Recent content by jmlibunao

So what would be my initial velocity for #2? 0?

Don't worry about that. I know how to solve differential equations :) After solving this I'm going to be setting the velocity to 0, right? Which one, u or v, or does it not matter?

Thanks for your reply! I forgot that the kinematic equations are only applicable for constant acceleration. What differential equation are you talking about? dv/dt = k(2u - v) ?? If this is right/wrong, then can you guide me through it? I'm just really having a hard time with this problem.

I put the complete problem. That is everything given. :\ Impulse is change in momentum right? At the start momentum is 0 since there was no movement. After the release from rest, then you would have momentum so shouldn't there be impulse?

Homework Statement A particle of mass m kg is traveling in a horizontal straight line with a velocity u m/s. It is brought to rest by means of a resisting force of magnitude km(2u - v), where v is the velocity of the particle at any instant and k is a positive constant. Find the distance...

Thanks for your reply for #2! For #3, when you release the system from rest, the 8-kg mass would go down while the 5-kg mass would go up, thus there would be a speed/velocity that would cause this movement. Following that, it's also asking for the impulse experienced by the string. I think...

It's the #1 question. So I computed my k = 200. To compute for u can I use the following equation? (1/2)(200)(0.2^2) + (1/2)(4 kg)(u^2) = (1/2)(4 kg)(10) The velocity u is just after the impact so the string would still be stretched at that moment so I included (1/2)(200)(0.2^2) on the...

Homework Statement Two particles of masses 8 kg and 5 kg are connected to the two ends of a light inextensible string which passes over a fixed smooth pulley. Initially each of the two particles are held at a position which is 5 m above a horizontal ground. The objects are then released from...

Okay, thanks for clearing that up! So I can just use kx = mg and just solve for k. But what about the modulus of elasticity? You say I don't need it but the problem is asking for it. Okay, now I remember. The potential energy stored in the string would be E = (1/2)kx^2, the same equation...

I'm sorry, I don't quite follow this. How would I be able to get the force per unit extension by considering the initial equilibrium? The modulus of elasticity is being asked by the problem. I can't recall a formula that gives the potential energy when hanging from a string. I can't use P =...

Wouldn't u be the initial velocity to be used when solving for #2 and #3? Once the string breaks you would have a new system (free fall) where in A would have an initial downward velocity of u while B would have an initial upward velocity of u as well. Is my logic correct? For #3, what I'm...

Yeah, but at the moment the string breaks the initial velocity would be u, right?

Homework Statement A light elastic string has a natural length 1 m. One end of the string is attached to the fixed point O and a particle P of mass 4 kg is suspended from the other end of the string. When hanging in equilibrium, P is 1.2 m below O. #1 Find the modulus of elasticity of the...

Okay. So I can get the respective displacements of each particle by using df = di + vt + (1/2)(a)(t^2), where v = u, and t = u/3.33. What I got for A was df = di - (u^2)/3.33 + (u^2)/2 [I subtracted them since A would be going down] while for B df = di - (u^2)/3.33 + (u^2)/2 [Added them since B...

I would use the kinematic equation vf = vi + at with my vi = 0 (since the system was released from rest) and my vf to be u. Solving for time, t, I'd get t = u/3.33 Wouldn't the acceleration the moment the string breaks still be 3.33 m/s^2 ??