Elastic collisions in two dimensions 5C

Transcription

1 Elastic collisions in two dimensions 5C 1 No change in component of velocity perpendicular to line of centres. So component of velocity for A=6sin10 Since B is stationary before impact, it will be moving along the line of centres. 6cos10 = 4w v w v= 6cos10 (1) 1 w+ v= 6cos10 = 3cos10 () Adding equations (1) and () gives: 3w= 9 cos10 w= 3cos10 Substituting in equation () gives: v= 0 So, after the impact, A has velocity 6sin10 =1.04ms 1 (3 s.f.) perpendicular to the line of centres, and B has velocity 3cos10 =.95ms 1 (3 s.f.) parallel to the line of centres. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 No change in component of velocity perpendicular to line of centres. So component of velocity for A=4sin30 = Since B is stationary before impact, it will be moving along the line of centres. 4 4cos30 = 4v+ w w+ v= 8cos30 (1) 1 4 w v= 4cos30 = cos30 () 3 3 Adding equation (1) and equation () gives: w= 8+ cos 30 w= = Substituting in equation () gives: v= v= A has speed () = = 08 7 = = 4 39 ms 1 9 A is moving at arctan = arctan = arctan = 46.1 (3 s.f.) to the line of centres B has speed ms 1 along the line of centres Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.

3 3 No change in component of velocity perpendicular to line of centres. So component of velocity fora=5sin45 = 5 Since B is stationary before impact, it will be moving along the line of centres. 3 5cos45 = 3v+ 4w 15 4w+ 3 v= (1) 1 5 w v= 5cos45 = () 4 Adding equation (1) and 3 equation () gives: w= + w= 4 8 Substituting in equation () gives: v= v= = A has speed = = = 5 5 = ms 1 A is moving at arctan B has speed 45 8 ms 1 along the line of centres =arctan7=81.9 (3 s.f.) to the line of centres Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 4 No change in component of velocity perpendicular to line of centres. So for A, the component perpendicular to the line of centres is vsinθ. mvcosθ =mv 1 mv vcosθ =v 1 v (1) v 1 +v =evcosθ v 1 =evcosθ v () Substituting for v 1 in equation (1) gives: vcosθ= ( evcos θ v ) v = evcosθ 3v = vcos θ(e 1) v= 3 1 vsinθ 3vsinθ tan(90 θ) = = = tanθ v vcos θ(e 1) So 1 3tanθ = tan θ e 1 e 1 tan θ 3 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 5 After impact: No change in component of velocity perpendicular to line of centres. mv=mv A +mv B v=v A +v B (1) va vb= v () 3 Adding equations (1) and () gives: 1 And by substitution vb = v va = v va = v 3 6 A has speed A is moving at 5 ( ) 6 v= v= 36 6 vms arctan = arctan = 50. (3 s.f.) to the line of centres. 5 B is moving along the line of centres with speed 1 1 ms. 6 v Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 6 a Perpendicular to the line of centres, component of velocity of A is usinα mucosα =mv+mw ucosα =v+w (1) w v=eucosα () Solving equations (1) and (): v=ucosα eucosα =ucosα(1 e) tanβ= usinα usinα = v ucosα(1 e) = tanα 1 e b The path of A has been deflected through an angle equal to β α tanα tanβ tanα 1 e tanα tan( β α) = = 1+ tanα tanβ 1+ tanα tanα 1 e tan α (1 e)tan α (1 + e)tanα = = 1 e+ tan α tan α+ 1 e (1+e)tanα Hence β α =arctan tan α+1 e Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 7 a After impact: No change in the components of velocity perpendicular to the line of centres. 1 cos60 3cos45 =v w v w=1 3 (1) v+ w= (3cos45 + cos60 ) v+ w= 1+ () 3 3 Solving equations (1) and (): w= 1+ + = + =.06ms (4 s.f.) v= v= = ms (3 s.f.) Kinetic energy lost in the impact= 1 ((3cos45 ) v )+ 1 1 ((cos60 ) w ) = 1 ((3cos45 ) )+ 1 1 ((cos60 ).06 ) = =.53J (3 s.f.) b Impulse on B =1(w+cos60 )=3.06Ns (3 s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 8 a After impact: No change in the components of velocity perpendicular to the line of centres. So after the collision the components of velocity perpendicular to the line of centres are 3 m s 1 and 4 m s 1. 3 m 3 cos45 m 5 = mw mv w v= 0 (1) 5 3 v+ w= 3 cos v+ w= () Solving equations (1) and () gives v=w= A has speed +3 = 13ms 1 B has speed +4 = 0= 5ms 1 b Total kinetic energy after impact= 1 m ( 13) + 1 m ( 5) = 33 mj Total kinetic energy before impact = m (3 ) + m 5 = m J Fraction lost= = Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 9 Line of centres parallel to j so no change in the components of velocity parallel to i 1+4 1= s 4 t s t=1 (1) 1 s+ t= (1 + 1) s+ t= 1 () Solving equations (1) and (): 3s=3 s=1 And by substitutiont=0 Velocity of A is 4i + j m s 1 Velocity of B is i m s 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 10 Line of centres parallel to i so no change in the components of velocity parallel to j =w 3v w 3v= 1 (1) 3 v+ w= (4 + 1) 4 v+ 4 w= 15 () 4 Solving equations (1) and (): 4w 1v= 4 16v=19 v= 19 41,w= After the impact, speed of A= =3.3ms 1 (3 s.f.) Speed of B= =3.5ms 1 (3 s.f) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10

