# What is the momentum of the pellet?

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1. Nov 7, 2016

### emily081715

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
i honestly am completely lost on this question, i assumed it would just be F0e(−t) but thats not correct. help?

2. Nov 7, 2016

### I like Serena

H
Hey Emily!

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?

3. Nov 7, 2016

### emily081715

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

4. Nov 7, 2016

### I like Serena

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:
$$\int Fdt=\int madt=mv=p$$

5. Nov 7, 2016

### emily081715

wouldn't the derivative be e(−t/τ) +F0e(−t/τ) ?

6. Nov 7, 2016

### I like Serena

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?

7. Nov 7, 2016

### emily081715

i was assuming there was also product rule too since fo is a constant

8. Nov 7, 2016

### I like Serena

Nope. No product rule for constants.
Or rather, the derivative of a constant is zero, causing the corresponding term to be zero as well.

9. Nov 8, 2016

### I like Serena

The derivative is:
$$F'(t) = F_0 \cdot -\frac 1\tau e^{-t/\tau}$$
This helps to find the integral, which is:
$$p(\tau) = \int_0^\tau F\,dt = F_0 \cdot -\tau e^{-t/\tau}\Big|_0^\tau$$

10. Nov 8, 2016

### emily081715

i solved the whole question already