Elastic collisions in one dimension 4D

Transcription

1 Elastic collisions in one dimension 4D 1 a First collision (between A and B) = u+ v u+ v= 11 (1) 1 v u e= = 5 1 v u= () Subtracting equation () from equation (1) gives: 3u=9 u=3 Substituting into equation (1) gives: 6+ v= 11 v= 5 [This result can be checked in equation ()] Second collision (between B and C) = x+ y x+ y= 13 (3) 1 y x e= = y x= (4) Adding equations (3) and (4) gives: 7 9 3y= y= = 4.5 Substituting into equation (4) gives: 4.5 x= 1 x=4 Solution:u=3,v=5,x=4,y=4.5 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 1 b First collision (between A and B) ( ) = 1.5u+ v 1.5u+ v= 11 (1) 1 v u v u e= = = v u= () Adding equation (1) to 1.5 equation () gives: 3.5v=14 v=4 Substituting into equation () gives: 4 u= u= [This result can be checked in equation (1)] Second collision (between B and C) = x+ y x+ y= 11 (3) 1 y x e= = y x= (4) Subtracting equation (4) from equation (3) gives: 1 7 3x= x= = 3.5 Substituting into equation (4) gives: y 3.5= 1 y=4 Solution:u=,v=4,x=3.5,y=4 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.

