Question

A dog chasing a cat has 0.55 times the kinetic energy of the cat, which has 0.45 times the weight of the dog. The dog speeds up by 0.91 m/s and then has the same kinetic energy as the cat. What was the original difference in speed of the dog and cat? In other words, what is v_sub_cat - v_sub_dog initially (before the dog speeds up)?

Answer 1

Answer:

The difference in speed of the dog and cat is 2.65 m/s

Explanation:

Given that,

Mass of cat = 0.45 times the mass of dog

K.E of dog = 0.55 times K.E of cat

We need to calculate the velocity of dog

Using given data

$\dfrac{1}{2}m_{d}v_{d}^2=0.55\times\dfrac{1}{2}m_{c}v_{c}^2$

$m_{d}v_{d}^2=0.55\times0.45\times m_{d}\times v_{c}^2$

$v_{d}^2=0.2475\times v_{c}^2$

$v_{d}=0.497v_{c}$....(I)

The dog speeds up by 0.91 m/s and then has the same kinetic energy as the cat.

We need to calculate the velocity of dog

Using conservation of kinetic energy,

$\dfrac{1}{2}m_{d}v_{d}^2=\dfrac{1}{2}m_{c}v_{c}^2$

$\dfrac{1}{2}m_{d}(v_{d}+0.91)^2=\dfrac{1}{2}m_{c}v_{c}^2$

$m_{d}(v_{d}+0.91)^2=0.45m_{d}v_{c}^2$

$v_{d}+0.91=0.67v_{c}$

$v_{d}=0.67v_{c}-0.91$...(II)

From equation (I) and (II)

$0.497v_{c}=0.67v_{c}-0.91$

$v_{c}=\dfrac{-0.91}{(0.497-0.67)}$

$v_{c}=5.26\ m/s$

Put the value in equation (I)

$v_{d}=0.497\times5.26$

$v_{d}=2.61\ m/s$

We need to calculate the difference in speed of the dog and cat

Using speed of dog and cat

$v_{c}-v_{d}=5.26-2.61$

$v_{c}-v_{d}=2.65\ m/s$

Hence, The difference in speed of the dog and cat is 2.65 m/s