Question

PLEASE HELP ASAP

A 1000 kg car moving at 5.8 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn to the positive x direction in 4.3 s, the inattentive operator drives into a tree, which stops the car in 220 ms. What is the magnitude of the impulse on the car (a) due to the turn and (b) due to the collision? What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision? (e) What is the angle between the average force in (c) and the positive x direction? (a) Number Units Units (b) Number (c) Number Units (d) Number Units (e) Number Units

Answer 1

Initial momentum vector (P1) = 1000*5.8 j = 5800j ( j is unit vector pointing towards y i.e. North)

Assuming it has same speed after turning

after turn momentum (P2) = 1000*5.8 i = 5800i ( i is unit vector pointing towards x i.e. East)  

(a) the magnitude of Impulse = |P2 - P1| = |5800i - 5800j | = 5800*sqrt(2) = 8202.43 N-s

(b) after collision momentum (P3) = 0\

the magnitude of Impulse = |P3 - P1| = |0- 5800i | = 5800 N-s

(c) Average force magnitude = |change in momentum|/time = 8202.43/4.3 = 1907.54 N

(d) Average force = change in momentum/time = 5800/0.220 = 26363.63 N

(e) Average force direction = P2 - P1 = 5800i - 5800j

angle = -atan(5800/5800) = -45 degrees with positive X axis

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