# Inelastic collsisions mass and acceleration changing with time

1. Jan 5, 2006

### *Alice*

I have tried to do the last part of the attached problem about inelastic collisions between a water drop and vapour, where the mass and acceleration of the drop is changing with time.

So, in order to find v and a wrt t I integrated the given equation.

However, the result I obtained is rather weird, so - if anyone sees what I do wrong - that would be great! Thank you!

v(t)

$$\frac {d(1 + alpha*T)v)} {dt} = g ( 1+ alpha*T)$$

$$(1 + alpha*T)*v = \in_0^T {g(1 + alpha*T)}dt$$

$$(1 + alpha *T)*v = gT + 0.5*alpha+T^2 + K$$

$$v= \frac {gT + .5*alpha*T^2} {1 + alpha*T} + v(initial)$$

a(t)

just differentiate v(t), giving

$$\frac {g(2 + 3*alpha*T + 2.5*alpha^2*T^2} {1 + alpha*T}$$

Thanks!

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Last edited: Jan 5, 2006
2. Jan 5, 2006

### LeonhardEuler

That capital T should be a t, otherwise the derivative is 0.
You lost a parenthases here. It should be:
$$(1 + \alpha *T)*v = g(T + 0.5*\alpha\times T^2) + K$$

3. Jan 5, 2006

### LeonhardEuler

I'm not sure how you got this, but it doesn't seem right. Did you use the quotient rule?

4. Jan 5, 2006

### *Alice*

hmmmm...ok...so, if I put in t instead of T, I get an equation with two unknowns:

$$v = \frac {gT + 0.5g*alpha*T^2} {(1 + alpha *t) + v(initial)}$$

how can one differentiate that?

before I used the following to find a:

(U´ * V - U* V´) / V^2 , as well as the derivate of v(initial) wrt t being g and than putting it all into one term. However, as I missed out g in v(t) (as you told me - thanks) , it should be 1.5 alpha^2 * T^2 instead of 2.5.

That small t still confuses me...but I'm sure you´re right

Last edited: Jan 5, 2006
5. Jan 5, 2006

### LeonhardEuler

This should be:
$$v = \frac {gT + 0.5g*\alpha*T^2} {(1 + \alpha *t)}+v_i$$
You can just use the rule you gave to differentiate this wrt to T to find the acceleration. Keep in mind that $v_i$ is a constant, not a variable, so just treat it as a number in the derivative.

Last edited: Jan 5, 2006
6. Jan 5, 2006

### LeonhardEuler

Wait, I'm sorry-it should be
$$v = \frac {gT + 0.5g*\alpha*T^2+K} {(1 + \alpha *t)}$$
You can't make K/(1+alpha*T) a new constant because it has that T in it. Sorry I missed that.

7. Jan 5, 2006

### *Alice*

cool thanks...so when one now differentiates v wrt T you get

$$a = \frac {g*(1 + alpha*T)} {1 + alpha*t}$$

ok, so T is the time in which the drop falls through the cloud. But what is small t? :uhh: It must be a constant as we integrate wrt to T and not t. If t was just big T a=g....which makes sense as at T the drop leaves the cloud again. But then...why can we exchange t with T now and not before in the expression for v(t).:grumpy:

8. Jan 5, 2006

### qbert

There seems to be a disconnect, somewhere between the immediate and larger
picture. To see what is going on, start a clock when the rain drop falls, it will
measure a time -- call it (t). The questions are, then, what are the velocity and
acceleration of the drop at t=T. (T is just a constant, it's the time the clock measures
when you get through the cloud).
To proceed, you use Newton's second dp/dt = F. Where p = p(t) = m(t)*v(t) and
F = F(t) = m(t)*g. This gives you the differential
equation $(\alpha t + 1) \dot{v} + \alpha v = (\alpha t + 1) g$.
This is a simple first order ODE that you solve for v(t). (That is you solve
for the velocity at an arbitrary time (t)). This, of course, is just an intermediate
step to get what you want, ie v(T).
To find a(T), you first have to differentiate v(t), to give a(t), then evaluate a(T).

9. Jan 6, 2006

### LeonhardEuler

You could have chnged it before, but you would have to be consistent. If you were to change the variable to T, then the expression would read:
$$\frac {d(1 + alpha*T)v)} {dT} = g ( 1+ alpha*T)$$
Where the derivative is with respect to T, otherwise the left hand side would be zero.