# Physics 2A Chapter 3 HW Solutions

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1 Phscs A Chapter 3 HW Solutons Chapter 3 Conceptual Queston: 4, 6, 8, Problems: 5,, 8, 7, 3, 44, 46, 69, 70, 73 Q3.4. Reason: (a) C = A+ B onl A and B are n the same drecton. Sze does not matter. (b) C = A B A and B are n the opposte drecton to each other. Sze matters onl n that A > B because C as a magntude can onl be postve. Assess: Vsualze the stuaton wth arrows. Q3.6. Reason: The acceleraton o the ball s due to gravt, so the acceleraton the ball eperences s alwas straght downward. (a) The veloct vector o the ball alwas has a component n the horzontal drecton snce t was thrown at an angle o 40. The horzontal component o the ball s veloct s constant throughout ts trajector. Snce the veloct vector alwas has a component n the horzontal drecton, t s never pontng entrel straght up or down. So there s no pont on the trajector where the acceleraton and veloct are parallel. Note that the ball were thrown straght up or down nstead o at 40, the ball s veloct and acceleraton would be parallel. (b) The acceleraton o the ball s alwas straght downward. For the veloct vector to be perpendcular to the acceleraton vector, the veloct must pont entrel n the horzontal drecton. The veloct vector alwas has a component n the horzontal drecton, as reasoned n part (a). It has a component n the vertcal drecton durng most o ts moton also snce the ball travels upward and downward n the vertcal drecton. There s one pont where the ball s not travelng up or down, and that s at the top o ts trajector. Ths s the onl pont where the veloct and acceleraton are perpendcular. Assess: The acceleraton due to gravt alwas ponts straght downward. See Fgure 3.30, whch shows the average veloct vectors and acceleraton along the trajector o a tossed ball. Q3.8. Reason: Runnng whle throwng a ball ncreases the dstance o the throw because t ncreases the horzontal component o the ball s veloct wthout changng the tme o lght. Relatve to the ground, the ball s horzontal veloct component s ncreased b an amount equal to the speed o the runner. Snce the veloct o the runner s purel horzontal, the vertcal veloct component o the ball s unchanged and so the tme o lght o the ball s unchanged. It s nterestng to note that snce, relatve to the ground, the ball has a greater horzontal component o veloct, the angle o the ball s veloct s hgher relatve to the person throwng the ball than relatve to the ground. However, nether relatve to the ground, nor relatve to the person throwng the ball wll a launch angle o 45 gve the mamum range. The angle 45 gves the mamum range or a ball launched b someone not movng because t gves the best compromse between havng a hgh tme o lght and a large horzontal component o veloct. But when the person s runnng, the horzontal component o veloct gets an advantage so the tme o lght can be greater at the epense o the horzontal component o veloct. Thus the best angle relatve to the person s greater than 45. Assess: Whle the best launch angle s no longer 45, t s stll true that runnng ncreases the range o the ball.

2 Q3.. Reason: The tme or an object to ht the ground does not depend on ts horzontal speed, but onl on ts heght and ntal vertcal speed. When the plot goes twce as ast, all that changes s the horzontal speed o the projectle. Thereore the tme o lght wll be the same n both cases,.0 s. However the dstance travelled horzontall wll be doubled snce the horzontal speed s doubled and the tme o lght s the same. At the doubled speed, the weght wll travel twce as ar, or 00 m. Assess: The above answer assumed no ar resstance. Actuall the greater speed o the aster projectle wll slghtl ncrease the tme o lght snce a aster object eperences more ar resstance. P3.5. Prepare: Acceleraton s ound b the method o Tactcs Bo 3.. Solve: The acceleraton vector at each locaton ponts drectl toward the center o the Ferrs wheel s crcular moton. Assess: As we wll learn later, ths acceleraton that s drected toward the center s called centrpetal acceleraton. P3.. Prepare: We wll ollow rules gven n the Tactcs Bo 3.3. Solve: (a) Vector d ponts to the rght and down, so the components d and d are postve and negatve, respectvel: d = d cos θ = (00 m) cos 45 = 70.7 m d = d sn θ = (00 m) sn 45 =7m (b) Vector v ponts to the rght and up, so the components v and v are both postve:

