Momentum and impulse of jump problem

[KevinFan]

Homework Statement

After winning a prize on a game show, a 72 kg contestant jumps for joy.
A.
If the jump results in an upward speed of 2.1 m/s, what is the impulse experienced by the contestant?
b. 0.0
c. 75.6 kg.m/s
d. 151.2 kg.m/s
e. 226.8 kg.m/s
B.
Before the jump, the floor exerts an upward force of magnitude mg on the
contestant. What additional average upward force does the floor exert if the contestant pushes down on the floor
for 0.36 s during the jump?
a. 0.0
b. 210.0 N
c. 420.0 N
d. 630.0 N

Homework Equations

P=mv
delta p= f*delta t

The Attempt at a Solution

For part A: Since the contestant is on the ground initially, he has zero momentum, and after he jump his final speed is 2.1m/s. P=mv, I calculated 151.2 kg*/s as his change in momentum. However, the answer for this part should be 75.6kg*m/s...
For part B I used the impulse divide by time to get the average force, and the value is 420N. The answer is 420 N but I thought that I should use mg substruct 420N to get the "additional force"

Any suggestions please?

 

[PeroK]

Homework Statement

After winning a prize on a game show, a 72 kg contestant jumps for joy.
A.
If the jump results in an upward speed of 2.1 m/s, what is the impulse experienced by the contestant?
b. 0.0
c. 75.6 kg.m/s
d. 151.2 kg.m/s
e. 226.8 kg.m/s
B.
Before the jump, the floor exerts an upward force of magnitude mg on the
contestant. What additional average upward force does the floor exert if the contestant pushes down on the floor
for 0.36 s during the jump?
a. 0.0
b. 210.0 N
c. 420.0 N
d. 630.0 N

Homework Equations

P=mv
delta p= f*delta t

The Attempt at a Solution

For part A: Since the contestant is on the ground initially, he has zero momentum, and after he jump his final speed is 2.1m/s. P=mv, I calculated 151.2 kg*/s as his change in momentum. However, the answer for this part should be 75.6kg*m/s...
For part B I used the impulse divide by time to get the average force, and the value is 420N. The answer is 420 N but I thought that I should use mg substruct 420N to get the "additional force"

Any suggestions please?

You're correct on part a). You can easily calculate a) from the answer to b).

What do you want to do with the answer of $420N$ for b)?

 

[zhouhao]

I think the answer for part A is 151.2kg*m/s and the 420N is the additional force as well as the net force.

 

[KevinFan]

What do you want to do with the answer of 420N420N420N for b)?

Since the question mentioned that the floor exerts an upward force of mg, and it is asking what ADDITIONAL upward force does the floor exert. By using the impulse expression, the average force can be found and the value is 420 N. I thought mg+ additional force =420N, therefore, I need to use mg-420N to find the additional force?

 

[zhouhao]

Since the question mentioned that the floor exerts an upward force of mg, and it is asking what ADDITIONAL upward force does the floor exert. By using the impulse expression, the average force can be found and the value is 420 N. I thought mg+ additional force =420N, therefore, I need to use mg-420N to find the additional force?

420N=upward force exerted by floor - mg ; If the floor exerts mg as upward force then there is no net force to push constestant.

 

[PeroK]

Since the question mentioned that the floor exerts an upward force of mg, and it is asking what ADDITIONAL upward force does the floor exert. By using the impulse expression, the average force can be found and the value is 420 N. I thought mg+ additional force =420N, therefore, I need to use mg-420N to find the additional force?

What direction is positive in this calculation?

 

[KevinFan]

the floor exerts an upward force of magnitude mg on the
contestant

 

[KevinFan]

What direction is positive in this calculation?

I think upward is positive

 

[PeroK]

I think upward is positive

So, the additional force on the man is downwards, which causes him to accelerate upwards?

$mg - 420N \approx -350N$

 

[KevinFan]

So, the additional force on the man is downwards, which causes him to accelerate upwards?

$mg - 420N \approx -350N$

I am afraid I am losing you now...
I thought the additional force is 285.6 N( mg- additional force=average force)

 

[PeroK]

I am afraid I am losing you now...
I thought the additional force is 285.6 N( mg- additional force=average force)

Let's suppose the additional force is $285N$. You have three forces on the man: 1) this additional force 2) The floor pushing up with $mg$ and 3) Gravity pushing down with $-mg$.

The net force on the man is, therefore, $285N + mg - mg = 285N$

So, the acceleration and impulse must be calculated using this force.

But, you've already calculated that you need a force of $420N$. Not $285N$.

 

[KevinFan]

Let's suppose the additional force is $285N$. You have three forces on the man: 1) this additional force 2) The floor pushing up with $mg$ and 3) Gravity pushing down with $-mg$.

The net force on the man is, therefore, $285N + mg - mg = 285N$

So, the acceleration and impulse must be calculated using this force.

But, you've already calculated that you need a force of $420N$. Not $285N$.

I see! Many thanks for your excellent explanation!