Elastic collisions in one dimension Mixed Exercise 1 u w A (m) B (m) A (m) B (m) Using conseration of linear momentum for the system ( ): mu m= mw u = w (1) 1 w e= = 3 u ( ) u+ = 3 w () Adding equations (1) and () gies: u=w u=w Substituting in equation () gies: w+=3w = w The ratio of the speeds before impact is u:=w:w=:1 as required. 0.5u P ( m) Q ( λm) P ( m) Q ( λm) Using conseration of linear momentum for the system ( ): 0.5mu = λm u= λ (1) 1 e= = u= 16 () 0.5u From equations (1) and (): u=16=λ λ= Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1
3 a Note that the boat moes in the opposite direction to the boy after the boy dies off. Before diing off V After diing off Boat + boy (M + m) Boat (M) Boy (m) Using conseration of linear momentum for the system ( ): 0= m MV V = m M b Let total kinetic energy of boy and boat after the die be KE 1 1 KE= MV + m 1 m 1 = M + m M m + mm = M m( m+ M) = M as required c The boat is large and heay, so there will be additional tilting/rolling motion. The boat is also on water, so gien waes, tides and currents it is unlikely to be at rest initially. 5 m s 1 3 m s 1 m s 1 P ( kg) Q ( kg) P ( kg) Q ( kg) Using conseration of linear momentum for the system ( ): 5+ ( 3)=+ =10 =.5ms 1 Loss of kinetic energy = initial kinetic energy final kinetic energy = 1 5 + 1 3 1.5 + 1 =50+9 1.5 =.5J Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.
5 a u w P (3m) Q (m) P (3m) Q (m) Using conseration of linear momentum for the system ( ): 3mu=3m+mw 3+ w= 3 u (1) w e= u w = eu () Subtracting equation () from equation (1) gies: =3u eu = u(3 e) b Substituting for in equation () gies: w=eu+ u(3 e) = eu+3u eu = 3u(e+1) Loss of kinetic energy = initial kinetic energy final kinetic energy = 1 3mu 1 3m 1 mw = m 3u 3 u (3 e) 16 = 3mu 3 9 u (1+e) 16 ( 16 (9 6e+e ) (3+6e+3e )) = 3mu 3 ( e ) = 3 8 mu (1 e ) c Impulse exerted on Q is change of momentum of Q = mw= 3mu(1+e) Ns Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3
6 a m s 1 8 m s 1 w P (0.07 kg) Q (0.1 kg) P (0.07 kg) Q (0.1 kg) Using conseration of linear momentum for the system ( ): 0.07 +0.1 ( 8)=0.07+0.1w 7+ 10w= 5 (1) 5 w e= = 1 ( 8) w = 5 () Adding equation (1) and 7 equation () gies: 17w= 5+35= 17 w= 1 Substituting in equation () gies: 1 =5 = 6 So the elocities after impact are 6 m s 1 and 1 m s 1 in the direction of the 100 g mass prior to the impact. b Let loss of kinetic energy in the collision be KE KE = initial kinetic energy final kinetic energy = 1 0.07 + 1 0.1 ( 8) 1 0.07 ( 6) + 1 0.1 ( 1) =(0.56+3.) (1.6+0.05)=.5J Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.
7 35 m s 1 0 m s 1 w P ( kg) Q (10 kg) P ( kg) Q (10 kg) Using conseration of linear momentum for the system ( ): 35+10 0=+10w + 10w= 70 (1) 3 w e= = 5 35 0 w = 9 () Adding equation (1) and equation () gies: 1w=70+18=88 w= Substituting in equation () gies: =9 =15 After the impact, assume that the particles moe at constant speed and use speed time = distance. Fie seconds after the impact the 10 kg mass moed a distance 5=10m It takes the kg mass a time of 10 to trael 10 m, i.e. 8 seconds. 15 The time that elapses between the 10 kg sphere resting on the barrier and it being struck by the kg sphere therefore =8s 5s=3seconds Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5
8 First consider impact of A with B, then of B with C, then of A with B again. Before the first collision V After the first collision w A (m) B (3m) C (3m) A (m) B (3m) C (3m) Using conseration of linear momentum for the system ( ):: V = + 3w + 3w= V (1) 3 w e= = w = 3 V () V Adding equations (1) and () gies: 7w=7V w=v Substituting in equation () gies: V =3V =0.5V Before the second collision 0.5V V After the second collision 0.5V x y A (m) B (3m) C (3m) A (m) B (3m) C (3m) Using conseration of linear momentum for the system ( ):: 3V = 3x+ 3 y x+ y= V (3) 3 y x e= = y x= 0.75 V () V Adding equations (3) and () gies: y=1.75v y=0.875v Substituting in equation () gies: 0.875V x=0.75v x=0.15v Ball A is now moing at 0.5V and ball B is moing at 0.15V so ball A will strike ball B for a second time. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6
