Recent content by DDarthVader

That typo is actually not a typo. And I'm trying to say that if the ##n=k## holds then I'll try to prove that ##n=k+1## also holds by doing ##(1+x)^k + (1^x)^{k+1}##. I can write it clearer in my language. But the main problem here is ##(1+x)^k + (1+x)^{k+1}##. Is this correct?

Hello! First of all I have like 5 exercises I don't quite understand so will it be a problem if I create 5 new topics in the next 24h? Homework Statement Prove, by using mathematical induction that if x+1 \geq 0 then (1+x)^n \geq 1+nx. Homework Equations The Attempt at a Solution Basic step...

Wow! Lots of stupid mistakes. Now I see what I did wrong! Thank you guys very much!

Oh, I don't know this formula. Thanks! About the (b), why is it wrong? This is what I did \frac{du}{dx}=\frac{d(\sqrt{5x+8})}{dx}\; \Rightarrow \frac{du}{dx}=((5x+8)^{1/2})' = \frac{1}{2}(5x+8)^{\frac{1}{2}-1}= \frac{1}{2}(5x+8)^{-\frac{1}{2}} =...

Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer. Homework Statement The integral is \int \frac{1}{\sqrt{5x+8}}dx and the exercise tell me to solve this by using (a) u=5x+8 and (b) u=\sqrt{5x+8} Homework Equations The Attempt at a Solution This...

This is what I did: F=ma_1 → mgsen\theta=ma_1(\theta) → -gcos\theta=v_1(\theta) and F=Ma_2 → Mgcos\theta sen\theta = Ma_2(\theta) → -gsen\theta cos\theta =v_2(\theta) Then I plugged these equations on the energy statement and it won't work. I think this is not the way...

As I wrote before ##-mv_{1f}=M_{2f}##. But I don't understand how I use it to solve the exercise.

If the horizontal position of the center of mass of the system remains at rest it means that ##v_{CM}=0## right?

I used both equations to try to calculate the final velocity 2 but my answer is way too different of my textbook. My textbook's answer ##v=\sqrt{2m^2gh(\frac{cos\theta^2}{(M+m)(M+m sin\theta^2)})}## and I got something like ##v=2\sqrt{\frac{mgh}{M(2mM-2)}}##. Lol, I'm terrible at solving...

Can I write ##K_i+U_i=K_f+U_f## for the system as a whole? If yes, in this case we would have ##0+mgh=\frac{m(v_{1})^2}{2}-\frac{M(v_{2})^2}{2}##. This is what I got trying to use the conservation of momentum ##mv_{1i}+Mv_{2i}=mv_{1f}+Mv_{2f}##. I think its incorrect but to me...

The kinetic energy that will be transferred to the system will come from the potential energy of the block. Trying to use ##K_i+U_i=K_f+U_F## where ##U_i=0## and ##K_f## is the kinetic energy of the whole system when the block hit the table and it's the difference in the kinetic energy of the...

But you guys are saying "what's the center of mass of the system" and "what's the velocity of the center of mass". This is what I calculated above right? Well, let me try again using the conversation laws. First of all I must know what is conserved. Well, there is no friction so the kinetic...

The only force that makes the wedge move is ##mgcos\theta sen\theta## so the velocity of the wedge is related to this force. *just a random thought* Check out my draw below. What I'm trying to say is: if we think about the block as a point (point P), we might use those things I wrote in the...

OK. The center of mass of the system has the coordinates ##(\frac{mx_1+Mx_2}{m+M},\frac{my_1+My_2}{m+M})##. Everything with 1 is the block and 2 the wedge. Now the velocity of the center of mass is ##v_{cm}=\frac{mv_1+Mv_2}{m+M}##. This isn't very helpful because I don't know ##x_1## etc...

Hello! I'm here again, now with a new exercise! :biggrin: Homework Statement A block of mass m is at rest over this wedge (it's a triangle) of mass M that is at rest over a table. There is no friction. The point P of the block has a height h from the table. Then the block starts to move. What's...