Compact car (M1 = 1,040 kg) traveling east SUV (M2 = 1,960 kg) traveling north A

Question

Compact car (M1 = 1,040 kg) traveling east

SUV (M2 = 1,960 kg) traveling north

After the collision the cars stick together and skid at an angle of 70 degrees

1a) Assuming the combined cars skid a total distance of 10 meters before coming to rest and that the coefficient of kinetic friction between the tires and the road during the skid was .80. Find the velocity of the two cars immediately after the collision.

1b) Assuming the collision was very fast, find the speed of both cars before the collision. If the speed limit was 13.4 m/s would either car get a ticket?

Problem one Consider two cors in a collision. Be fore they hi the compac+ car (m, 101 wostroveling eost while an Suu (m2 1,96o x) ws troveling north 20 rm Be fore Afer A Fler 4he collision the s stick together oncl Skid at on angle of = 70 Assumin the combined cars sKiJ a toto+ disence of io melers before coming to Cest, and that the coeficent Kinetic friction betuem the #res and the road during the ski wos Ax o. fid the velocit fihd the Ve locity of the tuo cars imediately aften the collision 0)) Assuming the collision was very fost find the Speedof both cor s be fore the collision, If the speed limit wo 13.1 " (30 mph) would either car get a ticket?

Explanation / Answer

Work done by the friction = change in KE

-fk*S= -0.5*m*u^2

fk is the kinetic frictional force = mu_k*(m1+m1)*g = 0.8*(1040+1960)*9.81 = 23544 N

S = 10 m

then 23544*10 = 0.5*(1040+1960)*u^2

then speed of the two cars immeadiately after the collision is u = 12.52 m/s

1B) using law of conservation of momentum then along east

m1*u1 = (m1+m2)*u*cos(70) = (1040+1960)*12.52*cos(70) = 12854.86

then initial velocity of Compact car is u1 = 12854.86/1040 = 12.36 m/s

along north is

m2*u2 = (m1+m2)*u*sin(70) = 3000*12.52*sin(70) = 35294.85

then initial velocity of SUV is u2 = 35294.85/1960 = 18.00 m/s

yes the SUV will get the ticket

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