 Can we measure the speed of this bullet using just this block and a string? Well, the answer is yes. It may sound insane, but but still let's see. Let's see how to do that So we have this bullet coming in coming towards this block and as it comes closer to the block It strikes a block like this and because of that the block and the bullet they start moving forward and they Move they move like this and at this at the maximum height point they come to a state of rest Now this is the whole situation. What's happening, but let's rewind. Let's let's come back So we have we have the bullet which is coming towards towards a block and we are interested in we are interested in figuring out The velocity the velocity of this bullet. Let's call that you let's say that the mass of the bullet is small M and the mass of this big block. This is capital M. This is capital M when the bullet strikes a block Like like this now both the bullet and the block they start moving forwards Let's say with a velocity of V and in this interaction in this collision when the bullet strikes a block There is one quantity which will be conserved which is conserved in every collision whether it is elastic or inelastic and That is momentum momentum of the bullet and the block system. That's really conserved So how can we express conservation of momentum mathematically? We can say we can say that the initial momentum We can say that PI initial momentum This is equal to the final momentum and we are talking about the bullet and the block system here So the initial momentum that is that is only the momentum of the bullet which is M into U This is equal to the final momentum that will be for both the bullet and the block So they become one the bullet stops up till this point They become one mass and that mass is small M plus capital M into V So you the speed of the bullet this is M plus capital M into V divided by small M Now over here we can start off by knowing some values Which is the mass of the bullet we can measure that the mass of the block We can also measure that but we don't know the speed with which both of them move We don't know this V, right? So we cannot really calculate Q. We would need to do something else So for that, let's see what else is happening We know after the collision the block moves forward both of them move forward and This they swing like a pendulum this swing like a pendulum they reach a certain height We can show that like this from the center of from the center of this block So they reach a certain height we can call that H and this is the height when the system of bullet and block They come to state of rest. So the final velocity here that is 0 So they start off with some velocity and they reach a point when their velocity is 0 At the highest point they have traveled a certain height of H So they did gain some energy, right? This is gravitational potential energy and where did this energy come from? Well, initially all they had was kinetic energy So kinetic energy all of this is being changed to gravitational potential energy and we can express this Mathematically we can say we can say that half into small m plus capital M V square This is equal to small m plus capital M into H This just gets cancelled off and V comes out to be equal to under root of 2 g H There will be there will be a G over here as well. Now you might be thinking that Well, I have seen examples where you throw a ball with a speed V It reaches a certain height of H and you can conserve energy over there Considering if we if you don't really consider air resistance and everything So the idea over there is that when there is no external work being done We can conserve energy between any two points, but here is there no external work being done? Let's think about it. So this is a bullet block and an earth system So the force of gravity is really an internal force which is doing some work and giving some gravitational potential energy But there is also one more other force. There is there is a force of tension, right? There is this force of tension. This is T. So isn't this doing any work? Isn't there an external force in the picture in the bullet and the block and the earth system? Isn't this tension doing a work? So how can we conserve energy? Well, it turns out that as the block moves this rope It's always perpendicular to the direction in which the bullet in the block system moves So somewhat like this the the curve the arc. Let's say the arc looks somewhat like this So this rope is always really perpendicular to it Which means that the tension the force of tension is always perpendicular to that direction in which the block in the bullet is moving and when the force when force is perpendicular to displacement We know that work done is always zero because f.d fd cos theta theta is 90 So work done due to the tension force. That's really zero. There is no external work done So we can conserve energy and when we do that we get we get this velocity We with which the block and the bullet started off that is under root of 2gh Now one other thing really we would just know the length of the string usually But there's a bit of a master we can calculate we can using some trigonometry We can figure out this height edge, but let's say you're using some camera You take a snapshot when the bullet in the block comes to a state of rest and because of that You're able to measure this height edge. So let's say we know the maximum height that the block in the bullet reaches Okay, so we know h we know g and when we substitute this v over here Then what do we get? Let's see you. This is equal to small m plus capital M into This is under root of 2gh divided by small m now Let's put in some numbers to get a to get a better feel of it. Usually bullet is 70 grams mass, let's say this is 1 kilogram and the height is 30 meters So when we replace replace these values over here and when you work out you you will come out to be approximately 458 meters per second and that is how one can measure the speed of a bullet using nothing But just one block and a piece of string