3 Before the first impact 6 m s 1 At rest At rest After the first impact u v At rest A (3m) B (5m) C (4m) A (3m) B (5m) C (4m) 3m 6= 3mu+ 5mv 3u+ 5v= 18 (1) Perfectly elastic means e = 1. So using Newton s law of restitution: v u e= 1= 6 v u= 6 () Adding equation (1) to 3 equation () gives: 8v=36 v=4.5 Substituting into equation () gives: 4.5 u=6 u= 1.5 Before the second impact 1.5 m s m s 1 At rest After the second impact 1.5 m s 1 x y A (3m) B (5m) C (4m) A (3m) B (5m) C (4m) 5m 4.5= 5mx+ 4my 5x+ 4y=.5 (3) y x e= 1= 4.5 y x= 4.5 (4) Adding equation (3) to 5 equation (4) gives: 9y=.5+.5=45 y=5 Substituting into equation (4) gives: 5 x=4.5 x=0.5 Solution: A 1.5ms, B 0.5ms, C 5ms Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 3 a Before the first impact u At rest At rest After the first impact w x At rest A (m) B (m) C (m) A (m) B (m) C (m) u= w+ x w+ x= u (1) x w e= u x w= eu () Adding equation (1) to equation () gives: x=eu+u x=0.5u(e+1) Substituting into equation () gives: 0.5u(e+1) w=eu w=0.5u(1 e) Before the second impact 0.5u(1 e) 0.5u(e + 1) At rest After the second impact 0.5u(1 e) y z A (m) B (m) C (m) A (m) B (m) C (m) 0.5 ue ( + 1) = y+ z y+ z = 0.5 ue ( + 1) (3) z y e= 0.5 u( e+ 1) z y= 0.5 uee ( + 1) (4) Adding equation (3) to equation (4) gives: z=0.5u(e+1)+0.5eu(e+1) z=0.5u(e+1)(1+e)=0.5u(1+e) Substituting into equation (4) gives: 0.5u(1+e) y=0.5eu(e+1) y=u(1+e)( e 0.5e)=0.5u(1+e)(1 e) Solution: 0.5u(1 e), 0.5u(1+e)(1 e), 0.5u(1+e) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 3 b A will catch up with B provided that 0.5u(1 e)>0.5u(1+e)(1 e), i.e. provided that > 1+ e Since e< 1 this condition holds and A will catch up with B resulting in a further collision. 4 a Before the first collision 4u u 3u After the first collision v w 3u A (m) B (m) C (m) A (m) B (m) C (m) 4u u= v+ w v+ w= u (1) w v e= 4u+ u w v= 6 eu () Adding equations (1) and () gives: w=u+6ue w=u(1+3e) B will collide with C if the speed of B after collision with A is greater than the speed of C, i.e. if w > 3u. This will occur if: u(1+3e)>3u 3e> e> 3 b Subtracting equation () from equation (1) gives: v=u 6eu v=u(1 3e) If e> then v < 0, and the direction of A is reversed by the collision with B. 3 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 5 Before first collision 4u u After first collision v u P (m) P (m) m 4u+3m ( u)=mv+3mu u=v+3u v= 0.5u Use this result to find the coefficient of restitution between particles P and Q. u v 1.5u e= = = 0.5 4u+ u 6u The second collision is between Q and the wall. Q rebounds from the wall with velocity u, as the coefficient of restitution between Q and 3 the wall is. 3 Second collision between P and Q 1 u u 3 After second collision x y P (m) P (m) Using conservation of linear momentum for the system ( ): 1 m u+ 3m u= x+ 3y 3 x+ 3y= 3 u (1) 1 x y e= = 1 4 u u u x y= u= () Adding equation (1) to 3 equation () gives: u 5u 5u 5x= 3u+ = x= Substituting into equation () gives: 5u 8 y= u 4 y=15u u = 14u 4 4 = 7u 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 6 a Before first collision 1u At rest After first collision v w P (m) P (m) m 1u=mv+3mw v+3w=1u (1) w v e= = 3 1u w v= 8 u () Adding equation (1) to equation () gives: 4w=0u w=5u Substituting into equation () gives: 5u v=8u v= 3u After the collision the speed of P is 3u and its direction is reversed, and the speed of Q is 5u. b Q then hits a wall and rebounds with speed 4 5u = 4u 5 Second collision between P and Q 3u 4u After second collision x y P (m) P (m) Using conservation of linear momentum for the system ( ): 3mu+ 1mu= mx+ 3my 3y+ x= 15 u (3) x y e= = 3x 3y= u 3 4u 3u (4) Adding equation (1) to equation () gives: 17u 4x= 17u x= 4 Substituting into equation (4) gives: 51u 4 3y=u y= 43u 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 7 a i Use v =u +as downwards with u=0,s=0.4 and a=g=9.8 to find the speed of approach for the first bounce: v =g 0.4=7.84 v=.8ms 1 Newton's law of restitution gives speed of rebound from floor as 0.7.8=1.96ms 1 Use v = u + as upwards with v=0,u=1.96 and a= g to find the height of the first bounce: 0=1.96 gs s= =0.196=19.6cm ii Use v =u +as downwards with u=0,s=0.196 and a=g=9.8 to find the speed of approach for the second bounce: v =g 0.196= v = 1.96ms 1 [This result can also be directly deduced] Following second collision with floor, the ball rebounds with speed =1.37ms 1 Use v u as 0= 1.96 = + upwards with v=0,u=1.37 and a= g gs 1.37 s= = = cm 19.6 b The ball continues to bounce (for an infinite amount of time) with its height decreasing by a common ratio each time. c Ratio of heights of successive bounces is = Total distance travelled = ( ) = (using the sum of an infinite geometric series) = 1.17m (3 s.f.) d The ball loses energy following every bounce, so an infinite number of bounces would be unrealistic. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 8 a Use v u as = + downwards with u=0,s=h and a=g v =gh v= gh Newton s law of restitution gives speed of separation from plane as e gh Let the height to which the ball rebounds after the first bounce be h 1 Use v =u +as upwards with v=0,u=e gh,a= g and s=h 1 0=gHe gh 1 h 1 =e H b Let the height to which the ball rebounds after the second bounce be h Before the second bounce, the ball drops from a height h 1 So using the result from part a, h =e h 1 So h =e h 1 =e (e H)=e 4 H c Let the total distance travelled by the ball be d, then d = H+ h + h = H+ e H+ e H+ 4 (1 ) = H+ e H + e + e + 4 e H(1 e e ) r= e, so is an infinite geometric series with first term a=e H and common ratio S = a 1 r = e H 1 e Therefore 4 e H = + ( ) = H+ 1 e H e + e H H+ e H d H e H e e (1 ) = = 1 e 1 e H(1 + e ) = 1 e Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 9 Initial position W 1 W d m m s 1 B (m) From O W, B travels at a speed of m s 1 d through a distance m. d distance d So the time taken is = = speed 4 B then rebounds with speed e ms 1. From W W1, B travels at a speed of e ms 1 through a distance d m. distance d So the time taken is = speed e B then rebounds with speed 1 ee 1ms. From W W, B travels at a speed of ee 1 1 ms 1 through a distance d m. So the time taken is distance d = speed ee 1 d d d d Therefore, the total time taken is + + = + + seconds. 4 e ee 1 e1 ee 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10

11 Challenge Consider the initial collision of particles P and Q: Before first collision m s 1 1 m s 1 After first collision v 1 v P (m) Q (m) P (m) Q (m) m m= mv + mv 1 1 v + v = 1 (1) e=0.5= v v 1 1 ( ) =v v 1 3 v v1 = 1.5 () Adding equations (1) and () gives: v =.5 v =1.5ms 1 Substituting into equation (1) gives: 1.5 v 1 =1.5 v 1 = 0.5ms 1 After first collision W 1 W 4 m 0.5 m s m s 1 P (m) Q (m) As P travels from O to W 1, the time taken is distance = = 8s speed 0.5 As Q travels from O to W, the time taken is distance = = 1.6s speed 1.5 Q rebounds with speed =0.5ms = 0.5 m s 1. Now let t be the time of the second collision, and suppose both particles collide at a distance d from W Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 11

12 Challenge continued Then for particle P: d=+0.5t And for particle Q: d=0.5(t 1.6) So +0.5t=0.5(t 1.6) 0.5t=+0.8=.8 t=11.seconds But it only takes 8 seconds for P to travel to W 1, so P will hit W 1 before colliding with Q for a second time. Therefore P hits W 1 before colliding with Q for a second time. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1