A 1140 kg car skidding due north on a level frictionless icy road at 178.56 km/h

Question

A 1140 kg car skidding due north on a level

frictionless icy road at 178.56 km/h collides

with a 1653 kg car skidding due east at

124 km/h in such a way that the two cars

stick together.

At what angle (?180 +180) East

of North do the two coupled cars skid off at?

Answer in units of degrees.

I got this part and the answer is 45.2 degrees

How much kinetic energy is lost in the colli-

sion?

Answer in units of J

A 1140 kg car skidding due north on a level frictionless icy road at 178.56 km/h collides with a 1653 kg car skidding due east at 124 km/h in such a way that the two cars stick together. At what angle (?180 +180) East of North do the two coupled cars skid off at? Answer in units of degrees. I got this part and the answer is 45.2 degrees How much kinetic energy is lost in the collision?

Explanation / Answer

v1 = 178.56*5/18 m/s = 49.6 m/s v2 = 124*5/18 = 34.44 m/s from momentum conservation in West-East direction; 1653*34.44 + 0 = (1653+ 1140)*Vx => Vx = 20.383 m/s from momentum conservation in North-South direction; 1140*49.6 + 0 = (1653+ 1140)*Vy => Vy = 20.245 m/s tan(theta) = Vx/Vy => theta = 45.195 degrees total velocity = sqrt[(20.383)^2 + (20.245)^2] = 28.729 m/s initial K.E. = 0.5*1140*(49.6)^2 +0.5*1653*(34.44)^2 = 2.383* 10^6 J final K.E. = 0.5*(1140 +1653)*(28.729)^2 = 1.153 *10^6 J K.E. lost = (2.383 -1.153)*10^6 J = 1.23 *10^6 J

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