Elastic collisions in two dimensions 5A

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1 Elastic collisions in two dimensions 5A 1 a e= 1 3, tanα = 3 4 cosα = 4 5 and sinα = 3 5 (from Pythagoras theorem) For motion parallel to the wall: 4 vcosβ = ucosα vcos β = u (1) 5 For motion perpendicular to the wall: vsinβ = eusinα vsin β = u = u () Squaring and adding equations (1) and () gives: 16 1 v cos β+ v sin β = u + u v (cos β + sin β) = u 5 v= 17 5 u b Dividing equation () by equation (1) gives: tanβ = 1u 5 4 u = 1 4 =0.5 5 β =14.04 ( d.p.) tanα = 3 4 =0.75 α =36.87 ( d.p.) Angle of deflection=α+β= =50.9 (1 d.p.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 For motion parallel to the wall: 7u cos β = u cos30 8 7u 3u cos β = (1) 8 For motion perpendicular to the wall: 7u sin β = eu sin30 8 7u eu sin β = () 8 Squaring and adding equations (1) and () gives: u cos β u sin β= 3 4 u e u u (cos β+sin β)=u 4 +e =3+e e = 1 16 e= 1 4 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.

3 3 a e= 3 5, tanα = 5 cosα = and sinα = 5 13 (from Pythagoras theorem) For motion parallel to the wall: 1 vcosβ = ucosα vcos β = u (1) 13 For motion perpendicular to the wall: vsinβ = eusinα vsin β = u = u () Squaring and adding equations (1) and () gives: v cos β+ v sin β = u + u v (cos β+ sin β) = u u 9 17u 3 17u v= = = Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 4 tanα = cosα = 1 5 and sinα = 5 (from Pythagoras theorem) For motion parallel to the wall: 3u cos β = u cos α 4 3u u cos β = (1) 4 5 For motion perpendicular to the wall: 3u sin β = eu sin α 4 3u eu sin β = () 4 5 Squaring and adding equations (1) and () gives: u cos β+ u sin β = u + eu e u (cos β+ sin β) = u = + e e = e= 64 8 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 5 Before the impact: After the impact: 1 The component of velocity parallel to the slope is 8sin30 = 8 = Perpendicular to the slope: v= e 8cos30 = 8 = 3 4 Therefore the speed immediately after impact = = 19ms 1 6 Before the impact: After the impact: The component of velocity parallel to the slope is 10sin0 Perpendicular to the slope: v= e 10cos0 = 10cos0 = cos0 5 Therefore the speed immediately after impact = (10sin0 ) +(4cos0 ) = 5.86=5.08ms 1 (3 s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 7 a Before the impact: After the impact: The component of velocity parallel to the slope is 5 sin45 =5 1 =5 Perpendicular to the slope: v= cos = = = Therefore the speed immediately after impact = = 31.5=5.59ms 1 (3 s.f.) b The impulse is perpendicular to the surface: I = mv m( 5 cos45 ) = ( 5) = = = 5.65Ns Before the impact: After the impact: tanα = 3 4 cosα = 4 5 and sinα = 3 5 (from Pythagoras theorem) The component of velocity parallel to the slope is 7.5sinα = =4.5 Perpendicular to the slope: v=e 7.5cosa=e =6e The speed immediately after impact is 5 m s 1, so 5 = (6 e) 5= e 4.75 e = = e= 0.36 ( s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 9 a Let the velocity v of the ball immediately after impact be pi + qj Parallel to the wall: q = 3 Perpendicular to the wall: p=e 5=.5 Sov=(.5i 3j)ms 1 b 1 1 Kinetic energy after impact= 0.8 ( ) = 6.1 Kinetic energylost= = 7.5 J Kinetic energy before impact = 0.8 (5 + 3 ) = 13.6 c Using the scalar product to determine the angle of deflection: cosθ = u.v u v cosθ = 5(.5)+( 3)( 3) = = = θ =98.8 (3 s.f.) The angle of deflection is 98.8 (3 s.f.). 10 a Let the velocity v of the ball immediately after impact be pi + qj Parallel to the wall: p = 3 Perpendicular to the wall: q=e 6= Sov=(3i j)ms 1 Speed= 3 + = 13ms 1 b Kinetic energybefore impact= 1 1 (3 +6 )=.5 1 Kinetic energy after impact= 1 13= 6.5 Kinetic energylost=.5 6.5= 16 J Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 11 a Let v = a + b, where a is parallel to the wall and b is perpendicular to the wall. 1 ( i+ j ) is a unit vector in the direction of the wall and 1 ( ) i + j is a unit vector perpendicular to the wall. Parallel to the wall: a= (4i+j). 1 (i+j) 1 (i+j)= (i+j)=3i+3j Perpendicular to the wall: b= 1 3 (4i+j). 1 (i j) 1 ( i+j)=1 3 1 (4 ) 1 ( i+j)= 1 3 ( i+j) Sov= 3i+ 3j i+ j= i+ j ms b Kinetic energybefore impact= 1 (4 + )= Kinetic energy after impact= + = Kinetic energy lost= 0 = J 9 9 Proportion of kinetic energy lost= 100= 8.89% (3 s.f.) 0 c Using the scalar product to determine the angle of deflection: u.v cosθ = u v ( 3) + ( 3) cosθ = = = θ = 4.8 (3 s.f.) The angle of deflection is 4.8 (3 s.f.). Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 1 For motion parallel to the cushion: 8 vcosβ = 8cos45 vcos β = (1) For motion perpendicular to the wall: vsinβ = e 8sin vsinβ = 8 = () 5 5 Squaring and adding equations (1) and () gives: v cos β + v sin β = v (cos β + sin β) = = v= 37.1 = 6.09ms 1 Dividing equation () by equation (1) gives: tanβ= = 5 =0.4 β=1.8 (3 s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 13 a For motion parallel to the cushion: vcosβ = ucos50 (1) For motion perpendicular to the cushion: vsinβ = eusin50 () Dividing equation () by equation (1) gives: eusin50 tanβ = = e tan 50 ucos50 Hence β is independent of u. b If the ball rebounds at right angles from its original direction: β =40 From part a: tanβ =tan40 =etan50 e= tan40 =0.704 (3 s.f.) tan50 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10

