A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 10 g bullet is fired into the block, in the face opposite the spring, and sticks. What was the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm?

Video transcript

Hey, everyone in this problem, we have a 1.6 kg piece of wood lying stationary on a frictionless bench that is mounted to a horizontal spring on one end and that spring is fixed to the wall. At the other end, we have a 25 g stone that is gonna be thrown parallel to the spring from a slingshot and enter a hollow in the piece of wood and get stuck. OK? If the maximum displacement in the aci motion of the piece of wood is 850 millimeters in the spring. Constant is 3200 Newton per meter. We are asked to find the stone speed when it hits the block. All right, we're given four answer choices all in meters per second. Option A is 2.45 times 10 to the exponent 3 m per second. Option B 38.3 option C 37.7 and option D 37.1. So let's just draw a diagram of what we have first. So we have our wall, we have our floor, we have our piece of wood that is attached to a spring. OK? And the other end of that spring is attached to the wall and we have the stone that is gonna be thrown towards that block. OK? And then sometime later, we're gonna get this max displacement and the spring is gonna be compressed as far as it will. And that stone is gonna be stuck in that wood. Yeah, let me get rid of these extra lines here. So it's not confusing. OK? That stone is gonna stick into that wood and we know that that this maximum displacement is 850 millimeters. And so that distance is 850 millimeters. Now let's do some conversion of our units. OK. The mass of the block um let's call it MW. The mass of the wood is 1.6 kg or told. Yeah, we have this stone, we'll call this MS. The mass of the stone is 25 g converting to kilograms. We're gonna multiply by 1 kg divided by 1000 g because we know that there are 1000 g in every kilogram. The unit of gram divides out what we're essentially doing is dividing by 1000 and we get 0.025 kg. We have this maximum displacement and we're gonna call this XF the final displacement is 850 millimeters and converting to meters, we multiply by 1 m divided by 1000 millimeters because we know again there are 1000 millimeters in every meter. The unit of millimeters divides out. We're essentially dividing by 1000 we get 0.85 m. And finally, our K value that spring constant we're told is 3200 newtons per meter. OK. So that's everything we're told in this problem. And we have two things to kind of consider. Uh we have pre collision where we have the stone moving in the block stationary and then we have post collision where that stone sticks into the block and they move together. OK. So let's look at the collision first when we have a collision. OK. This is frictionless. We have no nec external forces. We want to think about the conservation of momentum. Right now. Our call conservation of momentum tells us that the initial momentum before the collision is gonna be equal to the momentum after the collision. OK. So we have P knot is equal to P A. We're using subscript not to indicate before the collision. We're using subscript A to be immediately after the collision. Now, before the collision, we have two things to consider, we have the momentum initially of the stone P not S and we have the momentum initially of the wood block pow. And this is gonna be equal to P A that momentum. After now, we only have one momentum to consider here because they're gonna stick together. That stone is gonna stick in the wood. So it's gonna be one total mass that's moving at one speed. Altogether. Now, we're told initially that this wood is stationary. And so the momentum from the wood is actually gonna be zero. Can I recall that momentum is equal to mass multiplied by velocity? So if we write that out, we have MS V, not S plus MW V, not W is equal to that total mass MS plus MW multiplied by va that velocity after the collision. OK. And like I just mentioned, the velocity of the wood initially is zero. So that second term on the left hand side goes to zero, substituting in our values, we have a 0.025 kg multiplied by V OS. OK. Now V not S is what we're trying to find the velocity of the stone before the collision. This is gonna be equal to 0.025 kg. That's 1.6 kg all multiplied by va the velocity after the collision. Now, we have two unknowns here. We have this initial velocity of the stone and we have the velocity immediately after the collision. We wanna find VA S but for now, we can't do that without finding VA. So let's simplify a little bit. Let's write VNAT S in terms of VA. Then we're gonna have to switch to the post collision situation to find va so that we can substitute that back into this equation for V not S OK. We're gonna divide by 0.025 kg on both sides and then we're gonna simplify, we get VNA S is equal to 65 multiplied by VA. And we're gonna call this equation star and we're gonna come back to that once we know VA. So now switching to post collision, OK? And this is where that information about the maximum displacement is gonna come from. Now, post collision, we don't have, we're not gonna consider the conservation of momentum. Instead, we're gonna consider the conservation of mechanical energy. OK. We know that we have some kinetic energy because this block block is moving. We know that we have some potential energy from the spring. So let's consider that conservation of mechanical energy, we have K A plus U A is equal to KF plus UF OK. That kinetic energy immediately after the collision plus the potential energy immediately after the collision is going to equal the sum of the kinetic and potential energies at the end. Now recall that the kinetic energy is one half MV squared. The potential energy due to the spring is one half kx squared. OK. We don't have any potential energy to worry about gravity because we are on a um horizontal surface. So we don't have to worry about gravity here. And so substituting this in, OK, we have one half. Now, after the collision, the mass is gonna be the combined mass. OK? The stone is stuck inside of the wood. And so we have both of those masses to consider. So we have MS plus MW multiplied five va squared that velocity immediately after the collision, whoops plus one half KX A squared. The displacement after the collision squared, this is gonna be equal to one half MS plus MW VF squared plus one half KXF squared. OK. So we've just applied those formulas all the way across. We're using the subscript A to indicate immediately after the collision. And the subscript F to be that final time point where we have a maximum displacement. Now, immediately after the collision, that block is going to be still in its equilibrium position. OK? It has not moved just yet. And so X A that displacement is gonna be zero. So we're gonna have no potential energy due to the spring on the left hand side. And on the right hand side, this is at a maximum displacement. So at maximum displacement recall that the speed is going to be equal to zero. And so VF is gonna be equal to zero. The first term on the right hand side goes to zero and we're left with just two terms. OK? We have one half MS plus MW multiplied by VA squared. On the left hand side, we have one half KXF squared. On the right hand side, let's substitute in our values one half multiplied by 0.025 kg plus 1.6 kg. Va squared is equal to one half multiplied by 3200 newtons per meter. Multiplied by 0.85 m squared. OK. Now we have an equation with only one unknown that is va and we can solve simplifying on the left and right, we get 0.81 2 5 kg, multiplied by va squared is equal to 1156 Newton meters. Recall that a Newton me is equivalent to a kilogram meter per second squared. So Newton meter is gonna be kilogram meter squared per second squared. We're gonna divide that by this 0.8125 kg on the left hand side, which leaves us with a unit of meter squared per second squared. So we have va squared is equal to 1422.76923 m squared per second squared. And when we take the square root and we're gonna be left with meters per second. Now this question was asking us to find the speed, ok? Not the velocity. So what we care about is the magnitude. So when we take the square route, we're just gonna take the positive route again because we want that magnitude and we get that va it's gonna be 37.72 m per second. Now, you have to be careful. Remember that va is not what we're trying to calculate. We're trying to calculate the stone speed when it hits the block, ok? So we're trying to calculate that initial speed of the stone V not S So what we're gonna do is we're gonna substitute this va value we found 37.72 meters per second into our equation star to get that speed we are looking for. OK? We're called that we had venna S is equal to 65 VA and let me just draw up these vector symbols because we want just the speed. This is gonna be equal to 65 multiplied by 37.72 m per second. Then we get the Vena S is equal to 2451.8 m per second. And we can write this in scientific notation A S 2.45 times 10 to the exponent 3 m per second. And that is the speed of the stone, right as it hits that block. And that is what the question was asking us to find. If we compare this to our answer choices, we can see that the correct answer is option a 2.45 times 10 to the exponent 3 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.