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Quiz 1 This is a photograph of an apple and feather free fall in a evacuated chamber. The apple and feather are released from the top. Suppose the camera opened flash every 1 s. (1)Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s to t =2s, A3 is the distance from t = 2s to t = 3s …… what is the ratio of A1: A2: A3……. (2) If camera opened flash every 0.06s, is the ratio of the distance between every 0.06 s the same as above or change? Why?

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Answer: The ratio is 1:3:5 :7……(2n -1), here n = 1, 2, 3, Show it in general case, suppose the time interval is Δt The distance from 0 –one Δt, A1 = ½*g*(Δt) 2 (notice here I consider the distance, so I neglect (-) sign) The distance from Δt---2 Δt, A2 = ½*g*(2Δt) 2 -½*g*(Δt) 2 = ½*g*(Δt) 2 (2 2 -1) Using the same way the distance from 2Δt---3 Δt, A3 = ½*g*(Δt) 2 (3 2 - 2 2 ) A4 = ½*g*(Δt) 2 (4 2 - 3 2 ) ……. An = ½*g*(Δt) 2 (n 2 – (n-1) 2 ) = ½*g*(Δt) 2 (n 2 – n 2 +2n-1) = ½*g*(Δt) 2 (2n-1) Therefore the ratio should be 1:3:5…….(2n-1), the ratio is independent of Δt

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Quiz 2 In a projectile motion, if we measure the maximum height h and horizontal range R (1)How to get launch angle θ in terms of h of and R? (2) A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point. What is the angle of launch? What is the launch speed? ( you need show your work, no work, no points)

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We know The maximum height and horizontal range Using (1)/ (2), we can get If h = 72.3m, R = 111 m, plug into above formula, you will get θ=69 o Then plug into (1), you can get Vi = 40.3m/s

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Quiz3 (due day: march 9 th, Friday) 1.You may use this way to get a bullet speed. Suppose a bullet of mass 8.00 g is fired into a block of mass 2.50 Kg initially at rest at the edge of a frictionless table of height 1.00m. The bullet remains in the block, and after impact the block lands a distance 2.00 m from the bottom of the table, Determine the initial speed of the bullet. (review perfectly inelastic collision and projectile motion)

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Solution: Suppose the speed of bullet is Vb, bullet is shot inside the wood block, that is perfectly inelastic collision, so momentum need be conserved, suppose the speed of block with bullet is V1, mb Vb = (mb +mw) V1, (1) We know the height of table is 1m, and distance from the edge is 2m, From the projectile motion kinematics, we can get: Then we plug in (1), get Vb = 1388m.s

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In this problem, If we set m B is 1Kg (1)What is the speed the ball A just before impact the block ? (2.4m/s) (2) What is the speed V A and V B after collision?(-1.6m/s, 0.8m/s) (3) What is rebounding angle θ 3 ?(29.6 o ) Quiz 4: Rebounding pendulum ( see the example 10.9, page 286 ) Mb =1Kg

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Quiz5: Figure shows a contest in which a sphere, a cylinder, and circular hoop all of mass M and radius R( notice particle is not taking the race). are rolling down from rest at the same instant of time. ( please read text book P 366-367 first, then do your own work) (1)What is the speed of center of mass for each of the object when they reach the bottom (2)What is acceleration of the center of mass for each? (3) Which will win the downhill race? why Due day: March 28, Wed.

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If we choose the bottom of the ramp as the zero point of potential energy, from energy conservation Here C is a constant that depends on the object’s geometry. C = 1 for hoop, =1/2 for cylinder = 2/5 for sphere From above equation

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Continue: Since they have the same h and θ, the greater of c, the longer of time. Hoop’s c =1> cylinder’s c = ½, > sphere’s c = 2/5, that results in Sphere taking shortest time, cylinder second, hoop is the last

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Quiz 6: Two equal masses are attached to identical ideal strings, length L, next to one another. One mass is pulled 10 o and the other pulled only 5 o, then release. The masses are released simultaneously. Which mass reaches the vertical position( θ = 0 o ) first? Explain your reasoning. 1.(due day: April 11 th Wednesday)

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Answer: They reach the equilibrium point at the same time Reason: For the small angle pendulum, the period If they have the same L, they will have the same period. The maximum angle does not effect the period. And the time reaches the equilibrium position should be T/4, since two pendulum system have the same period, two masses should reach the equilibrium position at same time. Note: at equilibrium position θ(t) = 0, the function for this oscillation is ( if you pull out mass to the maximum and release, the phase constant φ=0)

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