 Friends, Myself Sandeep Javeri, Assistant Professor of the Department of Civil Engineering from the Washington Institute of Technology, Sholapur. In today's session, we are going to discuss problem on conservation of momentum. The learning outcome at the end of this session, students will be able to solve problem on conservation of momentum. Now let us consider the principle. The principle of conservation of momentum may be stated as the momentum is conserved in a system in which resultant force is 0. In other words, in a system if the resultant force is 0, initial momentum will remain equals to final momentum. Here we should note that this law is valid for entire system, not for single element. It is valid for entire system. Now let us consider the problem on law of conservation of momentum. The statement is a bullet having weight 0.5 Newton and moving with a velocity of 400 meter per second hits centrally a 30 Newton block of wood which is moving away at 15 meter per second and gets embedded in it. Then the velocity of the bullet after the impact and the amount of kinetic energy lost. So here the two bodies given one is bullet another is block. So they are moving with different velocity we are supposed to find out the velocity of bullet after the impact and also we are supposed to find out the amount of kinetic energy lost. Let us consider the solution for this even problem. So we have bullet is given which has velocity as 400 meter per second. So let us the velocity of bullet is even which is equals to 400 meter per second. And another body that is a wooden block is given having a velocity of say u2 is equal to 15 meter per second. The rate of bullet is given as 0.5 Newton. The rate of block is given as 30 Newton. Now after impact this bullet is going to embedded in the block and both bullet and block are moving with a common velocity say V. So here the total mass of bullet and total mass of bullet and block while considering the weight of the body or mass of the body. Now this is a 30 Newton that is a weight of block and this weight of bullet is 0.5 Newton. So according to law of conservation of momentum initial momentum is equals to final momentum. So if you consider this is a initial momentum and this part as final momentum. So we obtain the expression for initial momentum as m1 u1 plus m2 u2 which is equals to capital M that is total mass of block and bullet we have to consider into the common velocities V. We know the mass is obtained as weight divided by acceleration due to gravity. So here m1 is nothing but w1 upon g into u1 plus m2 is w2 upon g into u2 which is equals to now the capital M is replaced by w1 plus w2 divided by small g. So here w1 is nothing but the weight of bullet w2 is weight of block which is 30 Newton. So writing the values this is 0.5 divided by 9.81 plus 30 divided by 9.81. So this is multiplied with the velocity that is 430 divided by 9.81 multiplied by the velocity of block is 15 which is equals to w1 weight of bullet plus weight of block is w2 which is 30 divided by 9.81 multiplied by the velocity that is common velocity V. So here w1 is a weight of bullet that is 0.5 divided by g is 9.81 velocity of bullet is 400 and weight of block is w2 that is 30 Newton and acceleration due to gravity g is 9.81 and u2 that is the velocity of block as 15 meter per second. So in this equation the unknown is the common velocity so this common velocity V is equal to V is equal to we are getting 21.37 meter per second. Now let us consider so we are getting the value as 21.31 meter per second. Now let us find kinetic energy lost so it is obtained by using expression initial kinetic energy minus final kinetic energy. So initial kinetic energy that means we have to consider the motion of bullet as well as motion of block and final kinetic energy we have to consider combined effect of motion of block as well as bullet. So we know the kinetic energy is obtained by using the expression half mv square so initial kinetic energy will be equals to half of m1 u1 square plus half of m2 u2 square this is kinetic energy of bullet this is kinetic energy of block minus the combined effect of block and bullet. So it is half of mv square where mass is a combined weight of or combined mass of bullet and block and V is a common velocity so which is equals to half of w1 upon g into u1 square plus half of w2 upon g into u2 square minus half of w1 plus w2 upon g into v square. So by putting this value we are getting the expression half of 0.5 divided by 9.81 into 400 square plus half of 30 divided by 9.81 multiplied by 30 15 square this velocity of block is 15 minus half of 0.5 plus 30 divided by 9.81 into 21.31 square. So we are getting the kinetic energy lost is equals to 3715.47 joule which is equals to 3.715 kilo joule so this is a required answer that we obtained. So you are supposed to pause the video and answer this question this is the answer. So in this question a bullet having weight 1 Newton which is moving with a velocity of 500 meter per second which hits centrally a 50 Newton block of wood which is moving away at 10 meter per second and gets embedded in it find the velocity of the bullet after the impact. So here we are going to use the expression for law of conservation of momentum that is initial momentum is equal to final momentum by putting the values of velocities of block and bullet we can obtain the final velocity of the bullet after impact and which is obtained as 19.60 meter per second. So this is the reference that we use for the creation of this video. Thank you.