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I tried using 1/2(m1+m2)v2=(m1+m2)gy. I tried solving for y, but I am not getting the right answer. Can you explain where I am going wrong?

alright so I did do it right. The initial height of ball 1 equals the final height of ball 2 because the collision is elastic. So how do I approach part B?

so I set 1/2mv^2 equal to mgh, and solved for the velocity. I got 1.95 m/s. Am I doing this right?

How do I find the height of the first ball?

can you explain part A more. How do I apply conservation of energy and conservation of momentum?

[SOLVED] collision problem Homework Statement Two identical steel balls, each of mass 3.4 kg, are suspended from strings of length 27 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle theta = 74o with the vertical and let it...

[SOLVED] Conservation of Momentum Homework Statement Two blocks with masses m1 = 2.6 kg and m2 = 3.6 kg are at rest on a frictionless surface with a compressed spring between them. The spring is initially compressed by 60.0 cm and has negligible mass. When both blocks are released...

never mind I got it, thanks

What do I put in for m1 and m2?

so then I just take 125 and divide by -9.8?

can you explain to me how to do that?

So the initial velocity is 125 m/s. The accel is -9.8 m/s^2. I still need the distance right? And then I solve for the final velocity?

I am still confused, I don't understand how to figure out the height.

I'm guessing that you are to assume that all of the KE goes into the second half. Can you help me out, I'm lost?

[SOLVED] Horizontal Speed of bullet fragment Homework Statement A 14-kg shell is fired from a gun with a muzzle velocity 125 m/s at 33o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the...

alright, I got it, thanks!

Then once I get the acceleration, I just multiply by 2450 kg?

The acceleration is negative, right. So doesn't that make the force negative? The van reaches 49.27 mph at 11 sec. So what other point should I use to find the slope?

molecular mass of oxygen as a molecule is 32 grams because it is O_2. The molar mass is 16 grams for oxygen atom.

Homework Statement The graph shows the speed as a function of time for a minivan, coasting on neutral along a straight, level road, loaded with windsurfing equipment and towing a boat.The total mass is 2450kg (weight = 5402lbs). Find the size of the force due to wind and rolling friction when...

or do I just set mgh=204 and solve for h?

So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?

Can you show me how to set up the equations that I need?

So then can the lowest the height at Point P only be equal to the height of Point A which is 2r?

I haven't done this before so I am really confused right now. I am not sure what it tells about the mechanical energy at point P.

I understand that it is a constant, so does that mean KE+PE equals the answer from part A?

I am not sure what to do from here. That's where I am stuck at

So when I take the velocity from the centripetal acceleration, and plug it into KE=1/2mv^2, I get 114.63 J. Is this right so far?

I am not sure how to do that. Could you please explain it more to me?

For part C I don't even know where to start. mv^2/r must be greater than or equal to mg, right?

For part B, I just take the KE that I solved for and set it equal to 1/2mv^2 and solve for v. Then v^2/r=a, right?

Alright so I got part A. I found the PE at point P and subtracted the PE at point A. For part B, wouldn't the acceleration just be equal to g?

I don't know v at P...to get that do I set KE=2PE

I am confused on point A. How would I solve for KE if I don't have a velocity? How do I solve for that quantity?

How do I figure out a velocity to use in the equation?

is that where you use KE_1+PE_grav?

They are both constant? right?

Well what I tried to do was set 1/2mv^2=2mgh, and solve for a velocity. Then I tried to plug that into KE=1/2mv^2, but I don't think that is right. Is the total mechanical energy conserved?

Homework Statement A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P at rest a distance h = 52.0 m above the bottom of the loop of radius R = 18.0 m. A)What is the kinetic energy of the mass at the point A on the...

W_f=umgd KE=1/2mv^2 W_spring=1/2kx^2 How do you relate these three equations to get a speed? That is what I am struggling with.

Does the kinetic energy go against the friction force. So would it be F_fric=U_k*mgd?

Homework Statement A moving 1.3 kg block collides with a horizontal spring whose spring constant is 491 N/m.The block compresses the spring a maximum distance of 5.0 cm from its rest postion. The coefficient of kinetic friction between the block and the horizontal surface is 0.49.What is the...

Oh never mind I got part B. Can you help me with part C though?

Is it F_k=U_k *N?

Is it -U_k*mgd?

Alright, it accepted the answer. I am lost when it comes to Part B

So then the answer should be -.61375 J ? So then how do I do the same thing with friction for part B?

I tried it again and it said that it was wrong. We must have gone wrong somewhere...

It says that the answer for part A is wrong though. Doesn't the coefficient of the kinetic friction play into the problem somewhere?

So is that all you have to do for part A?