A bullet of mass m is fired with speed v0 toward a steel block of mass M which i
ID: 1672146 • Letter: A
Question
A bullet of mass m is fired with speed v0 toward a steel block of mass M which is hanging from a massless rod of length D. After a very short impact, the bullet bounces backward with the speed vf while the block swings up to a height h. The collision is not elastic. Treat both bullet and block as point particles (point masses). What is the rotational inertia of the block relative to point P, in terms of M, D, and numerical constants as needed. What is the angular momentum of the bullet relative to point P after the impact, in terms of m, vf, D, and numerical constants as needed. What is the angular speed omega f of the block after the impact, in terms of v0, vf, D, m, M, and numerical constants as needed. How high the block swings after the impact, in terms of M, D, wf, and numerical constants as needed. Answers: I = MD2, L = mDvf(-i), omega)f = m(v0 + vf)/MD h = (D omega f)2/2g. The answers are provided. Help me figure out this problem?Explanation / Answer
Part A is simple, its just using the formulaI=mr2, and since there's only one body at adistance d, its I=MD2 B: Angular momentum(L) is given by L=pD (where p is linear momentumand is equal to mv) velocity is vf, so momentum p=mvf and L=mvfD towards negative x axis (hence the -i-cap in the answer ) C: We know that in any system, if there's no external torque,angular momentum is conserved.. Just before collision: bullet: L=mDvo block: 0 (no velocity, so no momentum) Just after collision: bullet: L=-mvfD (calculated already, minus sign becauseits in the negative x direction) block:L=I = MD2 (where is theangular velocity we are trying to calculate) Since there's no torque, we can equate the two quantities: Li=Lf mDvo=-mDvf+MD2 MD=m(vo+vf) =m(vo+vf)/MD proved ! D: We know that energy is always conserved....So the final(potentialenergy) of the block, should be equal to all the rotationalenergy it gains (from the bullet's kinetic energy). We already know from the last part.. Ei=1/2I2 =1/2MD22 Final energy: Ef=Mgh (just the potential energy of the block) Equating the two (final and initial energies): 1/2MD22=Mgh h=D22/2g =(D)2/2g Done !
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