 College Physics for AP® Courses

# Chapter 8

1.

(a) $1.50 × 10 4 kg ⋅ m/s 1.50 × 10 4 kg ⋅ m/s size 12{1 "." "50" times "10" rSup { size 8{4} } "kg" cdot "m/s"} {}$

(b) 625 to 1

(c) $6 . 66 × 10 2 kg ⋅ m/s 6 . 66 × 10 2 kg ⋅ m/s size 12{6 "." "66" times "10" rSup { size 8{2} } "kg" cdot "m/s"} {}$

3.

(a) $8.00×104 m/s8.00×104 m/s size 12{8 "." "00" times "10" rSup { size 8{4} } " m/s"} {}$

(b) $1.20×106 kg·m/s1.20×106 kg·m/s size 12{1 "." "20" times "10" rSup { size 8{6} } " kg" cdot "m/s"} {}$

(c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be $−0.0100 m/s−0.0100 m/s size 12{ - 0 "." "0100""m/s"} {}$, which is probably not noticeable.

5.

54 s

7.

$9 . 00 × 10 3 N 9 . 00 × 10 3 N size 12{9 "." "00" times "10" rSup { size 8{3} } N} {}$

9.

a) $2.40×103 N2.40×103 N size 12{2 "." "40" times "10" rSup { size 8{3} } " N"} {}$ toward the leg

b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same.

11.

a) $800 kg⋅m/s800 kg⋅m/s size 12{"800""kg" cdot "m/s"} {}$ away from the wall

b) $1.20 m/s1.20 m/s size 12{1 "." "20""m/s"} {}$ away from the wall

13.

(a) $1.50×106N1.50×106N size 12{ - 1 "." "50" times "10" rSup { size 8{6} } N} {}$ away from the dashboard

(b) $1.00×105N1.00×105N size 12{ - 1 "." "00" times "10" rSup { size 8{5} } N} {}$ away from the dashboard

15.

$4 . 69 × 10 5 N 4 . 69 × 10 5 N size 12{4 "." "69" times "10" rSup { size 8{5} } " N"} {}$ in the boat’s original direction of motion

17.

$2 . 10 × 10 3 N 2 . 10 × 10 3 N size 12{2 "." "10" times "10" rSup { size 8{3} } N} {}$ away from the wall

19.

p = mv ⇒ p 2 = m 2 v 2 ⇒ p 2 m = mv 2 ⇒ p 2 2m = 1 2 mv 2 = KE KE = p 2 2m p = mv ⇒ p 2 = m 2 v 2 ⇒ p 2 m = mv 2 ⇒ p 2 2m = 1 2 mv 2 = KE KE = p 2 2m alignl { stack { size 12{p=mv drarrow p rSup { size 8{2} } =m rSup { size 8{2} } v rSup { size 8{2} } drarrow { {p rSup { size 8{2} } } over {m} } =mv rSup { size 8{2} } } {} # drarrow { {p rSup { size 8{2} } } over {2m} } = { {1} over {2} } mv rSup { size 8{2} } = ital "KE" {} # {underline { ital "KE"= { {p rSup { size 8{2} } } over {2m} } }} {} } } {}

21.

60.0 g

23.

0.122 m/s

25.

In a collision with an identical car, momentum is conserved. Afterwards $vf=0vf=0$ for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body.

27.

22.4 m/s in the same direction as the original motion

29.

0.250 m/s

31.

(a) 86.4 N perpendicularly away from the bumper

(b) 0.389 J

(c) 64.0%

33.

(a) 8.06 m/s

(b) -56.0 J

(c)(i) 7.88 m/s; (ii) -223 J

35.

(a) 0.163 m/s in the direction of motion of the more massive satellite

(b) 81.6 J

(c) $8.70×10−2m/s8.70×10−2m/s size 12{8 "." "70" times "10" rSup { size 8{ - 2} } "m/s"} {}$ in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen.

37.

0.704 m/s

–2.25 m/s

38.

(a) 4.58 m/s away from the bullet

(b) 31.5 J

(c) –0.491 m/s

(d) 3.38 J

40.

(a) $1.02×10−6 m/s1.02×10−6 m/s size 12{1 "." "02" times "10" rSup { size 8{ - 6} } " m/s"} {}$

(b) $5.63×1020J5.63×1020J size 12{5 "." "63" times "10" rSup { size 8{"20"} } J} {}$(almost all KE lost)

(c) Recoil speed is $6.79×10−17m/s6.79×10−17m/s size 12{6 "." "79" times "10" rSup { size 8{ - "17"} } "m/s"} {}$, energy lost is $6.25×109J6.25×109J size 12{6 "." "25" times "10" rSup { size 8{9} } J} {}$. The plume will not affect the momentum result because the plume is still part of the Moon system. The plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles.

42.

24.8 m/s

44.

(a) 4.00 kg

(b) 210 J

(c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy.

45.

(a) 3.00 m/s, $60º60º$ below $xx size 12{x} {}$-axis

(b) Find speed of first puck after collision: $0=m v1′sin30º−m v2′sin60º⇒ v1′= v2′sin60ºsin30º=5.196 m/s0=m v1′sin30º−m v2′sin60º⇒ v1′= v2′sin60ºsin30º=5.196 m/s$

Verify that ratio of initial to final KE equals one: $KE = 1 2 mv 1 2 = 18 m J KE = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 = 18 m J KE KE′ = 1.00 KE = 1 2 mv 1 2 = 18 m J KE = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 = 18 m J KE KE′ = 1.00$

47.

(a) $−2.26m/s−2.26m/s size 12{ - 2 "." "26""m/s"} {}$

(b) $7.63×103J7.63×103J size 12{7 "." "63" times "10" rSup { size 8{3} } J} {}$

(c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots.

