A 1600 kg sedan goes through a wide intersection traveling from north to south w

Question

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200 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 these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.48 m west and 6.55 m south of the impact point.

Part A

How fast was sedan traveling just before the collision?

Part B

How fast was SUV traveling just before the collision?

Explanation / Answer

given are

m1=1600kg

m2=2200kg

uk=0.750

d1=5.48m

d2=6.55m

distance travelled after the collision will be

d=sqt(5.48^2+6.55^2)

=8.5m

Work =Frce*distance

=uk(m1+m2)gd

=0.750*(1600+2200)*9.8*8.5

=238470.5J

Tan thetha=5.48/6.55

=39.5 degrees

k.E before collision

E=1/2mv^2

V=undrt(2E/m)

=undrt(2*238470.5/(2200+1600))

=11.2 m/sec

Velocity component after the collision is

Vx=11.2sin 39.5

=7.12m/sec

Vy=11.2cos 39.5

=8.64m/sec

Now momentum is conserved not K.E

Y direction

So 1600v1i=(2200+1600)Vfy

V1i=3800/1600*(8.64)

=20.52m/sec

X direction

2200*V1i=3800vfx

V1i=3800/2200*7.12

=12.29m/sec

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