How Does Force Calculation Error Occur in Momentum Problems?

[rootX]

Homework Statement

A 72 kg man jumps
a) If the jump results in an upward speed of 2.1 m/s, what additional upward force does the floor exert, if the person pushes the ground for 0.36 s?

Homework Equations

F= mv/t

The Attempt at a Solution

F_net = Impule/t

F_byFloor - W = F_net

f_byFloor = F_net + W
= (mv_f / t) + mg
= m ((v_f / t)+g)
= 1126.32 N

But, the answer in the book is 420 N, and it just ignored Weight of the person.
If my Force equation (F_byFloor - W = F_net) is right?

Homework Statement

A bee lands on one end of a floating 4.75 g Popsicle stick. After sitting at rest for a moment, it runs towards the other end with a speed of 3.80 cm/s relative to still water. The stick moves in the opposite direction at 0.12 cm/s relative to the still water. What is the mass of bee?

Homework Equations

mv_b + Mv_popsicle = 0

The Attempt at a Solution

I am confused about the velocities. Like, How the results would be changed if I use bee velocity relative to Popsicle. And, do I need to have velocities of both objects relative to some stationary thing in order to solve this problem?

 

[Dick]

Question 1) asks for 'additional upward force', not 'total upward force' implying you should ignore the weight. For 2) a consequence of the conservation of momentum is that the center of mass of SYSTEM doesn't accelerate unless acted upon by a force from OUTSIDE of the system. Consider the bee and the popsicle stick as a system.

 

[m2003]

I would suggest approaching these problems by breaking them down into smaller parts and analyzing each part separately. In the first question, we can start by defining the variables and identifying the forces at play. We have the mass of the man (m = 72 kg), the upward speed (v_f = 2.1 m/s), the time the man pushes the ground (t = 0.36 s), and the force exerted by the floor (F_byFloor). The forces at play are the force exerted by the man (F_net), the weight of the man (W = mg), and the force exerted by the floor (F_byFloor).

Using the equation F_net = mv_f/t, we can calculate the force exerted by the man, which is equal to the force exerted by the floor and the weight of the man combined. So, F_net = F_byFloor + W. Substituting in the values, we get F_net = (72 kg)(2.1 m/s)/0.36 s = 420 N. This is the force exerted by the man, but we need to find the additional upward force exerted by the floor. To do this, we can rearrange the equation to solve for F_byFloor, which gives us F_byFloor = F_net - W. Substituting in the values, we get F_byFloor = 420 N - (72 kg)(9.8 m/s^2) = 420 N - 705.6 N = 1126.32 N. So, the additional upward force exerted by the floor is 1126.32 N.

In the second question, we can start by defining the variables and identifying the forces at play. We have the mass of the bee (m_b), the mass of the Popsicle stick (m_popsicle = 4.75 g), the speed of the bee relative to the still water (v_b = 3.80 cm/s), and the speed of the Popsicle stick relative to the still water (v_popsicle = 0.12 cm/s). The forces at play are the force exerted by the bee (F_b), the force exerted by the Popsicle stick (F_popsicle), and the weight of the Popsicle stick (W = mg).

Using the equation mv_b + Mv_popsicle = 0,