A 73 kg swimmer dives horizontally off a 462 kg raft initially at rest. If the diver's speed immediately after leaving the raft is 5.54 m/s, what is the corresponding raft speed

Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Explanation:

Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.

And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.

(Final momentum of the diver) + (Final momentum of the raft) = 0

Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)

Mass of the diver = 73 kg

Diving velocity of the diver = 5.54 m/s

Momentum of the diver = 73 × 5.54 = 404.42 kgm/s

Momentum of the raft = (mass of the raft) × (velocity of the raft)

Mass of the raft = 462 kg

Velocity of the raft = v

Momentum of the raft = 462 × v = (462v) kgm/s

404.42 + 462v = 0

462v = -404.42

v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Hope this Helps!!!

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s