8 continued Before the third collision 0.5V 0.15V 0.875V After the third collision j k 0.875V A (m) B (3m) C (3m) A (m) B (3m) C (3m) Using conseration of linear momentum for the system ( ):: ( 0.5) V+ (3 0.15) V = j+ 3k j+ 3k = 1.375 V (5) 3 k j e= = 0.15V k j= 0.375 V (6) Adding equations (5) and (6) gies: 7k=1.75V k=0.5v Substituting in equation (6) gies: V j=0.375v j=0.1565v After three collisions the elocities are 0.1565V, 0.5Vand0.875V for balls A, B and C respectiely. In fractions, the respectie elocities are 5 V, 1 V and 7 V. 3 8 As 5 V < 1 V < 7 V there are no further collisions. 3 8 9 a Velocity of bullet after hitting the barrier =600 0.=0 m s 1 1 1 Kinetic energy lost= 0.06 600 0.06 0 = 907 J b Either heat or sound. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7
10 a u x y A (m) B (3m) A (m) B (3m) Using conseration of linear momentum for the system ( ): u=3y+x 3y+ x= u (1) y x e= u y x= eu () Adding equation (1) and equation () gies: 7y=u+eu y= 7 u(1+e) Substituting in equation () gies: 7 u(1+e) x=eu x= u+eu 7eu 7 = u 7 ( 3e) b Impulse = change in momentum of B So mu=3m 7 u(1+e) 1+e= 1 1 e= 1 6 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8
11 a kv V X A ( m) B ( λm) A ( m) B ( λm) Using conseration of linear momentum for the system ( ): mkv +λmv =λmx X = (λ+k)v λ X e= kv V ( λ+ k) V = (substituting for X) λ( kv V) λ+ k = λ ( k 1) λ+ k b As e < 1, < 1 λ( k 1) So λ+k<λk λ (as λ>0 and k>1) λ+k<λk λk λ>k λ(k )>k Since k>0 and λ>0, therefore k >0 k So λ > and k > k Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9
1 a Use = u + at downwards with u=0,t=1 and a= g=9.8 to find the elocity of the first ball before impact. This gies: = 9.8 9.8 m s 1 A (m) 1 A (m) 7 m s 1 B (m) B (m) Using conseration of linear momentum for the system ( ): 9.8m 7m= m + m 1 1 + =.8 (1) 1 1 e= = 9.8+ 7 =. () 1 Adding equations (1) and () gies: =7 =3.5ms 1 Substituting in equation () gies: 3.5 1 =. 1 = 0.7ms 1 Both balls change directions, the first moes up with speed 0.7 m s 1 and the second moes down with speed 3.5 m s 1. b 1 1 Kinetic energy before impact= m 9.8 + m 7 = 7.5m J Kinetic energyafterimpact= 1 m 0.7 + 1 m 3.5 =6.37mJ Percentage loss of kinetic energy= 7.5 6.37 7.5 =91.%=91% (s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10
13 a Stage one: particle falls under graity : Use = u + as downwards with u=0,s=8 and a= g =g 8=16g = 16g Stage two: first impact: The particle rebounds with elocity 1 16g = g Stage three: particle moes under graity : Let the height to which the ball rebounds after the first bounce be h 1 Use =u +as upwards with =0,u= g,a= g and s= h 1 0= g gh 1 h 1 =0.5m b Use =u+at upwards with reach the top of the bounce 1 0= 16g gt g t= = 0.319 g So the time taken to reach the plane again 1 = 0, u= 16 g and a= g to find the time it takes the particle to = 0.319= 0.6s ( s.f.) or s g c Speed of approach= g g 1 The speed of the particle after the second rebound = e g = = 0.78ms ( s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 11
1 Stage one: particle falls under graity : Use = u + as downwards with u=0,s= h and a= g =gh = gh Use s ut 1 = + at to find the time to the first bounce 1 h h= gt1 t1= g Stage two: particle rebounds from plane. The particle rebounds with elocity e gh Stage three: particle moes under graity until it hits the plane again : Use s= ut+ 1at to find the time from the first to the second bounce, u=e gh, s=0 and a= g 1 0= e ght gt t e gh h = = e g g Stage four: particle rebounds (again) from plane. Speed of approach=e gh, so speed of rebound = e gh Similar working finds that the time from the second bounce to the third bounce is And the time from the third bounce to the fourth bounce is t Let the total time taken by the particle be T, then h g 3 = e t 3 = e h g h h h h T e e e g g g g 3 = + + + + h h = + e + e + e + g g 3 ( ) The expression in the bracket is an infinite geometric series with a = e and r = e. Using the formula S = a 1 r = e, the expression for T can be simplified as follows 1 e T = g h e 1+ 1 e = 1 e+e 1 e h g =1+e 1 e h g Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1
