V x v 2 x v 2 y dv y dt g b m v y v 2 x v 2 y e dv x

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v x v 2 x v 2 y , dv y dt g b m v y v 2 x v 2 y (e) dv x dt 1 2 CAρv v x , dv y dt g 1 2 CAρv v y This question asks us to find the differential equation for a bead under- going projectile motion in a medium which exerts a drag force. We know that drag forces, like friction forces, always act to oppose the motion of the object. We also know that the drag force is proportional to the velocity of the object. The problem gives us the expression for ~ D , the length of the drag force vector. Any vector can be broken into components, so we can write ~ D D x ˆ x D y ˆ y, where D x and D y are the components of the vector in the ˆ x and ˆ y di- rections respectively. We also know that the length of such a vector is given by ~ D D 2 x D 2 y . Thus, by inspecting the expression ~ D b v 2 x v 2 y , we can infer that ~ D bv x ˆ x bv y ˆ y . The only forces in this problem are gravity and the drag force. Because Newton’s 2nd law related the sum of the forces on an object to the acceleration, or second time derivative of the objects position, we can construct a differential equation by simply summing the forces on the object. Again it is most convenient to break the problem into components before we sum the forces. Since gravity acts in the ˆ y direction, the sum of the forces in the ˆ y direction is 12
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F y ma y m dv y dt F g D y . Because we chose upwards to be the positive ˆ y , we know that F g mg . Now consider the bead with some positive v y , we know that the drag force will oppose the motion, because D y is proportional to v y we know that the sign must be negative to in order to make the drag force act downwards when the object moves upwards. So we can write F y m dv y dt mg bv y , which gives us our differential equation for the ˆ y direction. For the ˆ x direction, the only force acting is the drag, so we can again use Newton’s second law to write F x m dv x dt D x . Using the same argument, for positive v x we want a negative force, we choose the sign to be minus. m dv x dt D x bv x . Dividing both the ˆ x equation and ˆ y equation by m yields dv x dt b m v x dv y dt g b m v y , which is choice a. Good luck! 13
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