11 14 a tanα = 3 4 cosα = 4 5 and sinα = 3 5 tanβ= 5 1 cosβ=1 5 and sinβ = (from Pythagoras theorem) (from Pythagoras theorem) For motion parallel to the cushion: vcosβ=ucosα v= u =13 15 u 1 Kinetic energy before impact= mu Kinetic energy lost= mu mv = mu mu = mu 1 = mu Kinetic energy lost 56 Fraction of kinetic energy lost= = Kinetic energy before impact 5 b For motion perpendicular to the cushion: vsinβ=eusinα u 5 13 =eu 3 5 e= = 5 9 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 11

12 15 a Impulse = mv mu I= m(i+ j) m(5i j) I= m( 3i+ 4 j) 3 ( i 4j) I= 5m The impulse acts in the direction of the unit vector ( i+ j) = ( i+ j ) b Component of (5i j) in the direction of the impulse = (5i j). 1 5 ( 3i+4j) 1 5 ( 3i+4j)= 1 5 ( 15 8)1 3 ( 3i+4j)= ( 3i+4j) Component of (i + j) in the direction of the impulse = (i+j). 1 5 ( 3i+4j) 1 5 ( 3i+4j)= 1 5 ( 6+8)1 5 ( 3i+4j)= ( 3i+4j) By Newton s law of restitution: 3 = e e= Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

13 16 a Impulse = mv mu I=m(3i j) m(i+3j) I=(i 4j) The impulse has magnitude 17Ns in the direction parallel to the unit vector 1 ( 4 ) 17 i j b Component of (i + 3j) in the direction of the impulse 1 = (i+3j). 17 (i 4j) 1 17 (i 4j)= 1 17 ( 1) 1 17 (i 4j)= (i 4j) Component of (3i j) in the direction of the impulse 1 = (3i j). 17 (i 4j) 1 17 (i 4j)= 1 17 (3+4) 1 17 (i 4j)= 7 17 (i 4j) By Newton s law of restitution: = e e= = c 1 1 Kinetic energy after impact= (3 + 1 ) = 10 Kinetic energylost= 13 10= 3 J Kinetic energy before impact = ( + 3 ) = a For motion parallel to the wall: vcos(90 α) = 3vcosα sinα = 3cos α (as cos(90 α) = sin α) tanα = 3 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 13

14 17 b For motion perpendicular to the wall: vsin(90 α)=eusinα e= vcosα 3vsinα = 1 3tanα = For motion parallel to wall: vcos α = ucos (1) α For motion perpendicular to wall: vsinα = eusinα = 0.4usin α () Dividing equation () by equation (1) gives: tanα 0.4tan = α (3) Using the identity: tanθ+ tanφ tan( θ+ φ) = with θ = φ= α 1 tanθtanφ α tan tanα = α 1 tan Writing tanα as t, and substituting into equation (3) t t= 0.4 if t 0 1 t 1 t = 0.8 t = 0. So t= tanα = 0. α = 4.09 α = 48. (3 s.f.) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 14

15 Challenge Let angle of rebound from W 1 be β and let the point where W 1 meets W be O For motion parallel to wall W 1 : vcosβ = ucos α (1) For motion perpendicular to wall W 1 : vsinβ = eusin α () Dividing equation () by equation (1) gives: OQ tanβ = etanα etanα = 1 Using Pythagoras theorem: PQ =OQ +1 PQ =e tan α+1 So PQ= e tan α+1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 15