S11.1momentumPx = -mv, cos(90° - ot)- 2mv, cos(90° -ot)-m'v5m + 4m')lo sin 0t2(5m+4m)lp=p?+p, =, = my, sin (90° - ot)+ 2mv, sin (90° - ot)+ m'vC5m + 4m')l@ cos ot2p, cot ottan αp
( ) ( ) ( m m )l t p mv t mv t m v x B 5 4 sin 2 1 1 cos 90 2 2 cos 90 = − + = − − − − − p = px + py = (5m + 4m )l 2 2 2 1 ( ) ( ) ( m m )l t p m v t m v t m v y A 5 4 cos 2 1 1 sin 90 2 2 sin 90 = + = − + − + t p p x y tan = = cot §11.1 momentum
$11.1momentum2. impulsei=Ft2.1lmpulseofa constantforce:2.2ImpulseofvariableforceImpulseofelement:impulseofvariableforceinmicrotimeperiod; That is:di = FdtThen the impulse of the force in the time period (t, -t, )isi = ( Fdtunit: N.st
2. impulse 2.1 Impulse of a constant force： I Ft = 2.2 Impulse of variable force d d I F t = Impulse of element: impulse of variable force in micro time period; That is: Then the impulse of the force in the time period is: ( ) 1 2 t −t 2 1 t t I F t = d unit：N·s §11.1 momentum
$ 11.2 momentumtheorem1. Momentum theorem for particlesF=maNewton's second lawdi:F=ma=mdt..d(mv)= F.dtwhen m is constant-Differentialequationof themomentumtheoremThat is, the momentum increment of the particle is equal tothe elemental impulse of the force acting on the particle
1. Momentum theorem for particles Newton's second law F ma = = d mv F dt ( ) dv F ma m dt = = －Differential equation of the momentum theorem That is, the momentum increment of the particle is equal to the elemental impulse of the force acting on the particle. §11.2 momentum theorem when m is constant
$ 11.2 momentum theoremTime period: t, →t, Speed change: V, →>V,d(mv)= F.dtIntegrate the formula:mV, - mVi = f" F . dt = I - The integral fthe momentumtheoremThat is, in a certain time interval, the change in momentum oftheparticle is equal to the impulse of the force acting on the particleduring this period
1 2 t →t 1 2 Time period: Speed change: v →v Integrate the formula: d mv F dt ( ) = 2 1 2 1 t t mv mv F dt I − = = －The integral of the momentum theorem That is, in a certain time interval, the change in momentum of the particle is equal to the impulse of the force acting on the particle during this period. §11.2 momentum theorem
$11.2 momentumtheoremExample2 Heavy hammer Q = 300 N, free fall from height H = 1.5 m forging, asshown in figure, forging deformation occurs, for t = 0.01 s for hammer averagepressure of forgings.Solution: Take the hammer as the research object. Theforce acting on the hammer is the counterforceof thehforging after the q-hammer and the forging contact. Butthe reaction force of the forging is variable. Set anaverage reaction of n. time needed for hammer dropheight H T as follows:2H1g
Example2 Heavy hammer Q = 300 N, free fall from height H = 1.5 m forging, as shown in figure, forging deformation occurs, for t = 0.01 s for hammer average pressure of forgings. Solution: Take the hammer as the research object. The force acting on the hammer is the counterforce of the forging after the q-hammer and the forging contact. But the reaction force of the forging is variable. Set an average reaction of n. time needed for hammer drop height H T as follows: h g H T 2 = §11.2 momentum theorem