Recent content by TraceBusta

I'm having trouble setting up these equations. For conservation of energy, I have the initial KE, but final KE I don't know what the final kinetic energies of each ball is. I think the incoming ball is (1/2)m(11 m/s)^2*sin (30) but the final kinetic energy of the ball initially at rest I don't...

A ball moving at 11 m/s makes an off-center elastic collision with another ball of equal mass that is initially at rest. The incoming ball is deflected at an angle of 30° from its original direction of motion. Find the velocity of each ball after the collision. Ball moving initally at 11 m/s...

A 4.5 kg object with a speed of 4.0 m/s collides head-on with a 11 kg object moving toward it with a speed of 3.0 m/s. The 11 kg object stops dead after the collision. (a) What is the final speed of the 4.5 kg object? (b) Is the collision elastic? so i tried initial momentum = final...

Well I think conservation of momentum is needed here. Jay and Dave decide that the best way to protest the opening of a new incinerator is to launch a stink bomb into the middle of the ceremony. They calculate that a 5.5 kg projectile launched with an initial speed of 36 m/s at an angle of...

What if you want to find the center of mass of a triangle? By symmetry, locate the y-coordinate of the center of mass of an equilateral triangle of side length l=95cm located with one vertex on the y-axis and the others at (-l/2,0) and (l/2,0). does this also involve integrating?

I know how to find the center of mass of a 2 dimensional object like a piece of plywood or something like that, but when it comes to 3-D objects I'm clueless. All I know is Mx=x1m1+x2m2+...xnmn here is my problem, i don't know if i should split the pyramid into 4ths or not. The Great...

Awesome, thanks for your help. I figured out c) by doing -(a)-(b)=1/2 mv^2 and solved for v=1.11 m/s

Ok. I did Wf=(mu)mg*.0687m and got 4.0899J which is wrong. /edit I put a - sign and it was right. edit/

i think i could solve b) if i knew the answer to c) because then I could use 1/2kx^2-1/2mv^2=(b)?

for part a i put a - in the answer and it was right..thnks. i'm still confused as to continue on with parts b) and c)

with deltaE do you mean change from Kinetic energy to potential energy of the spring?

bump...anyone? I've tried to solve for c) by doing 1/2 kx^2=1/2mv^2 and solve for v=.4664 m/s but that is wrong.

The spring constant of a toy dart gun is 4950N/m. To cock the gun the spring is compressed 3cm. The 4g dart, fired straight upward, reaches a maximum height of 39.7362385321101m, measured from the compressed position of the spring. Determine the energy dissipated by air friction during the...

A moving 8.07 kg block collides with a horizonal spring whose spring constant is 372 N/m. The block compresses the spring a maximum distance of 6.87 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.752. (a) How much work is done by...

i set up the problem like this U=mg(h+x sin (theta)) Uspring=1/2 k x^2 so mg(h+x sin (theta)) = 1/2 kx^2? solving for x, x1=1.1331, x2=-1.0965 those are both wrong.