In just talked to someone about this apparently you use C_v to find the change in thermal energy. So it is just a matter of finding the two temperatures T=PV/nR then the change in temperature T_1-T_2. Then applying (delta)E_th=n*C_v*(delta)T
I'm stuck on this one myself but one thing I know is that the specific heat is the piece of knowledge missing here since the values for the specific heat one can look up are either c_p(for constant pressure) or c_v (for constant volume) and this problem has neither constant volume nor pressure.
I got it!
vp=[2mb/(mb+mp)]*vb
0.5*mvbf^2+mghf=0.5*mvbi^2+mghi
gL(1-costheta)=).5(Vo^2)(3364/16641)
Vo=6.17m/s
although when I actually did it there were a few more intermeadiate steps but I didn't feel it necessary to type out the algebra since square roots and fractions are hard to type.
I think the problem with my calculations was that I used a mass ratio which I think only applies to inelastic but here's what I did
0.5*mp*vp² = mp*g*L*(1 - cosθ)
vp = √[2*g*L*(1 - cosθ)]
0.5*mb*v0² = 0.5*mb*vb² + 0.5*mp*vp²
mb*v0² = mb*vb² + mp*vp²
mb*vb² = mb*v0² - mp*vp²
vb = √[v0²...
Homework Statement
A 29.0 g ball is fired horizontally with initial speed v0 toward a 100 g ball that is hanging motionless from a 1.10 m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle θmax = 50.0°
see image for...
I got the answer :) but for others what I did was
from the graph u can see that potential at highest point is 5J and at lowest point is 1J
mgh+ 1/2mU^2 = mgH + 1/2 mV^2 in this U = initial velocity = 0 and V = max velocity at lowest point and mgh is initial potential energy and mgH is...
The Question
Relevant equations
KE=0.5*m*v^2
T=2*pi*sqrt(m/k)
EE=0.5*k*x^2
KEi+UEi=KEf+UEf
I think that's all of them
attempt at a solution
I was thinking about just using the conservation of energy at the the equilibrium point, as the kinetic energy would be at a max there and...