Elastic collisions in two dimensions 5B

Transcription

1 Elastic collisions in two dimensions 5B a First collision: e=0.5 cos α = cos30 () sin α = 0.5 sin30 () Squaring and adding equations () and () gies: cos α+ sin α = 4cos 30 + sin 30 (cos α+ sin α)= = 3 4 = 3 =.80ms (3 s.f.) Diiding equation () by equation () gies: = 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.

2 b Second collision: e=0.5 cosβ = cos(90 α) = sin α (3) sinβ = 0.5 sin(90 α) = 0.5 cos α (4) Squaring and adding equations (3) and (4) gies: cos β+ sin β = sin α+ 0.5 cos α As = 3, by Pythagoras sinα = 3 and cosα = 3 3 (cos β+ sin β)= = = =ms Diiding equation (4) by equation (3) gies: tanβ = 0.5 = 3 = 3 β= 60 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

3 a First collision for motion parallel to the wall: cos0 = cos40 () = cos40 cos0 = 0.85ms (3 s.f.) b sin 0 = e sin 40 () Diiding equation () by equation () gies: tan 0 = etan 40 tan 0 e= = (3 s.f.) tan 40 c Second collision for motion parallel to the wall: cosα = cos70 (3) sinα = e sin 70 (4) Diiding equation (4) by equation (3) gies: tan0 tan70 = e tan 70 = tan 40 but tan 0 tan70 = So = = tan(90 40 ) = tan 50 tan 40 α = 50 Substituting into equation (3) gies: cos50 = cos70 cos40 cos50 = cos70 cos0 but cos50 = sin 40 and cos70 = sin 0 cos40 sin 0 So = cos0 sin 40 tan 0 = = tan ms (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 3 a First collision let the angle the sphere makes with the wall after the collision be α 0.3cosα = 0.5cos30 () cosα = = α = 9.7 (3 s.f.) After the impact with the first wall, the sphere moes at 9.7 to the wall b 0.3sinα = 0.5esin30 () Diiding equation () by equation () gies: = e tan 30 tan9.7 e= = = 0.6 (3 s.f.) tan30 tan30 c First collision let the angle the sphere makes with the wall after the collision be β and its speed after the collision be cosβ = 0.3cos(90 α) = 0.3sin α (3) sinβ = 0.3esin(90 α) = 0.3ecos α (4) Squaring and adding equations (3) and (4) gies: cos β + sin β = 0.3 sin α+ 0.3 e cos α (cos β+ sin β) = 0.3 ( cos α+ e cos α) = 0.059( ) = 0.04 Kinetic energy after second collision m = = = J (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 4 a First collision let the angle the sphere makes with the wall after the collision be α, its speed before impact be u and its speed after impact Kinetic energy before first collision = u So u = 9 u= 3m s cos α = 3cos60 =.5 () sin α = sin 60 =.5 3 () Squaring and adding equations () and () gies: cos α+ sin α = (cos α+ sin α)= =.46ms (3 s.f.) Diiding equation () by equation () gies: = 0.75tan60 = α = 5.4 (3 s.f.) b Second collision for motion parallel to the wall: cosβ = cos(90 α) = sin α (3) sinβ = 0.6 sin(90 α) = 0.6 cos α (4) Squaring and adding equations (3) and (4) gies: cos β + sin β = sin α cos α (cos β+ sin β) = (sin α cos α) = ( ) = Kinetic energy after second collision = m = = 4.6 J (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 5 First collision cosβ = ucos α () sinβ = eusin α () Diiding equation () by equation () gies: tanβ =e Second collision wcosθ = cos(90 β) = sin β (3) wsinθ = esin(90 β) = ecos β (4) Diiding equation (4) by equation (3) gies: e tanθ = tanβ = e e = = cotα = tan(90 α) So θ = 90 α, which means that after the collision with the second wall the sphere s path is parallel to its original path but in the opposite direction. Substituting for θ in equation (3) gies: wcosθ = wcos(90 α) = sinβ wsinα = sinβ So from equation () wsinα = sinβ = eusinα w= eu, the speed after the second collision Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 6 First impact: u cosα = ucos45 = () u sinα = eusin 45 = () 3 Squaring and adding equations () and () gies: cos α+ sin α =u + 6 = u 3 Diiding equation () by equation () gies: = α = 30 3 So = 80 β β = 60 Second impact: wcosθ = cos60 = (3) wsinθ = esin60 = (4) Squaring and adding equations (3) and (4) gies: w cos θ + w sin θ = w u = = 3 u w= = 3 3u ms 3 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 7 a First impact: cos α = 5cos30 () sin α = 0.8 5sin30 () Squaring and adding equations () and () gies: 3 cos α+ sin α = = = 4.77ms (3 s.f.) 4 Diiding equation () by equation () gies: 4 = 0.8 tan 30 = α = 4.8 (3 s.f.) 5 3 b Second impact: cosβ = cos(45 + α ) (3) sinβ = 0.8 sin(45 + α ) (4) Diiding equation (4) by equation (3) gies: tanβ= 0.8tan(45 +α)=0.8tan( )=.733 β= 65.3 (3 s.f.) From equation (3) cos65.9 = cos(45 +α) = 4.77 cos69.79 cos65.9 = 3.94ms (3 s.f.) c If e > 0.8 and e remains unchanged, (45 + α) and do not change; howeer, tan β increases, so β increases and (which depends on and cos β) increases. So the elocity of the sphere after the second collision would be greater and the angle it makes with the wall after the collision would be greater. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 8 a Since first collision causes no change to the i component, first collision must be with the wall parallel to this ector. Considering j components only for the first collision: =4e e= Considering only the i component changes for the second collision: 5=ep=p p=.5 ( u ) Loss of kinetic energy= m = m = m = m J 8 ((5 4 ) (.5 )) b Initial elocity, u = 5i 4j and final elocity, =.5i + j The coefficients of i and j hae the same ratio in both cases, but the signs of both are reersed: u and are therefore antiparallel. The sphere is deflected through a total angle of a First collision let the angle the sphere makes with the wall after the collision be α, and its speed after impact For motion parallel to wall: cos α =.5cos45 () For motion perpendicular to wall: sinα = 0.6.5sin45 () Diiding equation () by equation () gies: = 0.6 tan 45 5 = 0.6 cos α = 34 α = = 3.0 (3 s.f.) Substituting into equation () gies: 5 34 = 5 = 7 = =.06ms Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 9 b First collision let the angle the sphere makes with the wall after the collision be β, and its speed after impact For motion parallel to wall: cosβ= cos(60 α) (3) For motion perpendicular to wall: sinβ = 0.6 sin(60 α) (4) Diiding equation (4) by equation (3) gies: tanβ= 0.6 tan( )= β=8.4 cosβ= Substituting into equation (3) gies: = =.8997 (4 d.p.) Kinetic energy after second collision= = 0.80 J (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

