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Physics Olympiads Series Problem 5

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Problem #5

A bullet of 10 g of mass and 1000 m/sec of velocity travels upward colliding a block 0f 5 kg of mass at rest. The bullet passes through center of mass of the block. Assume that very short time is needed during the collision.

a) Find the velocity of the block after collision, if the velocity of the bullet as it leaves the block is 400 m/sec !

b) Find the maximum heigth of the block! c) Find the loss of origin kinetic energy in this collision! Use the acceleration due to gravity g = 10 m/s2. (Source: OSN 2008 District Selection /Indonesian Physics Olympiad)

 

Solution

a) The data bullet mass mp = 10 g = 0.01 kg block mass mb = 5 kg block velocity before collision vb = 0 m/s bullet velocity before collision vp = 1000 m/s bullets velocity after collision v'p = 400 m/s Using the conservation of linear momentum law we get: mpvp + mbvb = mpv'p + mbv'b (0.01)(1000) + (5)(0) = (0.01)(400) + (5)(v'b) 10 = 4 + 5v'b v'b = 6/5 = 1.2 m/s.

b) Using accelerated motion (negative acceleration) formulae: In the maximum heigth, the velocity of the block is 0 m/s (vt = 0 m/s). The initial velocity of the block vo = 1.2 m/s, and use the acceleration due to gravity g = 10 m/s2: (vt)2 = (vo)2 − 2gh 0 = (1.2)2−(2)(10)(h) 20h = 1.44 h = 0.072 meter c) The loss of origin kinetic energy: Δ E = Ep + Eb − {E'p + E'b} Δ E = 1/2 mp(vp)2 + 1/2 mb(vb)2 − {1/2 mp(v'p)2 + 1/2 mb(v'b)2} Δ E = 1/2 (0.01)(1000)2 + (0) −{1/2(0.01)(400)2 + 1/2(5)(1.2)2} Δ E = = 5000 – 800 – 3.6 = 4196.4 joule

Literature: www.tofi.or.id

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