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3 years ago

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A Honda Civic and an 18 wheeler approach a right angle intersection and then collide. After the collision, they become interlock

ed. If their mass ratios were 1: 4 and their respective speeds as they approached were both 13 m/s, find the magnitude and direction of the final velocity of the wreck. (Please do not worry no one was harmed in making of this question) A. 16.3 m/s at 79° B. 10.7 m/s at 79° C. 12.5 m/s at 59° D. 15.7 m/s at 59°

1 answer:

Goryan [66]3 years ago

8 0

Answer:

The concept of conservation of momentum is applied in the particular case of collisions.

The general equation ig given by,

$M_1V_1 + M_2V_2 = (M_1+M_2) * V_f,$

Where,

$M_2 = 4 M_1$

The crash occurs at an intersection so we must separate the two speeds by their respective vector: x, y.

In the case of the X axis, we have that the body $M_2$ has a speed = 0, this because it is not the direction in which it travels, therefore

$M_1* 13 = (M_1+M_2) * V_{fx} \\M_1*13 = (M_1+4M_1)*V_{fx}\\M_1*13=5M_1*V_{fx}\\Vx = \frac{13}{5}m/s$

The same analysis must be given for the particular case in the Y direction, where the mass body $M_1$ does not act with its velocity here, therefore:

$M_2* 13 = (M_1+M_2) * V_{fy},\\4*M_1* 13 = 5Ma * V_{fy} ,\\V_{fy} = \frac{52}{5}m/s ,$

We have the two components of a velocity vector given by $V_f = \frac{13}{5}\hat{i} + \frac{52}{5}\hat{j}$

Get the magnitude,

$V_f = \sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}$

$V_f = 10.72 m/s$

With a direction given by

$Tan^{-1} \frac{4}{1} = 75.96 \°$

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