In: Physics
Two cars collide at an intersection. Car A, with a mass of 1900 kg , is going from west to east, while car B , of mass 1400 kg , is going from north to south at 17 m/s . As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60 ∘ south of east from the point of impact. How fast were the enmeshed cars moving just after the collision? How fast was car A going just before the collision?
Using momentum conservation in x-direction:
Pix = Pfx
mA*Vixa + mB*Vixb = (mA + mB)*Vfx
here,
Vi = initial velocity of carA = ??
x-component of initial velocity of carA = Vixa
x-component of initial velocity of carB = Vixb = 0
mA = mass of car A = 1900 kg
mB = mass of car B = 1400 kg
then, Vfx = (mA*Vixa + mB*0)/(mA + mB)
Using given values:
Vfx = final velocity in east direction = (1900*Vixa + 1400*0)/(1900+1400)
Vfx = 19*Vixa/33 eq(1)
Now using momentum conservation in y-direction:
Piy = Pfy
mA*Viya + mB*Viyb = (mA + mB)*Vfy
Vfy = (mA*Viya + mB*Viyb)/(mA + mB)
here, Viya = 0
Viyb = -17 m/s
Using given values:
Vfy = final velocity in south direction = (1900*(0) + 1400*(-17))/(1900+1400)
Vfy = -7.21 m/sec.
given angle after collision = =
arctan(Vfy/Vfx) = -60 deg
then, Vfx = Vfy/tan(-60 deg)
Vfx = -7.21/tan(-60 deg)
Vfx = 4.16 m/s
therefore, their velocity immediately after collision = V = sqrt(Vfy^2 + Vfx^2)
V = sqrt(4.16^2 + (-7.21)^2)
V = 8.32 m/s
from eq(1),
4.16 = (19/33)*Vixa
Vixa = 4.16*33/19
Vixa = initial velocity of car just before the collision = 7.22 m/s
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