49.

(a) $5.36×105m/s5.36×105m/s$ at $−29.5º−29.5º$

(b) $7.52×10−13J7.52×10−13J size 12{7 "." "52" times "10" rSup { size 8{ - "13"} } J} {}$

51.

We are given that $m1=m2≡mm1=m2≡m size 12{m rSub { size 8{1} } =m rSub { size 8{2} } equiv m} {}$. The given equations then become:

$v 1 = v 1 cos θ 1 + v 2 cos θ 2 v 1 = v 1 cos θ 1 + v 2 cos θ 2$

and

$0= v1′sinθ1+ v2′sinθ2. 0= v1′sinθ1+ v2′sinθ2.$

Square each equation to get

$v 1 2 = v′ 1 2 cos 2 θ 1 + v′ 2 2 cos 2 θ 2 + 2 v′ 1 v′ 2 cos θ 1 cos θ 2 0 = v′ 1 2 sin 2 θ 1 + v′ 2 2 sin 2 θ 2 + 2 v′ 1 v′ 2 sin θ 1 sin θ 2 . v 1 2 = v′ 1 2 cos 2 θ 1 + v′ 2 2 cos 2 θ 2 + 2 v′ 1 v′ 2 cos θ 1 cos θ 2 0 = v′ 1 2 sin 2 θ 1 + v′ 2 2 sin 2 θ 2 + 2 v′ 1 v′ 2 sin θ 1 sin θ 2 .$

Add these two equations and simplify:

$v 1 2 = v′ 1 2 + v′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 cos θ 2 + sin θ 1 sin θ 2 = v′ 1 2 + v′ 2 2 + 2 v′ 1 v ′ 2 1 2 cos θ 1 − θ 2 + 1 2 cos θ 1 + θ2 + 1 2 cos θ 1 − θ 2 − 1 2 cos θ 1 + θ 2 = v ′ 1 2 + v ′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 − θ 2 . v 1 2 = v′ 1 2 + v′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 cos θ 2 + sin θ 1 sin θ 2 = v′ 1 2 + v′ 2 2 + 2 v′ 1 v ′ 2 1 2 cos θ 1 − θ 2 + 1 2 cos θ 1 + θ2 + 1 2 cos θ 1 − θ 2 − 1 2 cos θ 1 + θ 2 = v ′ 1 2 + v ′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 − θ 2 .$

Multiply the entire equation by $12m12m size 12{ { { size 8{1} } over { size 8{2} } } m} {}$ to recover the kinetic energy:

$1 2 mv 1 2 = 1 2 m v ′ 1 2 + 1 2 m v ′ 2 2 + m v ′ 1 v ′ 2 cos θ 1 − θ 2 1 2 mv 1 2 = 1 2 m v ′ 1 2 + 1 2 m v ′ 2 2 + m v ′ 1 v ′ 2 cos θ 1 − θ 2$

53.

$39 . 2 m/s 2 39 . 2 m/s 2 size 12{"39" "." 2"m/s" rSup { size 8{2} } } {}$

55.

$4 . 16 × 10 3 m/s 4 . 16 × 10 3 m/s size 12{4 "." "16" times "10" rSup { size 8{3} } "m/s"} {}$

57.

The force needed to give a small mass $ΔmΔm size 12{} {}$ an acceleration $aΔmaΔm size 12{a rSub { size 8{} } } {}$ is $F=ΔmaΔmF=ΔmaΔm size 12{F=Δ ital "ma" rSub { size 8{Δm} } } {}$. To accelerate this mass in the small time interval $ΔtΔt size 12{{}$ at a speed $veve size 12{v rSub { size 8{e} } } {}$ requires $ve=aΔmΔtve=aΔmΔt size 12{v rSub { size 8{e} } =a rSub { size 8{Δm} } {}$, so $F=veΔmΔtF=veΔmΔt size 12{F=v rSub { size 8{e} } { {Δm} over {} } } {}$. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so $Fthrust=veΔmΔtFthrust=veΔmΔt size 12{F rSub { size 8{"thrust"} } =v rSub { size 8{e} } { {Δm} over {} } } {}$, where all quantities are positive. Applying Newton’s second law to the rocket gives $Fthrust−mg=ma⇒a=vemΔmΔt−gFthrust−mg=ma⇒a=vemΔmΔt−g size 12{F rSub { size 8{"thrust"} } - ital "mg"= ital "ma" drarrow a= { {v rSub { size 8{e} } } over {m} } { {Δm} over {} } - g} {}$, where $mm size 12{m} {}$ is the mass of the rocket and unburnt fuel.

60.

$2 . 63 × 10 3 kg 2 . 63 × 10 3 kg size 12{2 "." "63" times "10" rSup { size 8{3} } "kg"} {}$

61.

(a) 0.421 m/s away from the ejected fluid.

(b) $0 .237 J 0 .237 J size 12{ - 0 "." "237"J} {}$.

1.

(b)

3.

(b)

5.

(a)

7.

(c) (based on calculation of $F= mΔv Δt F= mΔv Δt$)

9.

(c)

11.

(d)

13.

(b)

15.

(d)

17.

(b)

19.

(c)

21.

(b)

23.

(c)

25.

(b)

27.

(a)

29.

(c)

31.

(b)

33.

(a)

35.

(b)

37.

(a)

39.

(a)

41.

(d)

43.

(c). Because of conservation of momentum, the final velocity of the combined mass must be 4.286 m/s. The initial kinetic energy is . The final kinetic energy is , so the difference is −161 J.

45.

(a)

47.

(d)

49.

(c)

51.

(b)

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