(5%) Problem 20: A car with mass me = 1427 kg is traveling west through an No) -
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(5%) Problem 20: A car with mass me = 1427 kg is traveling west through an No) -intersection at a miagntude of velocity of = 13.5 ms when a truck of mass mt = 1546 kg traveling south at vt = 12.6 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of 05. E(ox) Otheexpertta.com 50% Part (a) Write an expression for the velocity of the system after the collasso, n terms of the variables given in the problem statement and the unit vectors i and j 50% Part (b) How far, in meters, will the vehicles slide after the collision? Grade SunamaryExplanation / Answer
(a) Since the two vehicles stuck together after collision ,this is a case of inelastic collision.We can apply the conservation of momentum.
Let the west and south directions are considered as positive along the co-ordinate axis.
The magnitude of velocity of the car is given as Vc=13.5 m/s=13.5i m/s(Since it is travelling in the west direction)
The magnitude of velocity of the truck is given as Vt=12.6m/s=12.6j m/s(Since it is travelling in the south direction)
Here i and j are unit vectors along x and y axis respectively.
The mass of the car and truck are Mc=1427kg and Mt=1546kg respectively.
Let the mass of both the vehicles after collision be M(=1427+1526=2953kg) and the velocity V.
MV=McVc + MtVt = 1427kg*13.5i m/s +1546*12.6j m/s = 2953*V
This is the required expression for velocity.
V=(19264.5/2953)i + (19479.6/2953)j = (6.52i + 6.6j) m/s
The magnitude of velocity V=square root of [ (6.52)2 + (6.6)2 ] = 9.28 m/s (ans)
(b) The kinetic energy of the system = 1/2MV2 = 1/2*2953*9.28*9.28=127082.36J
The maximum value of frictional force Fs = coefficient of friction*Normal reaction = 0.5*2953*9.8=14469.7 N
where the normal reaction=mass*acceletarion due to gravity
If S is the minimum stopping distance ,then the work done against the frictional force is W=Fs*S
By the work energy theorem,W=1/2MV2=FsS
S= (127082.36J) / (14469.7N) =8.783m
The vehicles slide upto 8.783m after the collision.
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