Solutions10 - Physics 140 - Discussion 10 - Solutions 8.13....

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Physics 140 - Discussion 10 - Solutions 8.13. I DENTIFY : The force is constant during the 1.0 ms interval that it acts, so . t = J F r r 2 1 2 1 ( ). m = - = - J p p v v r r r r r S ET U P : Let x + be to the right, so 1 5 00 m/s. x v = + . Only the x component of J r is nonzero, and 2 1 ( ). x x x J m v v = - E XECUTE : (a) The magnitude of the impulse is 3 3 (2 50 10 N)(1 00 10 s) 2 50 N s. J F t = ∆ = . × . × = . The direction of the impulse is the direction of the force. (b) (i) 2 1 . x x x J v v m = + 2 50 N s. x J = + . 2 2 50 N s 5 00 m/s 6 25 m/s. 2 00 kg x v + . = + . = . . The stone’s velocity has magnitude 6.25 m/s and is directed to the right. (ii) Now 2 50 N s x J = - . and 2 2 50 N s 5 00 m/s 3 75 m/s. 2 00 kg x v - . = + . = . . The stone’s velocity has magnitude 3.75 m/s and is directed to the right. E VALUATE : When the force and initial velocity are in the same direction the speed increases and when they are in opposite directions the speed decreases. __________________________________________________________________________________ 8.29. I DENTIFY : The horizontal component of the momentum of the system of the rain and freight car is conserved. S ET U P : Let x + be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the x -axis. E
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This note was uploaded on 02/15/2012 for the course PHYS 140 taught by Professor Uher during the Winter '10 term at University of Michigan.

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Solutions10 - Physics 140 - Discussion 10 - Solutions 8.13....

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