3. Two automobiles of equal mass approach an intersection. One vehicle is travel

Question

3. Two automobiles of equal mass approach an intersection. One vehicle is traveling with speed 13.0 m/s toward the east, and the other is traveling north with speed v Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0' north of east. The speed limit for both roads is 35 mph, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred Is he telling the truth? Explain your reasoning.

Explanation / Answer

Given

let m1 moving along East , m2 moving along north before collision with initial velocities u1,u2 before collision

after collision both stick together and move with velocity v alont 55 degrees north east direction

writing the momenta of both before and after collision , so that the conservation of momentum taken place

before collision  

__________________

m1

m1u1_x i + m1u1_y j

m2

m2u2_x i + m2u2_y j

total momentum is  

P_initial = m1u1_x i + m2u2_y j

After collision

__________________

(m1+m2) v cos theta i + (m1+m2) v sin theta j

equating the components

m1u1_x i + m2u2_y j = (m1+m2) v cos theta i + (m1+m2) v sin theta j

given u1_x= 13 m/s

m1u1_x = (m1+m2) v cos theta

and m1 = m2 = m

m*u1_x = (2m) v cos theta

v = u1_x/(2cos theta)

v = 13/(2cos55) m/s

v = 11.33 m/s

substituting the value of v so that we get m2 velocity before collision

m2u2_y = (m1+m2) v sin theta

u2 = 2*v sintheta

u2 = 2*11.33 sin 55 m/s

u2 = 18.56 m/s = 41.52 mph

we know that the conversion of speed from mph to m/s is 35 mph = 15.6464 m/s

but we got that the initial velocity of m2 is 18.56 m/s so nothward moving driver is not within the speed limit of 35 mph.

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