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

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

Step-by-step 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

`(1)/(2)m_(d)v_(d)^2=0.55*(1)/(2)m_(c)v_(c)^2`

`m_(d)v_(d)^2=0.55*0.45* m_(d)* v_(c)^2`

`v_(d)^2=0.2475* 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,

`(1)/(2)m_(d)v_(d)^2=(1)/(2)m_(c)v_(c)^2`

`(1)/(2)m_(d)(v_(d)+0.91)^2=(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)=(-0.91)/((0.497-0.67))`

`v_(c)=5.26\ m/s`

Put the value in equation (I)

`v_(d)=0.497*5.26`

`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