increases. In part (b) the impulse and initial momentum are in opposite directions and the velocity decreases.

Transcription

1 8IDENTIFY and SET U: p = K = EXECUTE: (a) 5 p = (, kg)( /s) = kg /s 5 p kg /s (b) (i) = = = 6 /s (ii) kg =, so T T SUV SUV, kg ( /s) 68 /s T SUV = T = = SUV kg EVALUATE:The SUV ust hae less speed to hae the sae kinetic energy as the truck than to hae the sae oentu as the truck 87IDENTIFY:The aerage force on an object and the object s change in oentu are related by Eq 89 The weight of the ball is w= g SET U:Let + x be in the direction of the final elocity of the ball, so x = and x = 5 /s EXECUTE: ( Fa) x( t t) = x gies x x (45 kg)(5 /s) x ( Fa) x = = = 56 N 3 t t s w = (45 kg)(98 /s ) = 44 N The force exerted by the club is uch greater than the weight of the ball, so the effect of the weight of the ball during the tie of contact is not significant EVALUATE:Forces tie exerted during collisions typically are ery large but act for a short 89IDENTIFY:Use Eq 89 We know the intial oentu and the ipluse so can sole for the final oentu and then the final elocity SET U:Take the x-axis to be toward the right, so x =+ 3 /s Use Eq 85 to calculate the ipulse, since the force is constant EXECUTE: (a) J x = px px J = F ( t t ) = ( + 5 N)(5 s) =+ 5 kg /s x x Thus p = J + p =+ 5 kg /s + (6 kg)( + 3 /s) = + 73 kg /s (b) x x x p 73 kg /s 6 kg x x = = =+ 8 kg /s (to the right) J = F ( t t ) = ( N)(5 s) = 6 kg /s (negatie since force is to left) x x px = Jx + px = 6 kg /s + (6 kg)( + 3 /s) = kg /s p kg /s 6 kg x x = = = 75 /s (to the left) EVALUATE: In part (a) the ipulse and initial oentu are in the sae direction and x increases In part (b) the ipulse and initial oentu are in opposite directions and the elocity decreases 86IDENTIFY:Apply conseration of oentu to the syste of you and the ball In part (a) both objects hae the sae final elocity SET U:Let + x be in the direction the ball is traeling initially A = 4 kg (ball) B = 7 kg (you) EXECUTE: (a) x = x gies (4 kg)( /s) = (4 kg + 7 kg) and = 568 /s (b) x = x gies (4 kg)( /s) = (4 kg)( 8 /s) + (7 kg) B and B = 3 /s EVALUATE:When the ball bounces off it has a greater change in oentu and you acquire a greater final speed 87 IDENTIFY:Apply conseration of oentu to the syste of the two pucks SET U:Let + x be to the right

2 EXECUTE: (a) x = x says (5) A = (5 kg)( /s) + (35 kg)(65 /s) and A = 79 /s (b) K = (5 kg)(79 /s) = 78 J K = (5 kg)( /s) + (35 kg)(65 /s) = 757 J and Δ K = K K = 3 J 88IDENTIFY:The x and y coponents of the oentu of the syste of the two asteroids are separately consered SET U:The before and after diagras are gien in Figure 88 and the choice of coordinates is indicated Each asteroid has ass EXECUTE: (a) x = x gies A = A cos3 + B cos 45 4 /s = 866A + 77B and 77 = 4 /s 866 B A = gies = sin3 sin45 and 5 = 77 y y A B Cobining these two equations gies 5 = 4 /s 866 and 93 /s Then (b) K A B = K = + A B A A 5 = (93 /s) = 7 /s 77 A B A = K + (93 /s) + (7 /s) = = = 84 K A B A (4 /s) ΔK K K K K K K = = = 96 96% of the original kinetic energy is dissipated during the collision EVALUATE:We could use any directions we wish for the x and y coordinate directions, but the particular choice we hae ade is especially conenient 833 IDENTIFY:The forces the two players exert on each other during the collision are uch larger than the horizontal forces exerted by the slippery ground and it is a good approxiation to assue oentu conseration Each coponent of oentu is separately consered SET U:Let + x be east and +y be north After the collision the two players hae elocity Let the linebacker be object A and the halfback be object B, so Ax =, A y= 88 /s, Bx = 7 /s and = Sole for x and By y EXECUTE: x = x gies A Ax+ B B x= ( A+ B) x y y + (85 kg)(7 /s) A Ax B Bx x = = = A + B kg + 85 kg = gies A Ay + B B y = ( A + B) y = x + y = 59 /s + ( kg)(88 /s) A Ay B By y = = = A + B kg + 85 kg tan y θ = = and x 496 /s 34 /s θ = /s 496 /s The players oe with a speed of 59 /s and in a direction 58 north of east EVALUATE:Each coponent of oentu is separately consered 837IDENTIFY:Since consered friction forces fro the road are ignored, the x and y coponents of oentu are

