Bubba has a 3.54 kilogram cat to keep rodents out of his barn. One day, the cat

Question

Bubba has a 3.54 kilogram cat to keep rodents out of his barn. One day, the cat is standing on a stationary 5.75 kilogram cart. The cart then starts to roll at 23.5 cm/s. The cat jumps off of the cart, in the direction that it is rolling, with a speed of 1.44 m/s relative to the cart.

a) What speed and direction does the cart move after the cat jumps off of it.

b) What is the change in mechanical energy as the cat jumps off of the cart?

2) Bubba has a 3.54 kilogram cat to keep rodents out of his barn. One day, the car is standing on a stationary 5.75 kilogram cart. The cart then starts to roll 23.5 cms. The cat jumps off of the cart, in the direction that it is rolling, with a speed of 1.44 m/s relarive to the cart. a) What speed and direction does the cart move after the cat jumps off of it? b) What is the change in mechanical energy as the cac jumps of the cart?

Explanation / Answer

Vitg = initial velocity of cat and cart combination relative to ground = 23.5 cm /s = 0.235 m/s

Vfcg = final velocity of cat relative to ground = 1.44 + 0.235 m /s = 1.675 m/s

Vftg = final velocity of cart relative to ground = ?

m = mass of cat = 3.54 kg

M = mass of cart = 5.75 kg

using conservation of momentum

(m + M) Vitg = m Vfcg + MVftg

(3.54 + 5.75) (0.235) = (3.54) (1.675) + (5.75) Vftg

Vftg = - 0.652 m/s

speed of cart relative to ground = 0.652 m/s

direction : in opposite direction of the direction the cat jumps in

b)

initial Total KE =KEi = (0.5) (m + M) V2itg = (0.5) (3.54 + 5.75) (0.235)2 = 0.257 J

final Total KE =KEf = (0.5) (m) V2fcg + (0.5) (M) V2ftg = (0.5) (3.54) (1.675)2 + (0.5) (5.75) (0.652)2 = 6.2 J

change in mechanical energy = KEf - KEi = 6.2 - 0.257 = 5.943 J

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