University Physics Al No. 7Impulse, Momentum, and Collisions Class Number ame I. Choose the Correct Answer An object is moving in a circle at constant speed v. The magnitude of the rate (A)is zero(B)is proportional to v. (C)is proportional to v2.(D)is proportional to y c) momentum of the object Solution The magnitude of the centripetal acceleration is constant for uniform circular motion, that constant Using the general statement of Newton s law of motion Fotal m F,=ma =m-ov 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 may change (B)The magnitude of p remains fixed, but its direction may change (C)The direction of p remains fixed, but its magnitude may change D) p remains fixed in both magnitude and direction Solution Using the general statement of Newton's law of motion Fta= -total, we can know the magnitude and/or the direction of p may change 3. If I is the impulse of a particular force, what is d/ /dt? (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, di=Fdt==F
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 ac , 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 F ma m t p total c 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 dI / dt 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 I F t r r r r = ⇒ = d d d d
4. A variable force acts on an object from 4 =0 to t,. The impulse of the force is zero. One can conclude that (A)Ar=0 and Ap=0 (B)A=0 but possibly4p≠0 (C) possibly A≠0but4p=0 (D) possibly△F≠0 and possibly A≠0 According to the impulse-momentum theorem Fdr =Ap, the impulse of the force is zero,the ch 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) 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 dt F=0, Protal=constant I. Filling the blanks 1. Fig. I shows an ap of force versus time during the collision of a F 58g tennis ball with a wall. This initial velocity of the ball is 32m/s perpendicular to 2 the wall; it rebounds with the same speed, also g the maximum contact force during the collision is 928N Solution Applying to the impulse-momentum theorem time(ms) I= Ap= Fdt →2×58×10×3(2+6)x10-3xFs
4. A variable force acts on an object from ti = 0 to f t . The impulse of the force is zero. One can conclude that （ C ） (A) ∆r = 0 r and ∆p = 0 r . (B) ∆r = 0 r but possibly ∆p ≠ 0 r . (C) possibly ∆r ≠ 0 r but ∆p = 0 r . (D) possibly ∆r ≠ 0 r and possibly ∆p ≠ 0 r Solution: According to the impulse-momentum theorem F t p r r = ∆ ∫ f i t t d ，the impulse of the force is zero， the change of the momentum ∆p = 0 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 0, constant , d d total total total = = = F p 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 Fmax , the maximum contact force during the collision is 928N . Solution: Applying to the impulse-momentum theorem ∫ I = ∆p = Fdt v v v So max 3 3 2 (2 6) 10 2 58 10 32 2 d F p mv F t SF × + × ⇒ × × × = ∆ = = = − − ∫ v v Fmax = 928 N 0123456 Fmax 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 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 Solution
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 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
