callmelako017
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Hello, I need help. Due right now


A 1383-kg car moving south at 11.2m / s is struck by a 1732-kg car moving east at 31.3m/s. After the collision, the cars are stuck together. How fast and in what direction do they move immediately after the collision? Define the system as the two cars. ​

Respuesta :

Snor1ax

Answer: See below

Explanation:

[tex]\\ \mathrm{Given:} \\\mathrm{Mass \ of \ first \ car}$\left(m_{1}\right)=1383 \mathrm{~kg}$ \\ Velocity $\left(\overrightarrow{V_{1}}\right)=-11.2\ {\math} \mathrm{m} / \mathrm{s}$\\ Mass of second car -$\left(m_{2}\right)=1732 \mathrm{~kg}$\\ Velocity $\left(\vec{v}_{2}\right)=31.3 {\math} \mathrm{m} / \mathrm{s}$[/tex]

[tex]m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}=\left(m_{1}+m_{2}\right) \vec{v} \\1383(-11.2 {\math})+1732(31.3 {\math}) \\=(1383+1732) \vec{v} \\-15489.6 {\math}+54211.6 {\math}=3115 \vec{v} \\ \vec{v}=(17.4 {\math}-4.972 {\math}) \ \mathrm{m/s}[/tex]

[tex]\text { So magnitude } \vec{|v|} =\sqrt{(17.4)^{2}+(-4.972)^{2}} \\ =\sqrt{327.605} \\ =18.099 \mathrm \ {m/s} \\[/tex]

[tex]\text { Direction } \\\theta=\tan ^{-1}\left(\frac{-4.972}{17.4}\right) \\\theta=-15.945^{\circ}\end{gathered}[/tex]

Therefore, both cars move with a velocity of 18.099 m/s in the direction of 15.945° downward from the x-axis (east)

Answer:

Explanation:

Apply conservation of momentum in N-S direction:

Initial momentum: 1383*(-11.2) + 1732*(0) = -15490

Final momentum: (1383 + 1732) * Final N-S velocity = -15490

Final velocity = 1549/(1383+1732) = -4.97 m/s (-ve means South)

Repeat the same for E-W direction:

Initial momentum: 1383*(0) + 1732*(31.3) = 54211.6

Final momentum: (1383 + 1732) * Final E-W velocity = 54211.6

Final velocity = 54211.6/(1383+1732) = 17.4 m/s (+ve means East)

Combining, the final velocity = sqrt (-4.97^2 + 17.4^2) = 18.10 m/s

at direction of arctan(17.4/4.97) = 74 degree South of East.

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