Mixed Exercise 5. For motion parallel to the wall: For motion perpendicular to the wall:

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1 Mixed Exercise 5 For motion parallel to the wall: 4u cos β u cos45 5 4u u cos β () 5 For motion perpendicular to the wall: 4u sin β eu sin45 5 4u eu sin β () 5 Squaring and adding equations () and () gives: 6 5 u cos β+ 6 5 u sin β 4 u + 4 e u 6 5 u (cos β+sin β) u (+e ) 3 5 +e e 7 5 e 7 5 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2 a The component of velocity parallel to the slope is usinα Perpendicular to the slope: v eucosα Therefore the speed immediately after impact +..33ms (3 s.f.) b Impulse mv m( ucos α) (. ( 4.8)) 3Ns Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

3 3 a Let v a + b, where a is parallel to the wall and b is perpendicular to the wall. ( i+j) is a unit vector in the direction of the wall and (i+j) is a unit vector perpendicular to the wall. Parallel to the wall: a ( 4i j). ( i+ j) ( i+ j) (4 ) ( i+ j) ( i+ j) Perpendicular to the wall: b [( 4i j). ( i+ j)] ( i+ j) ( 4 ) ( i+ j) 3 ( i+ j ) 3 5 So v ( i+ j) + ( i+ j) i+ j b Kinetic energy after impact Kinetic energy lost J 8 Kinetic energy before impact (4 ) 5 c Using the scalar product to determine the angle of deflection: u.v cosθ u v 5 ( 4 ) + ( ) 5 cosθ θ 8 (3 s.f.) 6 4 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 4 5 a If the component of the initial velocity perpendicular to the wall is u and that of the final velocity is v, then the coefficient of restitution, e is given by: v e u Since neither i nor j component is unchanged, wall is not along one of the axes. Impulse vector, I is perpendicular to the wall: I mv mu I m(i j) m(6i+ 3 j) I m( 4i 5 j) The component of each velocity perpendicular to the wall is given by the scalar product of the velocity and the impulse vector: uu.im(6i+3j).( 4i 5j)m(6 4)+(3 5) 39m vv.im(i j).( 4i 5j)m( 4)+( 5)m e m 39m 39 First collision: e0.5 For motion parallel to the wall: u cosα cos30 () For motion perpendicular to the wall: usinα esin30 () Squaring and adding equations () and () gives: u cos α+u sin α 4cos 30 +4e sin 30 u (cos α+sin α) u 3.5 u.80ms (3 s.f.) Dividing equation () by equation () gives: tanα 0.5 tan 30 3 α 6. (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 5 b Second collision: e0.5 For motion parallel to the wall: vcosβ ucos(90 α) usin α (3) For motion perpendicular to the wall: vsinβ 0.5 usin(90 α) 0.5ucos α (4) Squaring and adding equations (3) and (4) gives: v cos β+v sin β u sin α+0.5u cos α As tanα 3, by Pythagoras sinα 3 v (cos β+sin β)u vms 3 Dividing equation (4) by equation (3) gives: tanβ 0.5 tanα 3 3 β60 and cosα 3 3 So the ball travels parallel to the original direction, but in the opposite direction. c The resistance to the motion due to the floor not being smooth will result in the final speed of the ball being slower, but the angle of motion will remain the same. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 6 a First collision For motion parallel to the wall: 0cosα 5cos60 () For motion perpendicular to the wall: 0sinα 5e sin60 () From equation (): 0 cosα 5cos cosα By Pythagoras theorem tanα and sinα 5 8 Dividing equation () by equation () gives: tanα e tan 60 tanα 39 3 e 0.7 (3 s.f.) tan Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 6 b Second collision: e0.5 For motion parallel to the wall: v cos β 0cos(90 α ) 0sin α (3) For motion perpendicular to the wall: v sin β 0 e sin(90 α ) 0 e cos α (4) Squaring and adding equations (3) and (4) gives: v cos β+v sin β 0 sin α+0 e cos α 39 v (cos β+sin β) v 8.0ms (3 s.f.) Dividing equation (4) by equation (3) gives: e tanβ tanα β 30 So the ball travels parallel to the original direction, but in the opposite direction. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 7 a First collision: e0. For motion parallel to the wall: v cos α 3cos5 () For motion perpendicular to the wall: v sin α 0. 3sin5 () Squaring and adding equations () and () gives: v cos α+v sin α 9cos sin 5 v (cos α+sin α) v.7307ms (5 s.f.) Dividing equation () by equation () gives: tanα 0.tan α 5.38 (5 s.f.) Loss of kinetic energy m(u v ) 0.5( ) J (3 s.f.) b Second collision: e0.4 For motion parallel to the wall: vcosβ v cos(80 α ) (3) For motion perpendicular to the wall: v sinβ 0.4v sin(80 α ) (4) Dividing equation (4) by equation (3) gives: tanβ0.4tan(80 α)0.4tan( )0.4tan β 55.6 (3 s.f.) Substituting in equation (3) gives: v cos cos v.8ms (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 8 a Let velocity of B immediately after the collision be v Using conservation of momentum: 4m(i+3j)+m(3i j)4m(3i+j)+mv 9 vi( )+j(4 3 4 ) i+3j b Impulse on A 4m(3i+j) (i+3j)4m(i j) (i j) is a unit vector parallel to the line of centres. 3 3 Tangent perpendicular to radius sin α, so cos α and tanα 3 Initial components of velocity of A are ucosα parallel to the line of centres, and usinα perpendicular to the line of centres. Using conservation of momentum: mucosαmv+mw v+wucosα () where v is the velocity of A along the line of centres and w the velocity of B along the line of centres immediately after the collision. Using Newton s law of restitution: w veucosα () Solving equations () and (): v ucosα eucos α, 3 ucos α( 3) u 3 u 3 v u usinα 6 tan( θ+ α) 3 v u 3 3 tan( θ+ α) tanα tanθ tan(( θ+ α) α) + tan( θ + α)tanα (+ ) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 0 Using conservation of momentum: mucosα mv+mw v+wucosα () Using Newton s law of restitution: w veucosα () Solving equations () and (): vucosα eucosα As tanα 3 4 cosα4 5 and sinα3 5 v ucosα( e) u( e) 5 tan(θ+α) usinα usinα 3 v u( e) ( e) 5 tan( θ+ α) tanα tanθ tan(( θ+ α) α) + tan( θ + α)tanα 3 3 ( e) 4 6( e) ( e) + 9 ( e ) e 7 8e Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

