1work =?
W=F.S=FScos(theta)
21erg , 1eV,1 cal,1kWh how many joule
1erg=10^-7J 1eV=1.6*10^-19J 1calorie=4.2J 1kWh=3.6*10^6J
3work done using work energy theorem
wext=∆K.E=(1/2)m((v2)^2-(v1)^2)
4work done in ping the chain
MgL/2(n^2)
5work done by conservative force formula
W=-∆U
6elastic potential energy
U=(1/2)Kx^2=1/2 (Fx)= 1/2 (F^2/k)
7Force and potential relationship
F=-dU/dx
8stable equilibrium
(d^2U/dx^2)>0
9unstable equilibrium
(d^2U/dx^2)<0
10neutral equilibrium
(d^2U/dx^2)=0
11kinetic energy
K.E=(1/2)mv^2=p^2/2m=(1/2)pv
12p and K.E relation
p=√(2mK.E)
13percentage change in kinetic energy
∆K.E/K.E *100% =(Pf^2-Pi^2/Pi^2)*100%
14percentage change in momentum
(∆p/p)*100%=(√K.Ef -√K.Ei/√K.Ei)*100%
15conservation of mechanical energy
Uf+Kf=Ui+Ki
16conservation of energy for spring mass system
k(x(max))^2 =kx^2+mv^2 , vmax=x(max)√(k/m)
17vertical circular motion Tension
T=mv^2/R+mgcos(theta) v is the velocity at that point
18vertical circular motion Tension with Va the point at the lowest point(special case theta =0°,90°,180°
T=m(Va)^2/R+3mgcos(theta)-2mg T(0°)=mva^2/R+mg T(90°)=mva^2/R-2mg T(180°)=mva^2/R-5mg
19critical velocity (at highest point)
vc>or = √(Rg)
20to complete and full circle from lowest point velocity?
V=√(5gR)
21velocity at lowest point to reach theta=90°
v=√(2gR)
22Va=√(5gR) T at 0° ,90°,180°
T(0°)=6mg T(90°)=3mg T(180°)=0
23power(average, instantaneous,)
average=work/time instantaneous=Fvcos(theta) =F.v 1hp=746watt
24constant power relationship with (x,v,a,t)
v=√(2Pt/m) x proportional to t^3/2 v proportional to t^1/2 accleration proportional to t^-1/2
25conservation of momentum
m1u1+m2u2=m1v1+m2v2
26coefficient of restitution (e)
e=|v2-v1|/|u1-u2|
27loss of kinetic energy with (e)
∆K=m1m2/2(m1+m2) *(u1-u2)^2*(1-e^2)
28perfectly elastic collision in 1D (v1,v2)
v1=(m1-m2)u1+2m2u2/(m1+m2) v2=(m2-m1)u2+2m1u1/(m1+m2)
29for velocity to be interchanged what is the condition of their mass
m1=m2
30if lighter mass colides with heavy target
v1=-u1 ,v2=0
31If heavier r mass colides with lighter target
v1=u1 v2=2u1
32(∆K/K) retained=?
(∆K/K) retained= (m1-m2/m1+m2)^2
33(∆K/K) transferred/lost=?
(∆K/K) transferred/lost=4m1m2/(m1+m2)^2
34(∆K/K) transferred/lost=? if m1=nm2
4n/(n+1)^2
35final velocity of perfectly inelastic collision
v=m1v1+m2v2/(m1+m2)
36elastic collision in 2D conservation of momentum
x direction: m1u1=m1v1cos(theta1)+m2v2cos(theta2) y direction : 0=m1v1sin(theta1)-m2v2sin(theta2) here theta1+theta2=90°
37elastic collision in 2D conservation of momentum (kinetic energy
(1/2)m1u1^2=(1/2)m1v1^2+1/2m2v2^2
38rebound of ball after collision with ground (height , velocity,timetaken, total distance travelled by ball
vn= e^n v° , hn=e^2n h° ,d=h°[1+e^2/1-e^2], T=(1+e/1-e)*√(2h°)