- #1

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## Homework Statement

A cat is chasing a mouse in a straight line. The cat's mass, m

_{c}, = 5.0kg. The mouse's mass, m

_{m}, 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)mv

^{2}

## The Attempt at a Solution

Solution for Problem 1.

E

_{kc}= (100)E

_{km}

(1/2)m

_{c}v

_{c}

^{2}= (100)(1/2)m

_{m}v

_{m}

^{2}

m

_{c}v

_{c}

^{2}= (100)m

_{m}v

_{m}

^{2}

v

_{c}

^{2}= [(100)m

_{m}v

_{m}

^{2}] / m

_{c}

v

_{c}

^{2}= [ (100)(0.035kg)v

_{m}

^{2}] / 5.0kg

v

_{c}

^{2}= (7/10) v

_{m}

^{2}

v

_{c}= (7/10)

^{(1/2)}v

_{m}

v

_{c}≅ 0.83 v

_{m}

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?

v

_{c}= v

_{m}

[(2E

_{kc})/m

_{c}]

^{(1/2)}= [(2E

_{km})/m

_{m}]

^{(1/2)}

E

_{kc}/m

_{c}= E

_{km}/m

_{m}

E

_{kc}= m

_{c}E

_{km}/ m

_{m}

E

_{kc}= {m

_{c}(1/2m

_{m}v

_{m}

^{2})} / m

_{m}

E

_{kc}= (1/2) m

_{c}v

_{m}

^{2}

E

_{kc}= 2.5kg v

_{m}

^{2}

E

_{kc}= 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!