MarenCurtis 0 Posted February 20, 2018 Share Posted February 20, 2018 I am confused about problem number seven part b in this pdf. http://aplusphysics.com/ap1/Problems/AP1 Momentum.pdf I read the given answer, and I am still confused. I don't know how you are supposed to get 0 m/s for the velocity of the block. Why do you use negative impulse for the block when you use positive impulse for the sphere? When do you use negative impulse? Why don't you use the mass of the entire system, which would be the block and the ball? This has to do with it being elastic/inelastic, but I don't know how you can tell from an impulse graph. Help?? Quote Link to post Share on other sites

FizziksGuy 92 Posted February 23, 2018 Share Posted February 23, 2018 Hi Maren, The positive/negative impulse implies direction. By Newton's 3rd Law, if the block applies an impulse to the ball of 10 N*s, the ball must apply an impulse back to the block of 10 N*s in the opposite direction, therefore the block experiences an impulse of -10 N*s. Since its mass is 0.5 kg, its change in velocity must be -20 m/s. If it's going 20 m/s in the positive direction, and experiences a change of -20 m/s, that leaves it at 0. Think of it this way. As the block travels to the right (positive direction) and hits the ball, it pushes the ball to the right, so the ball receives a positive impulse. Likewise, the ball must push the block to the left, so the block receives a negative impulse. This is not a realistic problem, but from a theoretical standpoint it walks you through some of the key points of momentum. The area under a force-time curve gives you the impulse (calculated in part A). In part B, you use the impulse-momentum theorem to find the change in the ball's velocity. In part C, you use Newton's 3rd Law coupled with the impulse momentum theorem to find the velocity of the block after the collision. And in part D, you utilize the definition of a elastic vs. inelastic collisions. Quote Link to post Share on other sites

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