(a)Use conservation of momentum mv=(m+ Mv'-r'=m m+ (b)Assume the position which the pendulum is at rest is zero potential energy position, use CWE M+m M+m (M+m)y2=(M+m)gh→v=√2sh→v M+m、2gh 10×10-+25×√2×981×10=352ms 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 vi =(15m/s)i+(30m/s)j and viB =(10m/)i+(.0m/s).After the collision,vA=(60m/s)i+(30m/s)j. The final velocity of B is 4i+5j(m/s) Solution Applying the conservation of the momentum, we have mAVAi+mBB=mAAf+mBBS So the final velocity of B is m1B-m,4)=二(下a-下4)+ →fs2(15i+30+6-30)-10+5j=4+5(m/) I. Give the solutions of the following problems 1. A massive anchor chain of length /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 ude of the normal fo the chain is 3 mgy/ Solution Choose the anchor chain on the ground as the study object, which have length y and mass m=y. The chain on the ground get the gravitation m'g, the force N acted by the floor and impulse force Tacted by the falling chain. So
(a) Use conservation of momentum m M mv mv m M v v + = ( + ) '⇒ '= (b) Assume the position which the pendulum is at rest is zero potential energy position, use CWE theory. gh m M m v m M m (M m)v' (M m)gh v' 2gh v ' 2 2 1 2 + = + + = + ⇒ = ⇒ = (c) 2 9.81 10 352 m/s 10 10 10 10 2.5 2 3 3 × × × = × × + = + = − − gh m M m 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 v i j iA ˆ (30m/s) ˆ = (15m/s) + r and v i j iB ˆ (5.0m/s) ˆ = (−10m/s) + r . After the collision, v i j fA ˆ (30m/s) ˆ = (−6.0m/s) + r . The final velocity of B is (m/s) 4ˆ 5 ˆ i + j . Solution: Applying the conservation of the momentum, we have A Ai B Bi A Af B Bf m v m v m v m v v v v v + = + So the final velocity of B is (m/s) 10ˆ 5 ˆ 4ˆ 5 ˆ ) 15ˆ 30 ˆ 6ˆ 30 ˆ ( 3 2 ( ) ( ) 1 v i j i j i j i j v v v m m m v m v m v m v Bf Ai Af Bi B A A Ai B Bi A Af B Bf ⇒ = × + + − − + = + = + − = − + v v v v v v v v 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 m' g ，the force N acted by the floor and impulse force T acted by the falling chain. So Origin l y i ˆ Fig.5
yg yg Choose dm as the study object: dm When it reaches the ground its speed is: v=v2 Using impulse-momentum theorem: (dmg -T' )dt=-dyv2gy Since dmg <<T, we can neglect it, 2 Use Newton's third law T=T, so T=.. 2gy Then we have the magnitude of the normal force of the ground on the chain is N=.yg +2gy 2. A fire hose fixed to a support delivers a horizontal stream of water at a speed vo and a mass flow rate B kilograms per (a) The water stream is directed against a wall along a perpendicular to the wall Fig 6 st.(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 mo. 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 At, and the mass of the water which act on the wall is Am= BAt. We choose Am as the study object, which get the average force from the wall is f. Using impulse-momentum theorem. ( the direction of right is i)
yg T l m yg T N N l m + − = 0 ⇒ = + Choose dmas the study object: y l m dm = d When it reaches the ground its speed is: v = 2gy Using impulse-momentum theorem: y gy l m (dmg −T')dt = − d 2 Since dmg << T' , we can neglect it, gy l m v gy l m gy t y l m T 2 2 2 d d '= = = Use Newton’s third law T = T' , 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 v0 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 m0. 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 ∆m = β∆t . 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 ˆ ） 0 v r (a) 0 v r v m r (b) Fig.6
Byoi(Unparallel to vo) Using Newton third law, we can have the magnitude of the force acted on the wall by the water b)The momentum of the water in length As is P=Ap, the momentum of the water in unit P Bi (c) Regard the water and the car as a whole system, and choose it as the study object dPi=(mdv +v dm ) dP=0 mdv=(vo-v)dm 1. Using conservation of momentum, at any time t m=(m-m)→v=(1-) Use the result in(c), we can have V →ln="=ln (1 3. A rocket at rest in space, where there is virtually no gravity, has a mass of 2.55 x10 kg, of which 1.81x10kg 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 dm 327×103×480=1.57×10°(N) (b)The mass of the rocket after the engine burn
