2. A stream of water impinges on a stationary"dished"turbine blade as shown in Fig. 2. The speed of the water is u, both before and after it strikes the curved surface of the blade and the mass of water striking the blade per unit time is constant at the value u. orce exerted by the water on the blade is 2up Solution Fig 2 Using the impulse-momentum theorem: I=Ap=FeAt △m-(-△m)=2△mu=Fa△ 2△mu Thus force exerted by the water on the blade is F =2l △ 2 3. A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle 0 of 33.0 and then rebounds with the same speed and angle(Fig 3). It is in contact with the wall for 10.4 ms(a)The impulse was experienced by the wall is_2. 17NS 6 (b) The average force exerted by the ball on the wall is _2.09x10-N Solution Using the impulse-momentum theorem: I= Ap= F At So the impulse was experienced by the wall is I=Ap,=mvsin8-(-mvsin0)=2mvsin8=2 x0.32 6.22 xsin33=2.17NS The average force exerted by the ball on the wall is =2.09×102N △At104×10 4. The muzzle speed of a bullet can be determined using a device called a ballistic pendulum, shown in Figure 4. A bullet of mass m moving at speed v encounters a large mass M hanging vertically as a pendulum at rest. The mass M absorbs the bullet. The hanging mass(now consisting of M+ m) then swings to some height h above the initial position of the pendulum as shown. The initial M The muzzle speed v of the bullet is Fig 4 m+M 10.0 cm, M=2.50 kg, and m=10.0 g. The muzzle speed v of the bullet Solution2. A stream of water impinges on a stationary “dished” turbine blade, as shown in Fig.2. The speed of the water is u, both before and after it strikes the curved surface of the blade, and the mass of water striking the blade per unit time is constant at the value µ. The force exerted by the water on the blade is 2uµ . Solution: Using the impulse-momentum theorem: I p F t = ∆ = ave∆ v v v So mu mu mu F t ∆ − (−∆ ) = 2∆ = ave∆ Thus force exerted by the water on the blade is uµ t mu Fave 2 2 = ∆ ∆ = 3. A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle θ of o 33.0 and then rebounds with the same speed and angle (Fig.3). It is in contact with the wall for 10.4 ms. (a) The impulse was experienced by the wall is 2.17Ns . (b) The average force exerted by the ball on the wall is 2.09×102 N . Solution: Using the impulse-momentum theorem: I p F t = ∆ = ave∆ v v v So the impulse was experienced by the wall is = ∆ = sin − (− sin ) = 2 sin = 2× 0.32× 6.22×sin 33 = 2.17 N ⋅s o I px mv θ mv θ mv θ The average force exerted by the ball on the wall is 2.09 10 N 10.4 10 2.17 2 3 = × × = ∆ = − t I Fave x 4. The muzzle speed of a bullet can be determined using a device called a ballistic pendulum, shown in Figure 4. A bullet of mass m moving at speed v encounters a large mass M hanging vertically as a pendulum at rest. The mass M absorbs the bullet. The hanging mass (now consisting of M + m) then swings to some height h above the initial position of the pendulum as shown. The initial speed v′ of the pendulum (with the embedded bullet) after impact is m M mv + . The muzzle speed v of the bullet is gh m M m 2 + . If h = 10.0 cm, M = 2.50 kg, and m = 10.0 g. The muzzle speed v of the bullet is 352m/s . Solution: h Fig.4 M m v r g. 2 θ m θ Fig.3 u r u r − Fig. 2