University Physics AI No. 7 Impulse, Momentum, and Collisions Class Number Name I．Choose the Correct Answer 1. An object is moving in a circle at constant speed v. The magnitude of the rate of change of momentum of the object （ C ） (A) is zero (B) is proportional to v. (C) is proportional to v 2 . (D) is proportional to v 3 . Solution: The magnitude of the centripetal acceleration is constant for uniform circular motion, that is constant 2 == r v a c , Using the general statement of Newton’s law of motion t p F total total d d v v = , we have 2 2 d d v r v mmaF t p ctotal total ∝=== v v 2. If the net force acting on a body is constant, what can we conclude about its momentum? (A) The magnitude and/or the direction of p r may change. （ A ） (B) The magnitude of p r remains fixed, but its direction may change. (C) The direction of p r remains fixed, but its magnitude may change. (D) p r remains fixed in both magnitude and direction. Solution: Using the general statement of Newton’s law of motion t p F total total d d v v = , we can know the magnitude and/or the direction of p r may change. 3. If I r is the impulse of a particular force, what is tI d/d r ? （ C ） (A) The momentum (B) The change in momentum (C) The force (D) The change in the force Solution: According the definition of the impulse, F t I tFI r r rr =?= d d dd . 4. A variable force acts on an object from 0= i t to f t . The impulse of the force is zero. One can conclude that （ C ） (A) 0=?r r and 0=?p r . (B) 0=?r r but possibly 0≠?p r . (C) possibly 0≠?r r but 0=?p r . (D) possibly 0≠?r r and possibly 0≠?p r Solution: According to the impulse-momentum theorem ptF r r ?= ∫ f i t t d，the impulse of the force is zero， the change of the momentum 0=?p r . 5. A system of N particles is free from any external forces. Which of the following is true for the magnitude of the total momentum of the system? （ B ） (A) It must be zero. (B) It could be non-zero, but it must be constant. (C) It could be non-zero, and it might not be constant. (D) The answer depends on the nature of the internal forces in the system. Solution: According to the general statement of Newton’s law of motion constant,0 , d d total total total == = pF t p F r r v v II. Filling the Blanks 1. Fig.1 shows an approximate representation of force versus time during the collision of a 58g tennis ball with a wall. This initial velocity of the ball is 32m/s perpendicular to the wall; it rebounds with the same speed, also perpendicular to the wall. The value of max F , the maximum contact force during the collision is 928N . Solution: Applying to the impulse-momentum theorem ∫ =?= tFpI d v v v So max 3 3 2 10)62( 3210582 d2 F StFmvp F × ×+ =×××? ===? ? ? ∫ v v N928 max =F 0123456 F max force (N ) time (ms) Fig. 1 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 μ. The force exerted by the water on the blade is μu2 . Solution: Using the impulse-momentum theorem: tFpI ave ?=?= v v v So tFmumumu ave ?=?=???? 2)( Thus force exerted by the water on the blade is μu t mu F ave 2 2 = ? ? = 3. A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle θ of o 0.33 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 2.09×10 2 N . Solution: Using the impulse-momentum theorem: tFpI ave ?=?= v v v So the impulse was experienced by the wall is sN17.233sin22.632.02sin2)sin(sin ?=×××==??=?= o θθθ mvmvmvpI x The average force exerted by the ball on the wall is N1009.2 104.10 17.2 2 3 ×= × = ? = ? t I F xave 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 Mm mv + . The muzzle speed v of the bullet is gh m mM 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 θ m Fig.3 u r u r ? Fig. 2 (a) Use conservation of momentum Mm mv vvMmmv + =?+= '')( (b) Assume the position which the pendulum is at rest is zero potential energy position, use CWE theory. gh m mM v m mM vghvghmMvmM 2'2')(')( 2 1 2 + = + =?=?+=+ (c) m/s3521081.92 1010 5.21010 2 3 3 =××× × +× = + = ? ? gh m mM v 5. Two objects, A and B, collide. A has mass 2.0 kg, and B has mass 3.0 kg. The velocities before the collision are jiv iA ? )m/s30( ? )m/s15( += r and jiv iB ? )m/s0.5( ? )m/s10( +?= r . After the collision, jiv fA ? )m/s30( ? )m/s0.6( +?= r . The final velocity of B is )m/s( ? 5 ? 4 ji + . Solution: Applying the conservation of the momentum, we have BfBAfABiBAiA vmvmvmvm vvvv +=+ So the final velocity of B is )m/s( ? 5 ? 4 ? 5 ? 10) ? 30 ? 6 ? 30 ? 15( 3 2 )()( 1 jijijijiv vvv m m vmvmvm m v Bf BiAfAi B A AfABiBAiA B Bf +=+??++×=? +?=?+= v vvvvvvv III. Give the Solutions of the Following Problems 1. A massive anchor chain of length l is held vertical from its top link so that its lowest link is barely in contact with the horizontal ground. The chain is dropped. When the top link has fallen a distance y (see Figure 5), show that the magnitude of the normal force of the ground on the chain is 3 mgy/l. Solution: Choose the anchor chain on the ground as the study object, which have length y and mass y l m m =' . The chain on the ground get the gravitation gm'，the force N acted by the floor and impulse force T acted by the falling chain. So Origin l y i ? Fig.5 Tyg l m NNTyg l m +=?