Respuesta :
Answer: If one person get's out of the boat, the speed will be 6.67m/s
Explanation:
The velocity of the boat is 5m/s, and we know that the weight of each person is around 60 kg, and we know that weight is equal to mass times the gravity acceleration, so the mass of each person is around 60/9.8 = 6.12kg
Now, the momentum of the boat initially is around:
P = 5m/s*M
and after one person drops, the momentum is:
P = v*(M - 6.12kg)
Where M is the total initial mass of the boat.
Now, the momentum conserves, so the momentum must be equal before and after one person gets out, so we have that:
5m/s*M = v*(M - 6.12kg)
now we need to solve this equation for v, that is the velocity of the boat.
v = 5m/s* (M/(M - 6.12kg))
Now, we do not know the actual mass of the boat, but we can think that M is equal to the mass of the 4 people inside the boat, then we have M = 4*6.12kg
v = 5m/s*(4*6.12/(3*6.12)) = 5m/s*(4/3) = 6.67m/s
So the final speed of the boat is 6.67m/s
Answer:
v_f = 6.67 m/s
Explanation:
Given:-
- The boat moves downstream with velocity, v_i = 5 m/s
- The number of people on boat, n = 4
- The average weight of all people , M = 60 kg
Find:-
Then what is the speed of the boat after one person gets off the boat?
Solution:-
- Considering the boat and people on board as a system. There are no external forces acting on this system that could accelerate the system. We can safely say that the momentum of the system is conserved.
- We will apply conservation of linear momentum for the boat and the people on board. Assuming boat to be massless.
P_i = P_f
( M )*v_i = ( M - M/n )* v_f
Where, P_i : initial momentum
P_f : final momentum - after one person gets off
Subtracting the mass of one person ( M/n) we can determine the speed of the boat with 3 people.
( 60 )*5 = ( 60 - 60/4 )* v_f
v_f = 300 / 45
v_f = 6.67 m/s
