1. f. The cart is initially moving in the negative x-direction, so $p_{\text{i}x}=-20\text{kg}\text{m/s}$. After it bounces, $p_{\text{f}x}=-10\text{kg}\text{m/s}$. Thus $\text{Δ}p=(10\text{kg}\text{m/s})-$ $(-20\text{kg}\text{m/s})=30\text{kg}\text{m/s}$.

2. b. The clay ball goes from $v_{\text{i}x}=v$ to $v_{\text{f}x}=0$, so $J_{\text{clay}}=\text{Δ}{p}_x=-mv$. The rubber ball rebounds, going from $v_{\text{i}x}=v$ to $v_{\text{f}x}=-v$ (same speed, opposite direction). Thus $J_{\text{rubber}}=\text{Δ}{p}_x=-2mv$. The rubber ball has a larger momentum change, and this requires a larger impulse.

3. Less than. The ball’s momentum $m_{\text{B}}{v}_{\text{B}}$ is the same in both cases. Momentum is conserved, so the total momentum is the same after both collisions. The ball that rebounds from C has negative momentum, so C must have a larger momentum than A.

4. c. Momentum conservation requires $\left(m_1+m_2\right)\times v_{\text{f}}=m_1{v}_1+m_2{v}_2$. Because $v_1>v_2$, it must be that $\left(m_1+m_2\right)\times v_{\text{f}}=m_1{v}_1+m_2{v}_2>m_1{v}_2+m_2{v}_2=\left(m_1+m_2\right)v_2$. Thus $v_{\text{f}}>v_2$. Similarly, $v_2<v_1$, so $\left(m_1+m_2\right)v_{\text{f}}=m_1{v}_1+m_2{v}_2<m_1{v}_1+m_2{v}_1=\left(m_1+m_2\right)v_1$. Thus $v_{\text{f}}<v_1$. The collision causes $m_1$ to slow down and $m_2$ to speed up

5. Right end. The pieces started at rest, so the total momentum of the system is zero. It’s an isolated system, so the total momentum after the explosion is still zero. The 6 g piece has momentum 6v. The 4 g piece, with speed $-2v$, has momentum $-8v$. The combined momentum of these two pieces is $-2v$. In order for $P$ to be zero, the third piece must have a positive momentum $(+2v)$ and thus a positive velocity.

6. e. Momentum is conserved, so the total momentum of the two pieces must be the initial $2î\text{kg}\text{m/s}$. $\vec{p}{}_1=2ĵ\text{kg}\text{m/s}$, so $\vec{p}{}_2$ has to be $(2î-2ĵ)\text{kg}\text{m/s}$.

1. 75 m/s

3. $3\text{kg}\text{m/s}î$

5. $1.0\times {10}^3\text{N}$

7. 80 m/s to the left

9. $-0.50\text{m/s}î$

11.

1. 4500 N

2. $2.3\times {10}^4\text{kg}\text{m/s}î$

3. $2.3\times {10}^4\text{kg}\text{m/s}î$

4. 15 m/s

13. 0.2 s

15. 4.8 m/s

17. $v_0$

19. $3.2\times {10}^2\text{m}\text{/}\text{s}$

21. 7.6 cm in the direction Brutus was running

23. 0.86 m/s, 2.9 m/s

25.

1. $0.048v_0$

2. 95%

27. 0.20 m/s

29. 13 s

31. $-3\text{kg}\text{m/s}î+6\text{kg}\text{m/s}ĵ$

33.

1. 12 s

2. $42\text{°}$ north of west

35. 1800 m/s

37. 1000 m/s

39.

1. 6.4 m/s

2. $F_{\text{avg}}=612(F_{\text{G}}){}_{\text{B}}$

41. $9.3\times {10}^2\text{N}$

43. 12t N

45.

1. $6.7\times {10}^{-8}\text{m}\text{/}\text{s}$

2. $2.2\times {10}^{-10}\%$

47. 31.5 m/s

49.

1. $v_{\text{bullet}}={\displaystyle \frac{m+M}{m}}\sqrt{2{\mu }_{\text{k}}gd}$

2. $4.4\times {10}^2\text{m}\text{/}\text{s}$

51. $4.0\times {10}^2\text{m}$

53. 99.98 m/s, downward

55. 1.5 m/s

57.

1. $v_m={\displaystyle \frac{m+M}{m}}\sqrt{4gL}$

2. $v_m={\displaystyle \frac{m+M}{2m}}\sqrt{4gL}$

59. 20 m/s downward

61. 16 m/s

63. 1.2 km/s

65. 0.54 m/s

67. Ball A: 5.0 m/s, ball B: 13 m/s, both to the right

69.

1. $-1.4\times {10}^{-22}\text{kg}\text{m/s}$

2. and c. $-1.4\times {10}^{-22}\text{kg}\text{m/s}$ in the direction of the electron

71.

1. 0.90 m/s, $45\text{°}$ opposite the initial direction

2. 23%

73. 4.2 m/s, $17\text{°}$ above the $+x\text{-axis}$

75.

1. 3200 m/s

2. 4200 m/s

81. 5.7 m/s

83. 7.9 m/s

85.

1. $v_{\text{max}}=v_{\text{ex}}\text{ln}\left({\displaystyle \frac{m_{\text{R}}+m_{\text{F}0}}{m_{\text{R}}}}\right)-{\displaystyle \frac{g}{R}}(m_0-m_{\text{R}})=v_{\text{ex}}\text{ln}\left({\displaystyle \frac{m_{\text{R}}+m_{\text{F}0}}{m_{\text{R}}}}\right)-{\displaystyle \frac{g}{R}}(m_{\text{F}0})$

2. 6900 m/s, 4500 m/s