Physics 4A Solutions to Chapter 9 Homework

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1 Physics 4A Solutions to Chapter 9 Homework Chapter 9 Questions:, 10, 1 Exercises & Problems: 3, 19, 33, 46, 51, 59, 86, 90, 100, 104 Answers to Questions: Q 9- (a) ac, cd, bc (b) bc (c) bd, ad Q 9-10 a, c, e, : the sum o the momenta ater explosion does not equal the momentum beore explosion Q 9-1 (a) positive (b) positive (c) and 3 Answers to Problems: P 9-3 We use Eq. 9-5 to locate the coordinates. (a) By symmetry x com = d 1 / = (13 cm)/ = 6.5 cm. The negative value is due to our choice o the origin. (b) We ind y com as y com my + m y ρ Vy + ρ V y = = m + m ρv + ρ V i com, i a com, a i i com, i a a cm, a i a i i a a 3 3 ( 11 cm / )( 7.85 g/cm ) + 3( 11 cm / )(.7 g/cm ) = = 8.3 cm g/cm +.7 g/cm (c) Again by symmetry, we have z com = (.8 cm)/ = 1.4 cm. P 9-19 (a) The change in kinetic energy is

2 ( ) ( ) i 100 kg 51 km/h 41 km/h ( ) ( ) ( ) Δ K = mv mv = = ( ) ( )( ) kg km/h 10 m/km 1 h/3600 s = J. (b) The magnitude o the change in velocity is Δ v = v + v = 41 km/h + 51 km/h = 65.4 km/h ( ) ( ) ( ) ( ) so the magnitude o the change in momentum is Δp m Δv i F1000 m/km = = b H G I 100 kg gb65. 4 km / hg K J = s/h (c) The vector Δp points at an angle θ south o east, where F HG I F KJ = H G I K J = 1 vi 1 41 km / h θ = tan tan 39. v 51 km / h kg m / s. P 9-33 We use coordinates with +x rightward and +y upward, with the usual conventions or measuring the angles (so that the initial angle becomes = 15 ). Using SI units and magnitudeangle notation (eicient to work with when using a vector-capable calculator), the change in momentum is J =Δ p= p p = = i ( ) ( ) ( ) (a) The magnitude o the impulse is J = Δ p= 5.86 kg m/s = 5.86 N s. (b) The direction o J is 59.8 measured counterclockwise rom the +x axis. (c) Equation 9-35 leads to 5.86 N s J = F Δ t = F = s 3 avg 5.86 N s avg N. 3 We note that this orce is very much larger than the weight o the ball, which justiies our (implicit) assumption that gravity played no signiicant role in the collision. (d) The direction o F avg is the same as J, 59.8 measured counterclockwise rom the +x axis.

3 P 9-46 Our +x direction is east and +y direction is north. The linear momenta or the two m =.0 kg parts are then p1 = mv1 = mv 1 j where v 1 = 3.0 m/s, and e j e j p = mv = m v i + v j = mv cosθ i + sin θ j x y where v = 5.0 m/s and θ = 30. The combined linear momentum o both parts is then P= p + p = mv ˆj + mv cosθ ˆi + sinθˆj = mv cosθ ˆi + mv + mv sinθ)ˆ j ( ) ( ) ( ( ) ( )( )( ) ˆ ( ) ( )( ) =.0 kg 5.0 m/s cos 30 i +.0 kg 3.0 m/s m/s sin 30 ˆj ( ˆ ˆ) = 8.66i + 11 j kg m/s. From conservation o linear momentum we know that this is also the linear momentum o the whole kit beore it splits. Thus the speed o the 4.0-kg kit is ( 8.66 kg m/s) ( 11 kg m/s) P Px + Py + v = = = = 3.5 m/s. M M 4.0 kg P 9-51 In solving this problem, our +x direction is to the right (so all velocities are positive-valued). (a) We apply momentum conservation to relate the situation just beore the bullet strikes the second block to the situation where the bullet is embedded within the block. ( kg) v= ( kg)(1.4 m/s) v= 71 m/s. (b) We apply momentum conservation to relate the situation just beore the bullet strikes the irst block to the instant it has passed through it (having speed v ound in part (a)). which yields v 0 = 937 m/s. ( kg) v 0 = (1.0 kg)(0.630 m/s) + ( kg)(71 m/s) P 9-59 As hinted in the problem statement, the velocity v o the system as a whole, when the spring reaches the maximum compression x m, satisies

