its speed if it just has 100 times the kinetic energy?
The problem in part 1 is solved. The cat cannot catch the mouse. The cat's speed, whatever value in metres-per-second that may be, is only 83% of whatever the mouse's actual speed in metres-per-second may be. Part 2 cannot apply to Part 1 because they are totally distinct. In part two we've started with the basis that their speeds are equivalent. Any attempt to use that value will show that the cat's speed is equivalent to the mouse's speed. It's a massive circular argument. Or, I am incredibly stupid. Sure, I admit it. You're brilliant and smug, and I'm a dumb worm.

Prove to me how the results of two can be used to provide a better solution to Problem 1.

(Why is it that so many people on here reply with, "oh... you can't see it. To everyone else it's so obvious I won't bother to explain any more." Clearly not! Is that not the whole point of writing in and asking for help? These forums have really tanked.)
 
  • #6
34,043
9,883
It is not circular if you solve part 2 first and then use "lower energy => lower speed". You know "lower energy" from the solution of part 2 and the problem statement for part 1.
 
  • #7
99
11
You could think of it this way...

P1. If the velocities were equal then the cat would not catch the mouse, and the velocity terms would be equal and cancel. But what would be left must reamin true...
5 = 100*0.035
Because you know this statement is not true then you know that the speeds are not equal, furthermore you know that the mouse must have a greater velocity for it to be true. Thfr cat does not catch mouse.

P2. Given velocities are equal, if the factor of 100x the KE is not correct then what is?
5 = (x)*0.035
x = 142.86

The 140 answer is correct to the number of significant figures you have although not expressed in sci notation.
 

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