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4 years ago

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Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

time and collide. The cars lock together and move off at 35.8 km/h[E31.6 S]. What was the velocity of each car before they collided?

1 answer:

Contact [7]4 years ago

6 0

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

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See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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3 years ago

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Hope it helps!

8 0

4 years ago

NEED HELP ASAP<br><br>ONLY ANSWER IF YK THE ANSWERS

nalin [4]

Answer:

there yah go that's the answer

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3 years ago