3 (c) Vector v = v cos θ = (300 m/s) cos 0 = 80 m/s v = v sn θ = (300 m/s) sn 0 = 00 m/s a has the ollowng components: a = a cos θ = (5.0 m/s ) cos 90 = 0 m/s a = a sn θ = (5.0 m/s ) sn 90 =5.0 m/s Assess: The components have same unts as the vectors. Note the mnus sgns we have manuall nserted accordng to the Tactcs Bo 3.3. P3.8. Prepare: We can nd the acceleraton on the ramp and then use the acceleraton to nd the nal veloct o the car. Solve: (a) The mamum possble acceleraton s gven b the ormula a = g sn θ. Pluggng n the values o g and θ, we get a =. m/s. (b) The nal veloct can be obtaned rom the ormula ( v ) ( v ) = a Δ, usng ( v ) = 0 m/s and Δ = 6.8 m, the m m latter beng obtaned b convertng 55 t: 55 t = 55 t. 3.8 t 55 t = 55 t The soluton to the rst equaton 3.8 t s ( v ) = 8.6 m/s. Assess: As wth most ramp problems, ths one was best solved b usng a rotated coordnate sstem wth the -as along the ramp. P3.7. Prepare: We wll assume the ball s n ree all (.e., we neglect ar resstance). The trajector o a projectle s a parabola because t s a combnaton o constant horzontal veloct (a = 0.0 m/s ) combned wth constant vertcal acceleraton (a = g). In ths case we see onl hal o the parabola. The ntal speed gven s all n the horzontal drecton, that s, (v ) = 5.0 m/s and (v ) = 0.0 m/s. Solve: (a) (b) (c) (d) Ths s a two-step problem. We rst use the vertcal drecton to determne the tme t takes, then plug that result nto the equaton or the horzontal drecton. Δ = a ( Δ t ) Δ ( 0 m) Δ t = = =.0 s a 9.8 m/s

4 We we use the.0 s n the equaton or the horzontal moton. Δ = v Δ t = (5.0 m/s)(.0 s) = 0 m Assess: The answers seem reasonable, and we would get the same answers to two sgncant gures n a quck mental calculaton usng g 0m/s. In act, I dd ths beore computng the algebra so I would know how to scale the graphs. P3.3. Prepare: We wll appl the constant-acceleraton knematc equatons to the horzontal and vertcal motons as descrbed b Equatons 3.5. The eect o ar resstance on the moton o the bullet s neglected. Solve: (a) Usng (b) Usng v t t a t t = + ( ) ( ) + ( ), we obtan (.0 0 m) = 0 m + 0 m + ( 9.8 m/s )( t 0 s) v t t a t t = + ( ) ( ) + ( ), (50 m) = 0 m + ( v ) ( s 0 s) + 0 m v t = s ( ) = 78 m/s Assess: The bullet alls cm durng a horzontal dsplacement o 50 m. Ths mples a large ntal veloct, and a value o 78 m/s s not surprsng. P3.44. Prepare: Because A= A + A, and B = B + B, so the components o the resultant vector are E = A + 3B and E = A + 3B. The vectors A, B, and E = A+ 3B are shown. Solve: (a) Wth A = 5, A =, B = 3, and B = 5, we have E = and E =. (b) Vectors A, B, and E are shown n the above gure. (c) From the E vector, E = and E =. Thereore, the magntude and drecton o E are

