What is the momentum of the pellet?

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The discussion revolves around calculating the momentum of a pellet after a time interval equal to one time constant, given the force function F(t) = F0e(−t/τ). Participants express confusion about integrating this exponential function to find momentum, which is defined as p = mv. The key insight shared is that the integral of the force over time yields the momentum, and the derivative of the exponential function is crucial for this calculation. Ultimately, the integration leads to a solution for the momentum in terms of F0, τ, and the exponential constant e. The conversation highlights the importance of understanding integration and differentiation in solving physics problems.
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Homework Statement


A burst of compressed air pushes a pellet out of a blowpipe. The force exerted by the air on the pellet is given by F(t)=F0e(−t/τ), where τ is called a time constant because it has units of time.
What is the momentum of the pellet after an interval equal to one time constant has elapsed?
Express your answer in terms of the variables F0, τ, and exponential constant e.

Homework Equations

The Attempt at a Solution


i honestly am completely lost on this question, i assumed it would just be F0e(−t) but that's not correct. help?
 
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emily081715 said:

Homework Statement


A burst of compressed air pushes a pellet out of a blowpipe. The force exerted by the air on the pellet is given by F(t)=F0e(−t/τ), where τ is called a time constant because it has units of time.
What is the momentum of the pellet after an interval equal to one time constant has elapsed?
Express your answer in terms of the variables F0, τ, and exponential constant e.

Homework Equations

The Attempt at a Solution


i honestly am completely lost on this question, i assumed it would just be F0e(−t) but that's not correct. help?

Hey Emily! :oldsmile:

We have momentum p=mv.
Furrhermore we have force F=ma.
And speed v is the integral of acceleration a with respect to time.
Suppose we integrate the expression. What will we get?
 
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I like Serena said:
HHey Emily! :oldsmile:

We have momentum p=mv.
Furrhermore we have force F=ma.
And speed v is the integral of acceleration a with respect to time.
Suppose we integrate the expression. What will we get?
I'm completely lost how to integrate it though,F(t)=F0e(−t/τ) is not like a function I'm use to working with
 
emily081715 said:
I'm completely lost how to integrate it though,F(t)=F0e(−t/τ) is not like a function I'm use to working with

The exponential function is an odd one. Its integral looks very similar to its derivative.
What would the derivative be?

Oh, and for the record:
$$\int Fdt=\int madt=mv=p$$
 
I like Serena said:
The exponential function is an odd one. Its integral looks very similar to its derivative.
What would the derivative be?
wouldn't the derivative be e(−t/τ) +F0e(−t/τ) ?
 
emily081715 said:
wouldn't the derivative be e(−t/τ) +F0e(−t/τ) ?

How so?

Note that the derivative of ##e^x## is ##e^x##. That leaves applying the chain rule. How familiar are you with the chain rule?
 
I like Serena said:
How so?

Note that the derivative of ##e^x## is ##e^x##. That leaves applying the chain rule. How familiar are you with the chain rule?
i was assuming there was also product rule too since fo is a constant
 
emily081715 said:
i was assuming there was also product rule too since fo is a constant

Nope. No product rule for constants.
Or rather, the derivative of a constant is zero, causing the corresponding term to be zero as well.
 
The derivative is:
$$F'(t) = F_0 \cdot -\frac 1\tau e^{-t/\tau}$$
This helps to find the integral, which is:
$$p(\tau) = \int_0^\tau F\,dt = F_0 \cdot -\tau e^{-t/\tau}\Big|_0^\tau$$
 
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i solved the whole question already
 
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