4 m 1 v 1i + m v i = (m 1 + m )v. The change in kinetic energy o the system is thereore ( mv + mv ) 1 1 Δ K= ( m + m) v mv mv = mv mv ( ) 1 1 1i i 1 1i i 1 1i i m1+ m which yields ΔK = 35 J. (Although it is not necessary to do so, still it is worth noting that 1 mm 1 algebraic manipulation o the above expression leads to ΔK = dm1+ mi v rel where v rel = v 1 v ). Conservation o energy then requires 1 ΔK ( 35 J) kxm = ΔK xm = = = 0.5 m. k 110 N/m P 9-86 (a) We use Eq twice: v = m 1 m 1 + m v 1i = v 3 = m m + m 3 v = m 1 1.5m 1 (4.00 m/s) = 16 3 m/s m (16/3 m/s) = m 9 m/s = 7.11 m/s. (b) Clearly, the speed o block 3 is greater than the (initial) speed o block 1. (c) The kinetic energy o block 3 is K 3 = 1 m 3 v 3 = 1 3 m v 1i = K 1i. We see the kinetic energy o block 3 is less than the (initial) K o block 1. In the inal situation, the initial K is being shared among the three blocks (which are all in motion), so this is not a surprising conclusion. (d) The momentum o block 3 is p 3 = m 3 v 3 = 1 m v 1i = 4 9 p 1i and is thereore less than the initial momentum (both o these being considered in magnitude, so questions about ± sign do not enter the discussion).

5 P 9-90 (a) We ind the momentum p nr o the residual nucleus rom momentum conservation. p = p + p + p = ˆ+ ˆ + p 3 0 ( ni e v nr kg m/s ) i ( kg m/s) j nr Thus, p nr ˆ 3 = (1. 10 kg m/s) i + ( kg m/s) ˆj. Its magnitude is 3 = kg m/s kg m/s = kg m/s. p nr ( ) ( ) (b) The angle measured rom the +x axis to p nr is θ = = kg m/s kg m/s tan 8. (c) Combining the two equations p = mv and K m = m n r ) = 1 ( kg m/s) 6 ( ) p K = = = m kg mv, we obtain (with p = p n r and J. P (a) We use Fig. 9-1 o the text (which treats both angles as positive-valued, even though one o them is in the ourth quadrant; this is why there is an explicit minus sign in Eq as opposed to it being implicitly in the angle). We take the cue ball to be body 1 and the other ball to be body. Conservation o the x and the components o the total momentum o the two-ball system leads to: mv 1i = mv 1 cos θ 1 + mv cos θ 0 = mv 1 sin θ 1 + mv sin θ. The masses are the same and cancel rom the equations. We solve the second equation or sin θ : v F sinθ = sinθ1 = sin H G m/si 00. K J =. v m/s Consequently, the angle between the second ball and the initial direction o the irst is θ = (b) We solve the irst momentum conservation equation or the initial speed o the cue ball. v = v cosθ + v cos θ = (3.50 m/s)cos.0 + (.00 m/s)cos 41.0 = 4.75 m/s. 1 i 1 1

6 (c) With SI units understood, the initial kinetic energy is and the inal kinetic energy is 1 1 Ki = mvi = m( 475. ) = 113. m Kinetic energy is not conserved. c h m K = mv1 + mv = m (. 350) + (. 00) = 81.. P We treat the car (o mass m 1 ) as a point-mass (which is initially 1.5 m rom the right end o the boat). The let end o the boat (o mass m ) is initially at x = 0 (where the dock is), and its let end is at x = 14 m. The boat s center o mass (in the absence o the car) is initially at x = 7.0 m. We use Eq. 9-5 to calculate the center o mass o the system: x com = m 1x 1 + m x m 1 + m = (1500 kg)(14 m 1.5 m) + (4000 kg)(7 m) 1500 kg kg = 8.5 m. In the absence o external orces, the center o mass o the system does not change. Later, when the car (about to make the jump) is near the let end o the boat (which has moved rom the shore an amount δx), the value o the system center o mass is still 8.5 m. The car (at this moment) is thought o as a point-mass 1.5 m rom the let end, so we must have x com = m 1x 1 + m x m 1 + m = (1500 kg)( δx m) + (4000 kg)(7 m + δx) 1500 kg kg = 8.5 m. Solving this or δx, we ind δx = 3.0 m.

(a) The x coordinate of the center of mass is. 3.0 kg kg kg. (b) The y coordinate of the center of mass is

(a) The x coordinate of the center of mass is. 3.0 kg kg kg. (b) The y coordinate of the center of mass is 1. Our notation is as follows: x 1 0 and y 1 0 are the coordinates of the m 1 3.0 kg particle; x.0 m and y 1.0 m are the coordinates of the m 4.0 kg particle; and, x 3 1.0 m and y 3.0 m are the coordinates