Momentum and impulse of jump problem

• KevinFan
In summary: But, you've already calculated that you need a force of ##420N##.Yes, I have calculated that the force I need is ##420N##.
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"

KevinFan said:

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"

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)?

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.

PeroK said:
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?

KevinFan said:
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.

KevinFan said:
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 said:
the floor exerts an upward force of magnitude mg on the
contestant

PeroK said:
What direction is positive in this calculation?
I think upward is positive

KevinFan said:
I think upward is positive

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

##mg - 420N \approx -350N##

PeroK said:
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)

KevinFan said:
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
PeroK said:
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!

1. What is momentum and impulse?

Momentum is a measure of an object's motion, calculated by multiplying its mass and velocity. Impulse is the change in momentum of an object, caused by a force acting on it for a certain amount of time.

2. How does momentum relate to jumping?

When jumping, the momentum of a person or object is transferred from their initial stationary position to an upward motion. This is achieved by applying a force to the ground, causing an impulse and resulting in a change in momentum.

3. Why is momentum important in jump problems?

Momentum is important in jump problems because it helps determine the height and distance an object can jump, as well as the force needed to achieve the desired jump. It also helps explain the mechanics and physics behind the jump.

4. How does mass affect momentum in jumping?

The mass of an object affects its momentum in jumping because momentum is directly proportional to mass. This means that the greater the mass, the greater the momentum and force needed for the jump.

5. How can impulse be calculated in a jump problem?

Impulse can be calculated by multiplying the force applied to an object by the time it is applied for. This can be represented by the equation: Impulse = Force * Time. In a jump problem, the force would be the force applied to the ground by the person or object, and the time would be the duration of the jump.

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