Recent content by flamespirit919

Thank you! I wasn't sure which equation to plug the values into. Here's what I did $$k(Bcos(\omega t)-Acos(\omega t))=-m_1A\omega ^2cos(wt)$$$$k(Bcos(\omega t)+\frac{m_2B}{m_1}cos(\omega t)=m_2B\omega ^2cos(\omega t)$$$$kBcos(\omega t)(1+\frac{m_2}{m_1})=m_2B\omega ^2cos(\omega...

I tried setting them equal to each other $$-m_1A\omega ^2cos(\omega t)=m_2B\omega ^2cos(\omega t)$$ But I end up with $$-m_1A=m_2B$$ I don't know what I can do after this.

Whoops. They should be fixed now.

No, so would it be $$k(x_2-x_1)=m_1\frac{d^2x_1}{dt^2}$$$$-k(x_2-x_1)=m_2\frac{d^2x_2}{dt^2}$$ Assuming that ##m_1## is on the left and ##m_2## is on the right.

The spring would be stretched by 2 units. Would the equations then be $$-k(x_2-x_1)=m_1\frac{d^2x_1}{dt^2}$$$$-k(x_2-x_1)=m_2\frac{d^2x_2}{dt^2}$$

So would it be $$-k(x_1+x_2)=(m_1+m_2)(\frac{d^2x_1}{dt^2}+\frac{d^2x_2}{dt^2})$$$$-k(Acos(\omega t)+Bcos(\omega t)=(m_1+m_2)(-A\omega ^2cos(\omega t)-B\omega ^2cos(\omega t)$$ Which simplifies to $$k=\omega ^2 (m_1+m_2)$$ But, what can I do with this? Or did I go about this the wrong way?

Homework Statement Two masses ##m_1## and ##m_2## are joined by a spring of spring constant ##k##. Show that the frequency of vibration of these masses along the line connecting them is: $$\omega =\sqrt{\frac{k(m1+m2)}{m1m2}}$$ Homework Equations ##x(t)=Acos(\omega t)## ##\omega...

Thank you! I wish I had seen that sooner.

That makes sense, thank you very much.

The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.

I think I got it. So the momentum in the x and y-direction is $$v_1=v_1^{'}cos(\theta _1)+v_2^{'}cos(\theta _2)$$ $$0=v_1^{'}sin(\theta _1)-v_2^{'}sin(\theta _2)$$ I then used the following to find the magnitude of the momentum $$p^2=p_x^2+p_y^2$$ Doing this I got $$v_1=v_1^{'2}(cos^2(\theta...

I forgot to square the initial velocity. Were there any other mistakes?

Homework Statement Show that if an elastic collision between a mass and a stationary target of equal mass is not head-on that the projectile and target final velocities are perpendicular. (Hint: Square the conservation of momentum equation, using ##p^2=p\cdot p##, and compare the resulting...

So the force will be 5*10-5 N no matter the time frame, right? Thank you so much. It makes much more sense now.

So the spaceship will accumulate dust at ##5\times10^{-11}~kg/s##, but I'm still not quite sure how to figure out the acceleration. Do I use one of the kinematic equations?