89.2 Impulse-momentum theorem Solution: ( a)I=P-P:=4p 1.=DD一4,=m-m? =mv cosp+mv=11.6kg. m/s ,=P-Po=4 mv sin -0=4.0kg.m/s Magnitude I=12+12=12.3kg.m/s direction 0= tan =19 89.2 Impulse-momentum theorem (bF 12.3 8200N At0.0015 6=tan=19° Ap (c) According to the Newtons third law, the change in momentum of the bat is equal and opposite to that of the ball. Magnitude 4p=1=v12+1 2 =12.3kg.m/s direction a=180-tan=171
11 Solution : cos 11.6kg m/s ( ) = + = ⋅ = − = = − − f i x fx ix x fx ix mv mv I p p p mv mv φ ∆ = sin − 0 = 4.0kg⋅m/s = − = = − φ ∆ f y fy iy y fy iy mv I p p p mv mv 12.3kg m/s 2 2 = + = ⋅ x y Magnitude I I I o tan 19 1 = = − x y I I direction θ I p p p f i r r r r (a) = − = ∆ §9.2 Impulse-momentum theorem φ x y pi r pf r I r θ (b) 8200N 0.0015 12.3 = = = t I Fav ∆ r r o tan 19 1 = = − x y I I θ (c) According to the Newton’s third law, the change in momentum of the bat is equal and opposite to that of the ball. 12.3kg m/s 2 2 = = + = ⋅ x y Magnitude ∆p I I I o o 180 tan 171 1 = − = − x y I I direction α §9.2 Impulse-momentum theorem α φ x y pi r pf r I r θ bat p r ∆
89.2 Impulse-momentum theorem Example 3: Coal drops from a stationary hopper of height 2.0 m at rate of 40 kg/s on to conveyer belt moving with a speed of 2.0 m/s, find the average force exerted on the belt by the coal in the process of the transportation 89.2 Impulse-momentum theorem Solution: Choose the mass element Jy of the coal as the particle 4=P2x-P1x=P2=△ P1 2-n=4 4,=P2y-n1y=n1=△Mm △m2gh v=2.0m/s According to the impulse-momentum theorem i=Fdt=F4t=4p
12 h A v r Example 3: Coal drops from a stationary hopper of height 2.0 m at rate of 40 kg/s on to conveyer belt moving with a speed of 2.0 m/s, find the average force exerted on the belt by the coal in the process of the transportation. §9.2 Impulse-momentum theorem m gh p p p p mv p p p p mv y y y x x x 2 2 1 1 2 1 2 = ∆ ∆ = − = = ∆ ′ ∆ = − = = ∆ According to the impulse-momentum theorem I F t F t p av t t f i r r r r = = ∆ = ∆ ∫ d Solution: Choose the mass element of the coal as the particle v = 2.0m/s p1 r p2 r x y △m p p p r r r 2 − 1 =∆ α β §9.2 Impulse-momentum theorem
89.2 Impulse-momentum theorem ∠mv =yv=80(N) F=4n=B=4 va\ 4-a==q√2gh=1252(N) P F=√Fxa+F1a=149(N) PI The angle with respect ′a to the x axis a=tanf=tan1 125.2_57.4 rav 89.2 Impulse-momentum theorem the average force exerted on the belt by the coal F"=149(N) B=180°-574=1226 △m 2-=4 B
13 2 125.2 ( ) 80 ( ) 1 av 2 av q gh N t m t p t p F qv N t mv t p t p F y y x x = = = = = = = = = = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ 149 ( ) Fav = Fxav + Fyav = N o 57.4 80 125.2 tan tan-1 av -1 av = = = x y F F α p1 r p2 r x y △m p p p r r r 2 − 1 =∆ α β The angle with respect to the x axis §9.2 Impulse-momentum theorem o o o 180 57.4 122.6 149 ( ) = − = ′ = β F N the average force exerted on the belt by the coal p1 r p2 r x y △m p p p r r r 2 − 1 =∆ α β §9.2 Impulse-momentum theorem