A 1400 kg sedan goes through a wide intersection traveling from north to sooth w

Question

A 1400 kg sedan goes through a wide intersection traveling from north to sooth when it is hit by a 2300 kg SUV traveling from east to west The two. cars become enmeshed due to the impact and slide as one thereafter On-the-scene measurements show that the coefficient of kinetic friction between the tires of those cars and the pavement is 0 75, and the cars slide to a halt at a point 5 57 m west and 6 55 m south of the impact point How fast was sedan traveling just before the collision? Express your answer in meters per second to three significant ' figures, How fast was SUV travelling just before the collision? Express your answer in meters per second to three significant figure.

Explanation / Answer

  Let the velocity of the two after the collision is v m/s at an angle from West
=>The displacement from the point of collision(s) = [{s(w)}^2 + {s(s)}^2] = [(5.57)^2 + (6.55)^2]
=>s = 8.59 m
& tan = s(s)/s(w)
=>tan = 6.55/5.57 =
=> = 49.6 degree
=>By the work energy relation:-
=>W = KE
=>Ff x s = 1/2(m1+m2)v^2
=>µk x (m1+m2) x s = 1/2(m1+m2)v^2
=>v^2 = 2 x µk x s
=>v = [2 x 0.75 x 8.59]
=>v = 3.58 m/s
(a)

By the law of momentum conservation:-
=> In South direction :- P(initial) = P(final)
=>1400 x u = (m1+m2) x vsin [Let the initial velocity of sedan is u m/s]
=>1400 x u = (1400 + 2300) x 3.58 x sin49.6*
=>u = 7.20 m/s
(b)

By the law of momentum conservation:-
=> In West direction :- P(initial) = P(final)
=>2300 x u = (m1+m2) x vcos [Let the initial velocity of SUV is u m/s]
=>2300 x u = (1400 + 2300) x 3.58 x cos49.6
=>u = 6.13 m/s

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