The six diagrams show a bird\'s eye view of an intersection just before two cars
ID: 1343988 • Letter: T
Question
The six diagrams show a bird's eye view of an intersection just before two cars collide. In each crash, the cars remain joined together after the impact and skid to rest. Rank the accidents by the angle at which the wreckage initially skids, from smallest to largest. Let zero degrees represent a wreckage sliding directly to the right and measure angles increasing counterclockwise from there. E.g., if the smallest angle occurs in B, the next larger one in A, then C, and so on, enter BACDEF. For equal angles, enter in alphabetical order.
Q 20m/s - d t14 m/s Q 13m/s 14m/s o 15m/s 14m 600kg 1200 kg 14m/s 1000 kg oO 1200kg 600kg 900kg C) B) A) Q 10m/s Q 10m/s 7m/s 7m/1200kg o O 5m/s o 7m/81500kg 1200 kg 700kg 1200kg 1300kg 1500kg F) E) D)Explanation / Answer
using momentum conservation,
initial momentum = final momentum
(m1 vx )i + (m2 vy) j = (m1 +m2) v
v = (m1vx / m1+m2) i + ( m2 vy / m1+m2) j
angle = tan^-1 [ ( m2 vy / m1+m2) / (m1vx / m1+m2) ]
= tan^-1 [ m2vy / m1vx ]
A) @ = tan-1( 900 x 14 / 1000 x 5) = 68.36 deg
B) @ = tan-1( 600 x 15 / 600 x 13) = 49.09 deg
C) @ = tan-1( 1200 x 14 / 1200 x 20) =35 deg
D) @ = tan-1( 1500 x 7 / 700 x 7) = 65 deg
E) @ = tan-1( 1300 x 7 / 1500 x 10) = 31.24 deg
F) @ = tan-1( 1200 x 5 / 1200x 10 ) = 26.56 deg
F E C B D A
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