Conceptual physics homework, easy, but understanding

[knnox]

Homework Statement

Bronco Brown wants to put Ft =∆mv to
the test and try bungee jumping. Bronco
leaps from a high cliff and experiences
free fall for 3 seconds. Then the bungee
cord begins to stretch, reducing his speed
to zero in 2 seconds. Fortunately, the cord
stretches to its maximum length just short
of the ground below.

Homework Equations

∆mv at 3-s free fall is 3000 kg m/s
Then the question asks: ∆mv during the 2-s interval of slowing
down

The Attempt at a Solution

To my knowledge it is at final velocity leaving it at 3000 kg*m/s
Would it be 1500kg m/s or 3000kg m/s?
I think 3000, but I'm not sure the equation to understand it.
The equation I used gave me 1500 because of the slowing down.
Please help me understand this better.

 

[cepheid]

Welcome to PF knnox!

You are not communicating particularly well, I'm afraid. Nowhere in your problem statement does it actually state what you are supposed to work out. If the problem is to figure out "delta mv during the 2-second interval of slowing down", then that's fine, but I don't know how to check your work, because I don't see a mass stated for Bronco Brown anywhere.

 

[knnox]

Welcome to PF knnox!

You are not communicating particularly well, I'm afraid. Nowhere in your problem statement does it actually state what you are supposed to work out. If the problem is to figure out "delta mv during the 2-second interval of slowing down", then that's fine, but I don't know how to check your work, because I don't see a mass stated for Bronco Brown anywhere.

I apologize, it is 100kg, but looking back over this I now realize its just ask delta mv during the 2-s interval which is just 3000kg*m/s which is the same as impulse. Thank you for the response though and sorry for my lack of communication!

 

[cepheid]

I apologize, it is 100kg, but looking back over this I now realize its just ask delta mv during the 2-s interval which is just 3000kg*m/s which is the same as impulse. Thank you for the response though and sorry for my lack of communication!

You can figure out his speed after free-falling for three seconds using v = v₀ - gt, where v₀ = 0, since he starts from rest at the beginning of the fall. I get 29.43 m/s for that, which when multiplied by 100 kg gives you a momentum of 2943 kg*m/s. (I guess you rounded up to 3000, but this introduces error). After that, you need to use the impulse-momentum theorem:

FΔt = Δ(mv)​

Here, Δ(mv) will be -2943 m/s, since he starts with 2943 kg*m/s, and ends with 0 kg*m/s, so the change in momentum is 0 - 2943.

Now, given Δ(mv), you can solve for F in Newtons, given that the impulse occurred over a time interval of Δt = 2 seconds.

It sounds like you had it right (up to rounding error) in the first place.

 

[Nickie]

Your attempt at a solution is partially correct. The equation you used, ∆mv = Ft, is correct for this scenario. However, the value of ∆mv will depend on the force (F) applied by the bungee cord during the 2-second interval of slowing down.

To calculate the force, we can use the equation F = ma, where m is the mass of Bronco and a is the acceleration. During free fall, Bronco's acceleration is equal to gravity, which is approximately 9.8 m/s^2. However, during the 2-second interval of slowing down, his acceleration will be different. We can use the kinematic equation v = u + at, where v is the final velocity (zero), u is the initial velocity (3000 m/s), a is the acceleration, and t is the time (2 seconds). Solving for a, we get a = -1500 m/s^2. This negative sign indicates that Bronco is decelerating during this interval.

Now, we can plug this value of acceleration into the equation F = ma. Since we know the mass of Bronco is 3000 kg, we can calculate the force to be F = (3000 kg)(-1500 m/s^2) = -4500000 N. The negative sign indicates that the force is acting in the opposite direction of Bronco's motion, which makes sense since it is slowing him down.

Finally, we can use the equation ∆mv = Ft to calculate the change in momentum during this 2-second interval. Plugging in the calculated force and the time interval of 2 seconds, we get ∆mv = (-4500000 N)(2 s) = -9000000 kg m/s. The negative sign indicates that the change in momentum is in the opposite direction of Bronco's initial momentum (3000 kg m/s).

Therefore, the correct answer is that the change in momentum during the 2-second interval is -9000000 kg m/s. This means that Bronco's final momentum is 3000 kg m/s - 9000000 kg m/s = -6000000 kg m/s. This value is negative because it is in the opposite direction of his initial momentum.

In summary, the equation ∆mv = Ft is a very useful tool for understanding the relationship between force and change in momentum. In this scenario, the force