Respuesta :
a) Take into account that the force can be written as follow:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]where F is the force, Δp the change in the momentum and Δt the time interval in which the force is applied.
Moreover, consider that the change in the momentum can be written as follow:
[tex]\Delta p=m\Delta v[/tex]where m is the mass and Δv is the change in the velocity.
By replacing the previous expression into the formula for the force F, you can solve for Δv to determine the change in the velocity of Jamper due the force he applies to the boat:
[tex]\begin{gathered} F=\frac{m\Delta v}{\Delta t} \\ \Delta v=\frac{\Delta t\cdot F}{m} \end{gathered}[/tex]In this case, m=80.0kg, Δt = 0.30s and F = 800.0N.
Then, for the change in the speed you obtain:
[tex]\Delta v=\frac{0.30s\cdot800.0N}{80.0kg}=3.0\frac{m}{s}[/tex]It means that related to the boat (where you can consider that Jasper is at rest) the speed of Jasper is 3m/s. However, it is necessary to take into account the speed of the boat before Jasper jumps to the dock, which is 3.0m/s.
The speed of Jasper when he lands on the dock is then:
3.0m/s + 3.0m/s = 6.0m/s
The momentum is the product of mass and speed, then, the momentum of jasper is:
p = (80.0kg)(6.0m/s) = 480.0kg*m/s
b) Based on the previous calculations, you have that the Jasper's speed on the dock is 6.0m/s
