A bullet of mass $m = 0.0010 \textrm{ kg}$ embeds itself in a wooden block with mass $M = 0.999 \textrm{ kg}$, which then compresses a spring $(k = 140 \textrm{ N/m})$ by a distance $x = 0.050 \textrm{ m}$ before coming to rest. The coefficient of kinetic friction between the block and table is $\mu = 0.50$.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. There are three moments to consider in this question; one moment to consider is before there is any collision at all we have this bullet approaching this mass and we have to assume that the mass is attached to the spring, you know, we need more information... one thing I was wondering when I read this is is the mass, you know, located here somewhere in front of the spring and then it slides some distance before it even touches the spring, we wouldn't be able to answer that without more information so we have to assume that it's attached to the spring. Okay so before the collision, we have some momentum in the system and that's entirely just in the bullet because this block is not moving so there's no momentum there. So before the collision, our total momentum in the system is mass of the bullet times the bullet's speed. And then just after the collision, before there's any movement, the momentum is gonna be the same because there's no external forces being exerted yet; this is before we consider friction and before we consider the spring pushing on the block because just at this instant, before the block moving, neither of those forces exist; the spring force and the friction force only exist after the block starts moving so this is just before movement The instantaneous velocity that it's gonna have at that moment just before it moves is gonna be v prime and this block is gonna have mass of the block plus the mass of the bullet there and that momentum is gonna be m plus M times v prime. And then we can solve for v prime and it equals mass of the bullet times the bullet's incoming speed divided by mass of the bullet plus mass of the block, okay. Then we have to think about this moment in time after the block stops and there's some potential energy contained within the spring and there has been some energy dissipated by friction into heat and both of those energies have to equal the total kinetic energy that the block and bullet have together at this moment just after the collision. So some of the kinetic energy that the bullet has alone here is gonna be lost because kinetic energy is not conserved within inelastic collision but whatever kinetic energy is leftover just after the collision will be accounted for entirely by potential energy in the spring and energy dissipated as heat by the friction. So we'll say that with algebra here; we have the kinetic energy just after the collision is gonna be one-half total mass times its velocity after the collision prime squared and then that's gonna equal the total potential energy in the spring one-half times spring constant times the compression distance squared plus the force of friction times the compression distance so this is the energy dissipated by friction force times distance. So the force of friction is the coefficient of kinetic friction multiplied by the normal force and the normal force upwards is gonna equal the weight downwards because there's no vertical acceleration so the weight down is gonna be the total mass of bullet plus block times g and force of friction is gonna be μ K times that. So we make that substitution into our energy formula here— that's what I have shown in red— and let's multiply everything by 2 to get rid of any fractions and rearrange in factors here just a little bit. And then on the next line here, we'll substitute for v prime and write it in terms of v; that's what we did with this momentum business up here, we have a equation saying v prime is m times v over m plus M. So we have the speed, just after the collision, equals bullet's mass times its speed, before the collision, divided by the bullet plus block mass and we substitute that in here in red. And then we'll cancel one of these m plus M factors here leaving us with one leftover, it would normally be squared but since one of them cancels there, we have only m plus M to the power of 1 on the bottom and we have v squared and we have mass of the bullet squared on top and nothing was done on the right hand side there. And then we'll divide both sides by mass of the bullet squared and multiply both sides by m plus M and that's all that's been done here and take the square root of both sides. So we are taking the square root of this whole numerator and then dividing it by m to the power of 1 and then we plug in some numbers. And when you are working with your calculator here, you have to be really careful with brackets. So we have the square root of the sum of 0.0010 plus 0.999— I guess I could have, you know, noticed that that's just one but anyway, we are just plugging in all the stuff here— times the spring constant of 140 newtons per meter and then times the compression of the spring of 0.50 meters and we square that plus 2 times times 0.50—the coefficient of kinetic friction— times 9.8 meters per second squared times 0.05 compression of the spring and then times the sum of the bullet mass— 0.0010 kilograms— plus the block mass of 0.999 kilograms, take the square root of that whole thing; one tip for bracket matching is to notice how many open brackets there are and compare that to the number of closed brackets that you have and they should be equal so we have one, two open brackets here and then three and then four open brackets and we should have one, two, three, four closed brackets to match them all... that's just a quick way with counting to make sure your bracket matching is correct. If we had done this without doing that and you know, made a mistake of not putting that bracket in instead of getting the correct answer, 920 meters per second, we would get 29 and that's clearly not correct because bullet's don't go at a mere 29 kilometers an hour or meters per second because, you know, in kilometers per hour, that's only 104 kilometers an hour—I mean a car can go that fast— so certainly, this number's not realistic for bullet speed. There we go; 920 meters per second is the answer. The fractional change in kinetic energy before and after the collision is gonna be the kinetic energy immediately after the collision before there's any movement and minus the kinetic energy that the bullet has before colliding divided by the kinetic energy that the bullet has before colliding. So kinetic energy just after the collision is one-half times the total mass of bullet plus block times their combined speed just after the collision squared and then minus the kinetic energy before the collision divided by the kinetic energy before the collision. So it's one-half mv squared on the bottom can get divided by both terms in the numerator and the second term becomes a nice convenient 1 and the first term, halves cancel but otherwise, nothing else cancels and so we are left with m plus M v prime squared over mv squared. These v's don't cancel because they represent different speeds at different times; this is the velocity before collision and this is the velocity of the bullet plus block just after the collision. So we'll substitute for v prime and write it as mv over m plus M— that's what we got from conservation of momentum up here so that's shown in red here. And one of these m's will cancel with one of the m's there so that leaves just m to the power of 1 v squared over v squared makes 1 so they cancel entirely and then this m plus M cancels with one of the m plus M in the bottom leaving with m plus M to the power of 1 there. So we have 0.001 kilograms— mass of the bullet— divided by mass of the bullet plus mass of the block—0.999 kilograms— that makes 1 in the denominator. So we have 0.0010 kilograms minus 1 which makes 0.999 negative; negative means there's a loss in kinetic energy here. So the answer for our question, what is the fractional loss in kinetic energy? The answer is 0.999, most of it is lost.