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標題:
momentum
發問:
In an experiment, Trolley A has a mass of 3kg , the motion of the two trolleysare recorded on two paper tapes as shown below. The ticker-tape timer makes 50 ticks per second.Tape of trolley A has 8 points, in 5cmTape of trolley B has 5 points, in 4cma) (i) What is the total momentum after firing? ... 顯示更多 In an experiment, Trolley A has a mass of 3kg , the motion of the two trolleysare recorded on two paper tapes as shown below. The ticker-tape timer makes 50 ticks per second. Tape of trolley A has 8 points, in 5cm Tape of trolley B has 5 points, in 4cm a) (i) What is the total momentum after firing? (ii) Find the mass of trolley B (b) Assuming that the two trolleys take 0.5s to separate after the trigger is tapped, find the average force acting on trolley A and that on trolley B. (c) Explain why friction compensated runway cannot be used in this experiment. Name one set up which can replace the friction compensated runway. (d)A small object A moving with velocity u collides head-on with a stationary object B of equal mass. What will happen to A and B if the collision is (i) perfectly elastic, (ii) completely inelastic?
最佳解答:
(a) (i) The initial momentum of the 2 trolleys is zero. Hence the final momentum is also zero. (ii) Speed of trolley A after firing = 5/(7/50) cm/s = 35.7 cm/s Speed of trolley B after firing = 4/(4/50) cm/s = 50 cm/s Using conservation of momentum 3 x 35.7 = m x 50, where m is the mass of trolley B i.e. m = 2.14 kg (b) Change of momentum of trolley A = 3 x 35.7/100 Kg-m/s = 107.1 Kg-m/s Hence force = rate of change of momentum = 107.1/0.5 N = 214.2 N (c) "Friction compensated runway" can only compensate friciton for object moving down the runway. For object moving up the runway, friction cannot be compensated. Use an air-track. (d) (i) Use conservation of momentum m.u = mv + mv' where v and v' are the velocities of objects A and B respectively Since collision is elastic, kinetic energy is also conserved. i.e. (1/2)mu^2 = (1/2)mv^2 + (1/2)mv'^2 solve for v and v' gives v = 0 m/s and v' = u Hence, object B will move forward with velocity u, whereas object A will stop. (ii) SImilarly, when the collision is completely inelastic, the two objects will stick together after collision. Thus, we have, mu = 2m.v", where v" is the common velocity after collision i.e. v" = u/2 The two objects will move together with velocity u/2 after collision.
其他解答:
Change of momentum of trolley A = 3 x 35.7/100 Kg-m/s = 107.1 Kg-m/s 3 x 35.7/100 =1.071,, 係唔係唔洗除100?C8D74AB62542840B
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