Question

Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is

Answer 1

Answer:

the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.

Explanation:

Let the spring constant of the spring is k.

For clown A:

m = 40 kg

let the extension in the spring is y.

So, the spring force, F = k y

m g = k y

40 x g = k x y

y = 40 x g / k      ..... (1)

For clown B:

m' = 60 kg

Let the extension in the spring is y'.

So, the spring force, F' = k y'

m' g = k y'

y' = 60 x g / k      .....(2)  

Kinetic energy for A, K = 1/2 ky^2

Kinetic energy for B, K' = 1/2 ky'^2

So, K/K' = y^2/y'^2 K / K' = (40 x 40) / (60 x 60)     (from equation (1) and (2))

K / K' = 0.444

K = 0.444 K'

So the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.