11 0 First collision For motion parallel to wall: cosα =0cos30 () For motion perpendicular to wall: sinα = 0.7 0sin30 () Diiding equation () by equation () gies: = 0.7tan = cosα = α =.005 (3 d.p.) Substituting into equation () gies: = 0 = 349 First collision For motion parallel to wall: cosβ= cos(80 75 α)= cos8.994 (3) For motion perpendicular to wall: sinβ = 0.5 sin(80 75 α)= 0.5 sin8.994 (4) Diiding equation (4) by equation (3) gies: tanβ= 0.5tan8.994 = β = 76.9 (3 d.p.) Substituting into equation (3) gies: 349 cos 76.9 = cos8.994 = Loss of kinetic energy= m(u ) = ( ) = 77. J (3 s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

12 Challenge a i Vertical component of motion is not changed by collision with wall. Therefore time taken for ball to reach floor for the first time is time taken for ball to fall m from rest. s= ut+ at = 9.8 t t= = Horizontal component of motion is 0 m s before collides with wall and = m s after collision. So time taken to reach wall: s.4 t = = = 0. s 0 So time for which ball traelling at m s is t = = Distance traelled in this time = t = = 3.98 m (3 s.f.) The ball first bounces on the floor 3.98 m from the wall. ii Vertical component of speed immediately before it hits the floor: =u + as = 9.8 = 9.6 Vertical component of speed immediately after it hits the floor = Maximum height gien when ertical component is again zero, so: =u + as 0=( ) ( 9.8 s) s=0.6 = 0.36m The maximum height after this bounce is 0.36 m. b If the ball is hotter and e increases, the horizontal speed after hitting the wall will be greater and the ball will therefore trael further before hitting the floor. Similarly, the ertical component of the speed immediately after it hits the floor will be greater and the maximum height after the first bounce will also be greater. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.