A car with mass mc=1074 kg is traveling west through an intersection at a magnitude of velocity of vc = 15 m/s when a truck of mass mt=1593 kg traveling south at vt=10.8 ms 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 μk=0.5. a)Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j. b)How far, in meters, will the vehicles slide after the collision?

Question

A car with mass mc=1074 kg is traveling west through an intersection at a magnitude of velocity of vc = 15 m/s when a truck of mass mt=1593 kg traveling south at vt=10.8 ms 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 μk=0.5.  a)Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.   b)How far, in meters, will the vehicles slide after the collision?

Mary Herrera · Accepted Answer

Given information: The mass of the car (mc)=1047 kg The velocity of the car (vc)=15 ms=−15 i ms (consider east as +ve x direction or +ve i) The mass of the truck (mt)=1593 kg The velocity of the truck (vt)=10.8 ms=−10.8 j According to law of conservation of momentum the momentum in all the direction is conserved, considering the along the &quot;i&quot; direction we can write: mcvc+mtvt=(mc+mt)Vi→Vi=−(mcmc+mt)vc Now, consider the momentum conservation along the &quot;j&quot; direction: mcvc+mtvt=(mc+mt)Vj→Vj=−(mtmc+mt)vt Therefore, the velocity of the vehicles after the collsion is: V=−(mcmc+mt)vci^−(mtmc+mt)vtj^ By substituting the correspondingvalues, we get the velocity of the vehciles after the collision as: V=−6.04i^−6.45j^→|V|=8.836 ms After the collision the cinetic energy of the both vehicles combined will be utilized to do work against the dricion force (given μk=0.5), The friction force both the vehicles is given by: f=μk(mc+mt)g=0.5(1074+1593)9.81=13081.635 N Now, according to work energy theorem the we can write: 12(mc+mt)V2=f.d Where &quot;d&quot; is the distance travelled by the wreckage after the collision. Substituting the corresponding values, we get: d=1074+15932(13081.635)×(8,836)2=7.958 m Hence, the distance travelled by the wreckage after the collsion is 7.958 m.

A car with mass m_c = 1074\ kg is traveling west through an inte

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