3 SET U:Let object A be the subcopact and object B be the truck After the collision the two objects oe together with elocity Use the x and y coordinates gien in the proble = = x = (6 /s)sin 4 = 65 /s ; = (6 /s)cos4 = 46 /s EXECUTE: x = x gies A Ax= ( A+ B) x y y A B Ax x A = gies = ( + ) y + 95 kg + 9 kg = = (65 /s) = 95 /s 95 kg A By A B y + 95 kg + 9 kg = = (46 /s) = 9 /s 9 kg A B By y A Before the collision the subcopact car has speed 95 /s and the truck has speed 9 /s EVALUATE:Each coponent of oentu is independently consered A y B y d 857IDENTIFY: a dt d uch during that interal The thrust is F = ex dt SET U:Take to hae the constant alue kg + 7 kg = 8 kg d / dt is negatie since the ass of the MMU decreases as gas is ejected d 8 kg EXECUTE: (a) = a = (9 /s ) = 6 kg/s In 5 s the ass that is dt ex 49 /s ejected is ( 6 kg/s)(5 s) = 53 kg ex = Assue that d / dt is constant oer the 5 s interal, since doesn t change d (b) F = ex = (49 /s)( 6 kg/s) = 59 N dt EVALUATE:The ass change in the 5 s is a ery sall fraction of the total ass, so it is accurate to take to be constant 86 IDENTIFY and SET U:Use Eq 84: exln ( / ) = ( fired fro rest ), so / = ln ( / ) ex = Thus / / ex, / ex = e or / = e If is the final speed then is the ass left when all the fuel has been expended; of the initial ass that is not fuel 3 5 (a) EXECUTE: = c= 3 /s gies 5 (3 /s) /(/s) 66 / = e = 7 / is the fraction EVALUATE:This is clearly not feasible, for so little of the initial ass to not be fuel (3 /s)/( /s) (b) EXECUTE: = 3 /s gies / = e = 3 EVALUATE:3% of the total initial ass not fuel, so 777% is fuel; this is possible 869IDENTIFY:The x and y coponents of the oentu of the syste are consered Set Up: After the collision the cobined object with ass tot = kg oes with elocity Sole for and Cx Cy EXECUTE: (a) gies

4 y y Cx ( kg)( 5 /s) + (3 kg)( 5 /s)cos6 ( kg)(5 /s) = 5 kg = gies + + = Cy A Ay B By C Cy tot y = 75 /s Cx A Ay+ B By tot y (3 kg)( 5 /s)sin 6 = = =+ 6 /s 5 kg (b) = + = 77 /s Δ K = K K C Cx Cy C Δ K = ( kg)(5 /s) [ ( kg)(5 /s) + (3)(5 /s) + (5 kg)(77 /s) ] Δ K = 9 J EVALUATE: Since there is no horizontal external force the ector oentu of the syste is consered The forces the spheres exert on each other do negatie work during the collision and this reduces the kinetic energy of the syste 88IDENTIFY: The ipulse applied to the ball equals its change in oentu The height of the ball and its speed are related by conseration of energy SET U: Let +y be upward EXECUTE: Applying conseration of energy to the otion of the ball fro its height h to the floor gies = gh, where is its speed just before it hits the floor Just before it hits, it is traeling downward, so the elocity of the ball just before it hits the floor is y = gh Applying conseration of energy to the otion of the ball fro just after it bounces off the floor with speed to its axiu height of 9h gies = It is oing upward, so =+ g(9 h ) The ipulse applied to the g(9 h) ball is Jy = py py = ( y y) = g(9 h) + gh = 76 gh The floor exerts an upward ipulse of 76 gh to the ball EVALUATE: The ipulse increases when increases and when h increases The ball does not return to its initial height because soe echanical energy is dissipated during the collision with the floor 89IDENTIFY: Apply conseration of oentu to the syste consisting of Jack, Jill and the crate The speed of Jack or Jill relatie to the ground will be different fro 4 /s SET U: Use an inertial coordinate syste attached to the ground Let +x be the direction in which the people jup Let Jack be object A, Jill be B, and the crate be C EXECUTE: (a) If the final speed of the crate is, C x=, and Ax= Bx= 4 /s x = x gies A Ax+ B Bx + C Cx = (75 kg)(4 /s ) + (45 kg)(4 /s ) + (5 kg)( ) = and (75 kg + 45 kg)(4 /s) = = 356 /s 75 kg + 45 kg + 5 kg (b) Let be the speed of the crate after Jack jups Apply oentu conseration to Jack juping: (75 kg)(4 /s) (75 kg)(4 /s ) + (6 kg)( ) = and = = /s Then apply 35 kg oentu conseration to Jill juping, with being the final speed of the crate: x = x gies (6 kg)( ) = (45 kg)(4 /s ) + (5 kg)( ) (45 kg)(4 /s) + (6 kg)( /s) = = 5 /s 6 kg y

5 (c) Repeat the calculation in (b), but now with Jill juping first Jill jups: (45 kg)(4 /s ) + (9 kg)( ) = and = 33 /s Jack jups: (9 kg)( ) = (75 kg)(4 /s ) + (5 kg)( ) (75 kg)(4 /s) + (9 kg)(33 /s) = = 466 /s 9 kg EVALUATE: The final speed of the crate is greater when Jack jups first, then Jill In this case Jack leaes with a speed of 78 /s relatie to the ground, whereas when they both jup siultaneously Jack and Jill each leae with a speed of only 44 /s relatie to the ground