=?+ 0 Choose md as the study object: y l m m dd = When it reaches the ground its speed is: gyv 2= Using impulse-momentum theorem: gyy l m tTmg 2dd)'d( ?=? Since 'd Tmg << , we can neglect it, gy l m gyv l m gy t y l m T 222 d d ' === Use Newton’s third law 'TT = , so gy l m T 2?= Then we have the magnitude of the normal force of the ground on the chain is l mgy gy l m yg l m N 3 2 =+= 2. A fire hose fixed to a support delivers a horizontal stream of water at a speed v 0 and a mass flow rate β kilograms per second. (a) The water stream is directed against a wall along a perpendicular to the wall and is brought to rest. (see Figure 6(a))What the magnitude of the force of the water on the wall? (b) What is the momentum per unit length of the stream? (c) The stream now is directed into a hole in the back of a tank car (see Figure 2(b)) that is free to roll from rest (with no frictional work done). The tank car has an initial mass m 0 . Let the total mass of the tank car be m (tank car plus accumulated water). Find the x-component of the velocity of the tank car at time t. Solution: (a) Assume the time interval during the water act on the wall is t? , and the mass of the water which act on the wall is tm ?=? β . We choose m? as the study object, which get the average force from the wall is f. Using impulse-momentum theorem. (the direction of right is i ? ） 0 v r (a) 0 v r v r m (b) Fig.6 iv t imv fimvtfP ? ? ? 0 0 0 0 β?= ? ?? =???=?=? rrr (Unparallel to 0 v r ) Using Newton’ third law, we can have the magnitude of the force acted on the wall by the water is 0 vβ ( b) The momentum of the water in length s? is pp rr ?= , the momentum of the water in unit length is i v iv t s iv t m s imv s P ? ? ? ? 0 0 0 0 β β == ? ? ? ? = ? ? = ? r (c) Regard the water and the car as a whole system, and choose it as the study object. mvvvm vvv P ivvmiP wc x wcx )d-(d - 0d ? )dmd( ? d 0 0 =? ? ? ? ? ? = = += i. Using conservation of momentum, at any time t 0 0 00 )1()( v m m vvmmmv ?=??= ii. Use the result in (c), we can have v m m v m m vv v m m vv v m m vv v m m v )1(lnln dddd 0 00 0 00 0 ?=?= ? ? = ? ?= ? ∫∫ 3. A rocket at rest in space, where there is virtually no gravity, has a mass of kg1055.2 5 × , of which kg1081.1 5 × is fuel. The engine consumes fuel at the rate of 480 kg/s, and the exhaust speed is 3.27 km/s. The engine is fired for 250 s. (a) Find the thrust of the rocket engine. (b) What is the mass of the rocket after the engine burn? (c) What is the final speed attained? Solution (a) the thrust of the rocket engine is )N(1057.14801027.3 d d 63 ×=××== t m vF e (b) The mass of the rocket after the engine burn is )k(1035.12504801055.2 d d 45 0 gt t m mm ×=×?×=?×?= (c) The final speed is m/s1005.4 1055.2 1081.11055.2 ln1027.3ln0 3 5 55 3 ×= × ×?× ×?==?=? i f efif m m vvvv 4. A barge with mass kg1050.1 5 × is proceeding downriver at 6.20 m/s in heavy fog when it collides broadside with a barge heading directly across the river; see Fig. 7. The second barge has mass kg1078.2 5 × and was moving at 4.30 m/s. Immediate after impact, the second barge finds its course deflected by o 0.18 in the downriver direction and its speed increased to 5.10 m/s. The river current was practically zero at the time of the accident. What is the speed and direction of motion of the first barge immediately after the collision? Solution: The coordinate system is shown in figure. Apply conservation of momentum. ffii vmvmvmvm 22112211 vvvv +=+ So ji jiji vv m m vv fiif ? 01.1 ? 31.3 )] ? 18cos1.5 ? 18sin1.5( ? 3.4[ 105.1 1078.2 ? 2.6 )( 5 5 22 1 2 11 ?= +? × × += ?+= oo vvvv )m/s(46.301.131.3 ? 01.1 ? 31.3 22 1 =+=?= jiv f Suppose the angle between the first barge and x direction is θ . Then o 22.17)31.0arctan(31.0 31.3 01.1 tan ?=?=??= ? = θθ 5. A system of two spheres of identical mass m collide obliquely as shown in Figure 8. The collision is inelastic. Initially one of the masses is at rest. The momenta of the particles immediately after the collision are 1 p′ r and 2 p′ r . The angle between the paths of the masses after the collision is β. What is the kinetic energy of the system ? β 1 P′ r 2 P′ r Before collision After collision Fig.8 18 ° 1 2 Fig. 7 x y θ Solution: Apply the conservation of momentum )cos('cos' 211 θβθ ?+= PPP m P m P m P KEKEKE m P m vm mvKE PP if 22 ' 2 ' 222 1 )sin('sin'0 2 1 2 2 2 1 222 2 21 ?+=?=??=== ?+= θβθ Using the adding of the vectors m PP KE PP m PPP m KE PPPPP β β β cos'' )cos''2( 2 1 )''( 2 1 cos''2'' 21 21 22 2 2 1 21 2 2 2 1 2 1 ?=?? ??=?+=?? ++=