A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is μ_k .Find an expression for the bullet's speed v_bullet.

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Hey, everyone in this problem. A forensic science technician shoots horizontally a projectile of mass g into a wooden box of mass 3.5 kg placed on a rough horizontal table. The projectile sticks into the box after the collision, the box moves 1.75 m on the table. We're told that the kinetic friction coefficient between the box and the table is 0.25. And we are asked to calculate the speed of the projectile before the collision. We're given four answer choices. Option a 10.3 m per second. Option B 234 m per second. Option C 290 m per second and option D 687 m per second. So we want to find the speed of the projectile just before the collision, but we're really not given any additional information about what happens before the collision. So let's start with after the collision, see if we can figure out the speed of this projectile after the collision and work backwards. We're gonna write down the variables we need for our kinematic equations. We have V not and we're gonna call this va OK. When we say the initial velocity here, we mean the initial velocity of this post collision period. And so we're gonna call that va the velocity immediately after the collision. We know that the final velocity is gonna be zero m per second. OK. The box travels 1.75 m on that table and then it is going to come to rest. So that final velocity is zero m per second. We know that the horizontal displacement or that distance traveled is 1.75 m. We don't know the acceleration and we aren't given the amount of time this takes. So what we want to solve for here is the velocity of the projectile immediately after the collision. OK. So that we can use that to relate to before the collision. But in order to do that, we need three known values. And right now, we only have two, we are given the coefficient of kinetic friction, which we know is related to the forces. So let's use our free body diagram look at the forces. OK? We know through Newton's second law, the forces are related to the acceleration. So we can use that to try to calculate this acceleration. A so that we can calculate this after collision velocity. So let's draw everybody diagram of our box, it's on a horizontal surface. So we know that we're gonna have the normal force pointing upwards. We're gonna have the force of gravity pointing downwards. And the only horizontal force we have acting is a force of kinetic friction, which we're gonna call F K. And we know that it's kinetic friction because the box is moving. Now, according to Newton's law, recall that the sum of the forces and in this case, in the direction, it's gonna be equal to the mass multiplied by the acceleration. The only force way of acting is this coefficient of kinetic friction or sorry, the force of kinetic friction we're gonna take to the right to be positive that forces acting to the left. And so that is going to be negative F K, it is equal to the mass um multiplied by the acceleration A X. Now the force of kinetic friction recall is given by negative mu K multiplied by the normal force N and that is equal to the mass multiplied by the acceleration A. Now, in order to calculate the acceleration A X, we need to know what this normal force is. So let's switch over to the forces in the Y direction. And because that's where that normal force is acting, if we look at the sum of the forces in the Y direction, we know that this is gonna be equal to zero because this box is not moving up and down. OK. In terms of the vertical component, it is an equilibrium. Let's take up to be positive. The normal force is acting in the positive direction. So yet N, the force of gravity is acting in a negative direction. So we have N minus F G is equal to zero. It tells us that the normal force is equal to the force of gravity which we know is equal to mass multiplied by gravitational acceleration G. So we can use this in our force is in the X component, we have negative mu K multiplied by that normal which we found to be M G is equal to MA X. OK. We can divide by the mass M on both sides are left with the acceleration A X is equal to negative mu K multiplied by G. We're told that the friction coefficient is 0.25. So we get negative 0.25 multiplied by the acceleration due to gravity 9.81 m per second squared, which gives us an acceleration of negative 2.45, 25 m per second squared. OK? So now we found our acceleration in the horizontal direction. We're gonna put a blue box around that. So we don't lose track of it. And we can actually add that to this list of information we have about this post collision motion. So this is gonna be negative 2. m per second squared. Now we have three known values we wanna calculate this velocity after the collision. We're gonna choose the kinematic equation that contains the three variables we know plus the one we're looking for. That means we're gonna leave out the time T and that equation is going to be the following V F squared is equal to V not squared plus two A delta X. OK. That final velocity is zero, this is equal to va that velocity after the collision squared plus two, multiplied by negative 2.4525 m per second squared, multiplied by that distance 1.75 m, we can move everything over. So we're gonna have va squared on one side. The second term, this was negative, we had moved it to the other side. So we needed to add it. So it's going to be positive now. And if we work it out, it gives us 8. meter squared per second squared and taking the square root. Now we're gonna take the positive route here and remember when you take the square root, you get both the positive and the negative route. We've said that to the left is our negative direction because we found this negative acceleration that acceleration is slowing this block down. So if the acceleration is negative, that means that this block is moving to the right. And so it's gonna have a positive velocity. OK. So we're gonna take the positive route here. We get 2.92, 98 m per second. All right. So we found this velocity but we have to be careful. This isn't the velocity we were looking for. Remember we want to find the speed of the projectile before the collision just before the collision. What we found is the velocity after the collision of this block. So now we need to switch gears here, we have a collision. So let's think about the conservation of momentum. Recall that the initial momentum P is going to be equal to the final momentum P F. The initial momentum is made up of the momentum of the projectile, which we call P not P plus the initial momentum of the box P not B. And this is equal to the final momentum. And we're just gonna call it P F because that projectile sticks into the box. So after the collision, we only have one object, it's the box and projectile combined. Now recall that the momentum is equal to the mass multiplied by the velocity. And so we get MP multiplied by VA P plus M B multiplied by V not B is equal to the total mass after the collision, which is the mass of the projectile MP plus the mass of the box M B multiplied by V. And this is va the velocity after the collision. OK. Now, initially, the box is not moving right. And so the velocity V not B is zero, which means that the second term on the left hand side goes to zero, remaining on the left hand side, we have the mass of the projectile. Now, we're told that this is 15 g, we wanna convert this into kilograms. And so we have 15 g. We're gonna multiply that by one kg divided by 1000 g. OK? Because we know that there are 1000 g in every kilogram, the unit of gram will divide out. What we're essentially doing is dividing by 1000 to go from grams to kilograms. That is multiplied by VA P. This is equal to the sum of the masses. OK? So we have 15 and I'm just gonna simplify because we've already shown it written out here that conversion from grams to kilograms. So we have 15 divided by 1000 kg plus the mass of the box 3.5 kg. And this is multiplied by that velocity after the collision, which we found to be 2. m per second. All right. So we wanna isolate for this initial velocity of the projectile. OK. Simplifying on the right hand side, we get 3.515 kg multiplied by 2. m per second. And we're gonna divide by this mass from the left hand side to isolate V knot. And that is gonna be 0.15 kg. We can work this out on our calculator and we get that V not P is going to be equal to 686. m per second. And that is the velocity of the projectile before the collision. Ok. Now we were asked to find speed. So we wanna just take the magnitude of this velocity, the velocity is positive. And so the speed is going to be the exact same as the velocity. If we look back at our answer choices, we can see that they're rounded to the nearest meter per second. And the correct answer is gonna correspond with option D 687 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is μ_k .Find an expression for the bullet's speed v_bullet.

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A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is μk .Find an expression for the bullet's speed vbullet.

Verified video answer for a similar problem:

Key Concepts

Conservation of Momentum

Frictional Force

Work-Energy Principle

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is μ_k .Find an expression for the bullet's speed v_bullet.