11 After impact tanα sinα cosα tanβ sin β and cosβ Using Newton s law of restitution: v+ w ( cosα+.5cosβ) 3 v+ w v w Using conservation of momentum: ( cosα) ( 3.5cosβ)3w v 43w v Substituting for v gives: 43w ( w) 4 4+5w w0 v w Let speed of A after collision be v A ( ) +v v A sinα v A +4 v A 5 Let speed of B after collision be v B ( ) +w v B.5sinβ v B 3 v B.5 +0 After the collision, the speeds of A and B are 5 m s and.5 m s respectively. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

12 Using conservation of momentum: mu( i+8j)ms( i+j)+mr(i+4j) So u s+r and 8us+4r Adding 6u6r ur and s4u The red ball moves along the line of centres after impact, so line of centres is parallel to ( i + j). Component of initial velocity along this line 0 Component of final velocity along this line 4u For the white ball, component of initial velocity along this line is the scalar product of this vector and its motion: u( i+8j).( i+j)u ( )+(8 ) ( )0u And, similarly, component of final velocity along this line is: u(i+4j).( i+j)u ( )+(4 ) ( )u Using Newton s law of restitution: 4u+ue 0u 60e e 3 5 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

13 3 Angle between line of centres and direction of travel is θ where: 6a sinθ 5 a 3 5 cosθ 4 5 Perpendicular to line of centres the components of the initial speeds are: usinθ 0.6u usinθ.u These remain unchanged. Parallel to line of centres the components of the initial speeds are: ucosθ 0.8u ucosθ.6u And the components of the final speeds v T and v B along these lines are v (for the top sphere) and w (for the bottom sphere) Using conservation of momentum along the line of centres: mv+mwmucosθ+mucosθ v+w.6u+0.8u.4u () Using Newton s law of restitution: v w 3 (0.8 u+.6 u ) 0.6 u () 4 Solving equations () and () gives: v0.6u+.4u v.5u.5u+w.4u w0.9u So final speeds are: v T (0.6u) +(.5u) ( )u.6u 6 00 u v T u v B (.u) +(0.9u) ( )u.5u v B.5u Angle between paths is the difference between the angles each makes with the line of centres: tanα usinθ.u w 0.9u 4 and tanβ usinθ 0.6u 3 v.5u tanα tanβ Usingtan( α β) tanα tanβ Immediately after impact, the speeds of the spheres are u m s and.5u m s and the angle between their paths is arctan 4 3 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 3

14 Challenge a From earlier questions know that in a collision: tan(angle of departure) etan (angle of approach) For the first collision: 0 5 tanα etanθ 7 7 α (4 s.f.) Since this is less than 45, the bounce is internal (see diagram) and, for the second collision: tanβ etan(45 + α) tan(80.54 ) 3 β 7.57 Since this is greater than 45, the bounce is external (see diagram) and, for the third collision: tanγ etan(80 45 β) tan(63.43 ) γ 45 In other words, the ball has rebounded from the first wall and is now travelling parallel to the second wall so will not hit either again. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

15 Challenge b From earlier questions know that in a collision: vcos (angle of departure) ucos (angle of approach) For the first collision: v cosα cosθ v For the second collision: 74 vcosβ cos(45 + α) 49 v For the third collision: v 3 cosγ 0 49 cos(80 45 β) v v Initial kinetic energy mu Loss in kinetic energy mu mv 3 m(u v 3 ) mu ( v ) u v Proportion loss of kinetic energy mu u The percentage of kinetic energy lost is 94.6% (to 3s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5