11 11 Line of centres parallel to i so no change in the components of velocity parallel to j 1 1=w v w v=0 (1) 3 v+ w= ( + 1) 5 v+ 5 w= 9 () 5 Solving equations (1) and (): w= 9 w= and v= 5 5 As components of velocity unchanged parallel to j all kinetic energy lost is parallel to i Kinetic energy lost= = 48 5 =1.9J Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 11

12 1 a Let velocity of B immediately after the collision be v Using conservation of momentum: m(i+5j)+m(3i j)=m(3i+j)+mv v=i(+ 3 3)+j(5 1 )=5i+j v= 5 i+ 1 j b Impulse on A= m((3i+ j) (i+ 5 j)) = m( i+ 3 j) Therefore the line of centres is parallel to 1 ( i 3 j ) a Let velocity of B immediately after the collision be v Using conservation of momentum: 3m(3i 5j)+m(4i+j)=3m(4i 4j)+mv v=i( )+j( )=i j Speed of B is 1 + = 5ms 1 b 3m m m m = (3(34 3) + (17 5)) = (6+ 1) = 9m J Kinetic energy lost = ((3 + 5 ) (4 + 4 )) + ((4 + 1 ) 5) 14 a Let velocity of B immediately after the collision be v Using conservation of momentum: m(i+5j)+m(i j)=m(3i+4j)+mv v=i( + 3)+j( 5 4)=0 b B is brought to a halt in the collision, therefore the line of centres must be parallel to the original direction of motion of B, i.e. ( i j ) In this direction, speed of A before impact = (i+ 5 j). ( i j ) = ( 5) = 3 Speed of A after impact = (3i+ 4 j). ( i j ) = (3 4) = Speed of B before = Speed of B after = 0 Therefore the impact law gives e= = Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

13 15 After impact: Before collision, components of velocity of A are 1 m s 1 perpendicular to the lines of centres and m s 1 parallel to the line. The components of the velocity of B are 3 m s 1 perpendicular to the line, and m s 1 parallel to it. 1 =1 w v w v= (1) w+v=e(+)= w+v= () Solving equations (1) and (): w=,v=0 After the collision, the speed of A is 1 m s 1 and speed of B is 3 + = 13ms 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 13

14 16 Parallel to the line of centres, using conservation of momentum and the law of restitution gives: mucos45 =mv+mwandw v=eucos45 By subtracting: v= ucos45 (1 e) u (1 e) v= 4 u sin 45 So tanα = = 1 e θ = α 45 u (1 e) ( ) 4 tanα tan 45 1 e 1 1+ e 1+ e tanθ = = = = 1+ tanα tan e+ 3 e 1 e Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 14

15 Challenge 1 Tangent perpendicular to radius sinα = Initial components of velocity of A are ucosα parallel to the line of centres, and usinα perpendicular to the line of centres. mucosα=mv+mw ucosα=v+w (1) () w v=eucosα Subtracting equation () from equation (1) gives: v= ucosα eucosα 1 ucosα1 3 1 u u 3 v= = = 8 u usinα 4 tan( θ+ α) = = = v u tan( θ + α) tanα tanθ = = = = = 1+ tan( θ+ α)tanα Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 15