5 Assess: E = () + ( ) = E φ = tan = tan = 5. E Note that φ s the angle made wth the -as, and that s wh φ = tan ( E / E ) rather than whch would be the case φ were the angle made wth the -as. tan ( E / E) P3.46. Prepare: The vectors then obtan the magntude and the drecton o the vector D. A, B, and C are shown. We wll rst calculate the - and -components o each vector and Solve: (a) The vectors A, B, and C are drawn above. (b) The components o the vectors A, B, and C are A = (3 m) cos 0 =.8 m and A = (3 m) sn 0 =.03 m; B = 0m and B = m; C = (5 m) cos 70 =.7m and C = (5 m) sn 70 =4. 70 m. (c) We have D= A+ B+ C = D + D, whch means D =. and D = D = (.m) + (3.73 m) = 3.9 m θ = tan = tan 3.36 = 73. The drecton o D s south o east, 73 below the postve -as. P3.69. Prepare: We wll appl the constant-acceleraton knematcs equatons to the horzontal and vertcal motons o the tenns ball as descrbed b Equatons 3.5. A vsual overvew s shown as ollows. To nd whether the ball clears the net, we wll determne the vertcal all o the ball as t travels to the net. Solve: The ntal veloct s ( v ) = v cos 5 = (0 m/s) cos 5 = 9.9 m/s ( v ) = v sn 5 = (0m/s)sn 5 =.743m/s The tme t takes or the ball to reach the net s = + ( v) ( t t) 7.0 m = 0 m + (9.9 m/s)( t 0 s) t = 0.35s

6 The vertcal poston at t = 0.35 s s = + ( v) ( t t) + a ( t t) = + + = (.0 m) (.743 m/s)(0.35 s 0 s) ( 9.8 m/s )(0.35s 0 s).0 m Thus the ball clears the net b.0 m. Assess: The vertcal ree all o the ball, wth zero ntal veloct, n 0.35 s s 0.6 m. The ball wll clear b appromatel 0.4 m the ball s thrown horzontall. The ntal launch angle o 5 provdes some ntal vertcal veloct and the ball clears b a larger dstance. The above result s reasonable. P3.70. Prepare: We can use the equaton or vertcal moton at constant acceleraton to nd the tme o all and then use the tme to nd the nal veloct. Solve: Snce the water s launched horzontall, ts tme o lght and vertcal dsplacement are related b the equaton: Δ = gδ t. Solvng or the tme, we have Δ t = Δ g = = / (53 m) / 9.8 m/s 3.9 s The horzontal component o the veloct, v, s constant, but the vertcal component s gven b the equaton: ( v ) = ( v ) + a Δt. At the moment the water strkes the pool, the vertcal component s ( ) ( v ) = 0 m/s 9.8 m/s (3.9 s) = 3. m/s At the moment o mpact the veloct o the water s: (9.0 m/s, 3. m/s). The angle that the water makes wth the vertcal s gven b The water s allng at an angle o 6 wth the vertcal. φ = = tan ((9.0 m/s)/(3. m/s)) 6 Assess: Even though the water s launched at a arl hgh speed ( 9.0 m/s s about 0 m/hr ), t s close to the vertcal when t lands because t spends such a long tme n the ar durng whch tme the absolute value o v ncreases steadl.

7 P3.73. Prepare: Frst draw a pcture. In part (a) use tlted aes so the -as runs down the slde. The acceleraton wll be a = g sn θ. Part (b) s a amlar two-step projectle moton problem where we use the vertcal drecton to determne the tme o lght and then plug t nto then equaton or constant horzontal veloct. Use aes that are not tlted or part (b). Solve: (a) We use Equaton.3 wth ( ) = 0.0 m/s. v v g θ Δ ( v ) =a Δ o ( ) = ( sn ) = (9.8 m/s )(sn40 )(3.0 m) = 6.m/s (b) We use Equaton.4 to nd the tme or an object to all 0.4 m rom rest: Δ = 0.4m. Δ = a ( Δ t) Δ ( 0.4m) Δ t = = = 0.86 s g 9.8 m/s At last we combne ths normaton nto the equaton or constant horzontal veloct. Δ= v Δ t = (6.5 m/s)(0.86 s) =.8 m Assess: We reported the speed at the bottom o the slde to two sgncant gures, but kept track o a thrd to use as a guard dgt because ths result s also an ntermedate result or the nal answer. We also kept a thrd sgncant gure on the Δt as a guard dgt. The result o landng.8 m rom the end o the rctonless slde seems just a bt large because ths slde was rctonless and real sldes aren t, but t doesn t seem to be too ar out o epectaton, so our result s probabl correct.

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