We're approximating the area under the curve by a bunch of rectangles, but if the little Δ t 's are small enough that the force isn't changing much during that short time interval, the total area of our rectangles is approximately equal to the area under the curve. We can continue through the entire collision with the spring, and we see that the total area under the curve is equal to the total impulse (and the total change in the momentum, which is the sum of all the changes to the momentum). In the next time interval Δ t 2, we can again represent the impulse (and the change in momentum) as the area of the next rectangle shown on the diagram. So the area of the rectangle is equal to the impulse during Δ t 1 and also equal to the change in momentum Δ p x1 during that short time interval. But F x1 Δ t 1 can be thought of as the area of a rectangle shown on the diagram, whose base is Δ t 1 and height is F x1. The small impulse F x1 Δ t 1 makes a small change Δ p x1 in the momentum. In the first short time interval Δ t 1, the spring is only slightly compressed, and the force F x1 on the cart is small. Find the speed of the melon as it flies off the performer's head.F How would you expect these values to compare?įinding Impulse Using Area Under the Curve There is a more accurate way to determine the actual impulse on the cart. event: initial final mass/object/velocity mass/object/velocity 0 b. ![]() ![]() ![]() Draw a momentum conservation diagram for the stunt. The arrow passes through the melon and emerges with a speed of 18 m/s. An archer shoots a 50 g arrow at the melon with a speed of 30 m/s. A 2 kg melon is balanced on a circus performer's head. Calculate the impulse that each girl imparts to the other. event: initial final mass/object/velocity mass/object/velocity 0 0 + b. Show the effect of the push on both girls with a momentum conservation diagram. The taller girl pushes the shorter girl so that the shorter girl rolls away at a speed of 10 m/s. Two girls with masses of 50 kg and 70 kg are at rest on frictionless in-line skates. What was the velocity of the cart after the boy jumped? Modeling Instruction - AMTA 2013 2 U9 Momentum - ws 3 v3.1ĥ. How large an impulse did the boy give to the cart? d. event: initial final mass/object/velocity mass/object/velocity 0 + 0 b. Complete the momentum conservation diagram. A 70 kg boy, riding in the cart, jumps off so that he hits the floor with zero velocity. A 50 kg cart is moving across a frictionless floor at 2.0 m/s. How much force did the racket exert on the ball? 4. event: initial final mass/object/velocity mass/object/velocity 0 + b. Use a momentum conservation diagram to show the change in momentum of the ball. A tennis player returns a 30 m/s serve straight back at 25 m/s, after making contact with the ball for 0.50 s. ![]() CModeling Instruction - AMTA 2013 U9 Momentum-Ws 3 v3.1ģ. Find the speed with which the astronaut moves off into space. event: initial final mass/object/velocity mass/object/velocity 0 Momentum conservation equation: b. By pushing the tank away with a speed of 2.0 m/s, the astronaut recoils in the opposite direction. An astronaut of mass 80 kg carries an empty oxygen tank of mass 10 kg. event: initial final mass/object/velocity mass/object/velocity + 0 Momentum conservation equation: b. Each of the boxcars has a mass of 9000 kg when empty, and the loaded car contains 55,000 kg of lumber. An empty boxcar, coasting at 3 m/s, strikes a loaded car that is stationary, and the cars couple together. In a railroad yard, a train is being assembled. Name Date Pd Impulsive Force Model Worksheet 3: Conservation of Momentum I 1.
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