v i t mv i P f t mv i f ˆ ˆ 0 ˆ 0 0 0 = −β ∆ − ∆ ∆ = ∆ = − ∆ ⇒ = r r r (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 βv ( b) The momentum of the water in length ∆s is p p r r = ∆ , the momentum of the water in unit length is i v v i t s v i t m s mv i s P ˆ ˆ ˆ ˆ 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. m v v v m v v v P P i m v v i wc x x wc d ( - )d - d 0 ˆ ( d dm) d ˆ 0 0 ⇒ = ⎪ ⎩ ⎪ ⎨ ⎧ = = = + i. Using conservation of momentum, at any time t 0 0 0 0 ( ) (1 )v m m mv = m − m v ⇒ v = − ii. Use the result in (c), we can have v m m v m m v v v m m v v v m m v v v m m v ln ln (1 ) d d d d 0 0 0 0 0 0 0 = ⇒ = − − ⇒ = − = ⇒ − ∫ ∫ 3. A rocket at rest in space, where there is virtually no gravity, has a mass of 2.55 10 kg 5 × , of which 1.81 10 kg 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 3.27 10 480 1.57 10 (N) d d 3 6 = = × × = × t m F ve (b) The mass of the rocket after the engine burn is
dm △t=255×103-480×250=1.35×10k (c)The final speed is -"="-0=hn=-327×10lh255×105-181×10 =405×103m/s 4. A barge with mass 1.50x10 kg 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 2.78 kg and was moving at 4.30 m/s Immediate after impact, the second barge finds its course deflected by 18.0 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? The coordinate system is shown in figure. Apply conservation of momentum m,I+m2V2i=m,1 +m2v2j mI 2.78×10 15×10543j-(5 Isin 18i+51eos18 =33i-1.01j =B312-1017=331+10 Suppose the angle between the first barge and x direction is 6. Then tan 6= 0.31→6= arctan(-0.31)=-17 3.31 5. A system of two spheres of identical mass m collide obliquely as shown in Figure 8. The Before collision After collision collision is inelastic. Initially one of the masse is at rest. The momenta of the particles immediately after the collision are pi and P2.O The angle between the paths of the masses after the collision is B. What is the kinetic energy of 19 the system?
2.55 10 480 250 1.35 10 (k ) d d 5 4 0 t g t m m = m − × ∆ = × − × = × (c) The final speed is 4.05 10 m/s 2.55 10 2.55 10 1.81 10 0 ln 3.27 10 ln 3 5 5 5 3 = × × × − × − = − = = − × i f f i f e m m v v v v 4. A barge with mass 1.50 10 kg 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 2.78 10 kg 5 × and was moving at 4.30 m/s. Immediate after impact, the second barge finds its course deflected by o 18.0 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. i i f f m v m v m v m v 1 1 2 2 1 1 2 2 v v v v + = + So i j i j i j v v m m v v f i i f 01ˆ 1. 31ˆ 3. )] 1cos18 ˆ 5. 18 ˆ [4.3 ˆ (5.1sin 1.5 10 2.78 10 2ˆ 6. ( ) 5 5 2 2 1 2 1 1 = − − + × × = + = + − o o v v v v 3.31 1.01 3.46(m/s) 01ˆ 1. 31ˆ 3. 2 2 v1 f = i − j = + = Suppose the angle between the first barge and x direction is θ . Then o 0.31 arctan( 0.31) 17.22 3.31 1.01 tan = − ⇒ = − = − − θ = θ 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 P1 ′ r P2 ′ r Before collision After collision Fig.8 18° 1 2 Fig. 7 x y θ
Solution Apply the conservation of momentum P=P cos0+ P, cos(B-6 0=Pl 6+ P, sin(B-8) KE\nm22m2m→△KE=KE-KEP12,P2P 2m2m2m Using the adding of the vectors P=P+P+2P P2 cosp →AKE=(P+P-P2)=1 2mm,(-2P P, B) →△KE=、P'P'cosB
Solution: Apply the conservation of momentum ' cos ' cos( ) P1 = P 1 θ + P 2 β −θ m P m P m P KE KE KE m P m m v KE mv P P f i 2 2 ' 2 ' 2 2 2 1 0 ' sin ' sin( ) 2 1 2 2 2 1 2 2 2 2 1 2 = = = ⇒ ∆ = − = + − = θ + β −θ Using the adding of the vectors m P P KE P P m P P P m KE P P P P P β β β ' 'cos ( 2 ' 'cos ) 2 1 ( ' ' ) 2 1 ' ' 2 ' 'cos 1 2 1 2 2 2 2 2 1 1 2 2 2 2 1 2 1 ⇒ ∆ = − ⇒ ∆ = + − = ⋅ − = + +