15 Before first impact u After first impact w P (m) Q (8m) P (m) Q (8m) Using conseration of linear momentum for the system ( ): mu= m+ 8mw + 8 w= u (1) 7 w e= = 8w 8= 7 u () 8 u Subtracting equation () from equation (1) gies: 9=u 7u = 3 u Substituting in equation () gies: 8w+ 16u 3 =7u 8w= 5u 3 w= 5u Let e p be the coefficient of restitution between P and the ertical place. So P then hits the ertical plane with speed u 3 and rebounds with speed 3 ue p Before second impact of P and Q After second impact of P and Q ue 5 u 3 p x P (m) Q (8m) P (m) Q (8m) Using conseration of linear momentum for the system ( ): 5 muep + mu= 8mx x= uep + 5 u (1) 3 3 7 x 7 5 35 e = = uep u = x x= 1 uep u 8 8 3 8 () Subtracting equation () from equation (1) gies: 1ue p =5u+ 35 75 u= 8 8 u e = 75 p 96 = 5 3 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 13
16 Before firing V After firing Gun (M) Shell (m) Gun (M) Shell (m) Using conseration of linear momentum for the system ( ): m m MV = 0 V = (1) M Energy released: 1 1 E m MV = + () Substituting for V into equation () gies: E 1 1 m = m + M M ME= mm + m ME = m( M + m) = 17 a Using ME ms m( M + m) 1 = u + as downwards with u=0,s= H and a= g =gh = gh The ball rebounds with speed e gh Using = u + as upwards with u=e gh,s= h and a= g 0= ghe e h = e= H gh h H b The ball rebounds the second time with speed Using e gh = u + as upwards with u=e gh,s= h and a= g 0=gHe g h h = He = H h H h e H = Hh H = h H h h H h = = H = = H H H c The ball continues to bounce (for an infinite amount of time) with its height decreasing by a common ratio each time. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1
18 a Use F = ma to determine the acceleration of the sphere down the smooth slope. This gies: g gsin30 = a a= gsin30 = Use = u + as with u=0,s= and a=0.5g to find the speed of the ball when it reaches the horizontal plane: =g = g g w B ( kg) C (1 kg) B ( kg) C (1 kg) Using conseration of linear momentum for the system ( ):: g = + w + w= g (1) w e= 0.75= g w = 0.75 g () Adding equation (1) and equation () gies: 3w= g+1.5 g =3.5 g w= 7 6 gms 1 Substituting in equation () gies: 7 3 g = g 6 1 9 5 = g = g ms 1 1 1 Both B and C continue in the direction B was originally moing. b Energy lost in the collision = initial kinetic energy final kinetic energy 1 1 5 g 1 7 g = ( g) + 1 1 6 50g 98g 50g 98g g 7g = g + = g + = = J 1 7 1 7 1 1 c If e< 0.75 the amount of kinetic energy lost would increase as the collision would be less elastic. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 15
19 Initial position W 1 W d d m s 1 3 m s 1 A B Suppose point Q is at a distance x from wall W 1 Consider the motion of sphere A: Time taken for A to trael from point P to wall W 1 is distance d = = d speed Sphere A rebounds with speed 3 5 = 6 5 ms 1 Time taken for A to trael from wall W 1 to point Q is distance x 5 x = = speed 6 6 5 Consider the motion of sphere B: Time taken for B to trael from point P to wall W is distance d = speed 3 Sphere B rebounds with speed 3 5 3= 9 5 ms 1 Time taken for B to trael from W to point Q is ( d x) distance 3d x 5 3 15d 5x = = = speed 9 9 9 5 When A and B meet at Q, they hae been traelling for the same time, so 5x d 15d 5x d+ = + 6 3 9 18d+ 15x= 6d+ 30d 10x 5x= 18d 18d 57d x= and 3d x= 5 5 18d+ 15x= 6d+ 30d 10x Therefore the distance ratio W 1 Q:W Q= x:3d x= 18d 5 :57d 5 =18:57=6:19 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 16
Challenge Before string B C becomes taut u After string B C becomes taut 1 1 A (m 1 ) B ( m ) C (m 3 ) A (m 1 ) B ( m ) C (m 3 ) Using conseration of linear momentum for the system ( ): m u= m + m 3 1 3 1 m u= ( m + m ) 3 1 3 m3u 1 = ( m + m ) 3 Before string A B becomes taut 1 1 After string A B becomes taut A (m 1 ) B ( m ) C (m 3 ) A (m 1 ) B ( m ) C (m 3 ) Using conseration of linear momentum for the system ( ): m + m = m + m + m 1 3 1 1 3 ( m + m ) = ( m + m + m ) 1 3 1 3 1( m+ m3) m3u = = ( m + m + m ) ( m + m + m ) 1 3 1 3 Total kinetic energy= 1 (m +m +m ) 1 3 = 1 (m +m +m ) m 3 u 1 3 (m 1 +m +m 3 ) m = 3 u (m 1 +m +m 3 ) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 17