Two cars a and b have masses of 1200 kg and 1160 kg...

Two cars A and B have masses of 1200 kg and 1160 kg respectively. They collide on the icy pavement of an intersection. The direction of motion of each car after the collision is measured from snow tracks as shown. Driver A states he was going \( 45.0 \mathbf{~ k m} / \mathbf{h} \) just before the collision and that after the collision he applied the brakes so that he skidded 13.2 m before stopping. (Assume the coefficient of kinetic friction is \( \mathbf{0 . 3 3} \) between locked car wheels and the icy pavement, and use \( \mathbf{g}=\mathbf{9 . 8 1} \mathbf{~ m} / \mathbf{s}^{\mathbf{2}} \).) Find the speed of the car B before the collision in \( \mathbf{k m} / \mathbf{h} \) correct to three sig sigs. (Note the units!) \( v_{B 1}= \) km/h

Transcript

00:01 In this question we have given that the car b of mass 1900 kilograms was at the red light when it was hit from behind by a car a whose mass is 1 ,500 kilograms.

00:12 We have given that the skid mass are 7 .3 mt long and the percentage is 0 .65 then first we have to find the speed of the car a just before the collision.

00:24 So given that mass of car b is 1900 kilogram and mass of the 5, car is 1500 kilograms the street marks are that is 7 .3 meter long and coefficient of friction is 0 .65 so as we know that friction force acting on cars that is f is mu into total mass that is ma plus mb into g because both the cars are locked together so this friction of force is opposed to the motion so that is minus of m a plus m b into acceleration that is equal to mu into m a plus mb into g so from here we can say that the acceleration is minus new g which is now using third equation of motion we can say that the final velocity b square that is equal to initial velocity plus 2a into s so finally both car will stop so this is 0 and that is equals to u square plus 2 into minus nulg into s so from here we get the initial speed that is under root of 2 into 0 .65 into 9 .8 into s is 7 .3 and that will come out 9 .64 meter per second so this is the combined speed of both the car after the collision using conservation of momentum you can say that initial momentum of the system before the collision is equal to final momentum of the system so this is momentum of first car a into its speed plus momentum of second car and that is equal to m a plus mb into u so this is 1500 into va plus as second car is addressed and that is equals to 1500 plus 1900 into 9 .64 so from here we get the value of va that is 21 .85 meter per second this is the speed in meter per second so we can say that this is this will comes out 70 if we convert it in kilometer per hour so which is 78 .66 kilometers per hour...

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