 So friends we have taken a new problem to apply the basics that we have learned and to learn how we can you know step by step approach the problem. So here goes the problem there is a small m kept on a larger mass capital M. The length of capital M is small l alright. The coefficient of friction between small m and capital M is mu by 2 and coefficient of friction between capital M and the ground is mu ok. Here we need to find out how much time small m takes to cross the capital M right. Now what comes in your mind when you look at this problem. When we look at this problem the entire system is moving towards the right hand side right. It is moving with velocity v it just started. Now what is the direction of friction force on capital M. From the ground of course it is moving relatively forward. So from the ground there must be a friction in the backward direction on capital M right. What about friction between small m and capital M. Now small m was at rest initially and it has been pulled towards the right hand side because of the friction between capital M and small m. So that is the reason why friction force on small m will be forward direction right. So now let us take one by one these two masses and will represent all the forces and acceleration. If we take a small m mass what are the forces will be applied on it. There will be a gravitational force right its value will be mg and there will be normal reaction between capital M and small m n1 right and then there will be a friction mu times n1 okay. This force since this normal reaction is applied by capital M on small m the small m also applies normal reaction in opposite direction okay. So this will be a pair of n1 so this is n1 force right. There will be mg force downward on the bigger mass and there will be normal reaction from the ground on capital M okay. Now this friction force is applied by capital M on small m. So there must be a pair of this friction force on the left hand side. Let us call this mu times n1 okay. Now there will be another friction between capital M and the ground that friction force value will be mu times n2. So the coefficient of friction value between small m and capital M is mu by 2. So let us correct this this is mu by 2 into n1 okay and similarly here it is mu by 2 into n1 okay. Now let us write down Newton's second law of equation for small m and capital M. At times what happens student get confused with the fact that velocity is given and they start to visualize in terms of velocity but you should remember the Newton's second law of equation is for the acceleration it is not for velocity right. There is no direct relation between instantaneous velocity and acceleration. Acceleration is rate of change of velocity okay. So if you write the Newton's second law of equation for small m along horizontal direction I will get it as mu by 2 n1 is the only force along horizontal direction isn't it. This will be equal to mass time acceleration of small m fine and along vertical direction I will get n1 minus mg is equal to 0 since there is no acceleration in vertical direction all right. Now let us write down equation for capital M also along the horizontal direction what are forces you see there is mu by 2 n1 force backward direction minus mu times n2 force right. This will be equal to mass time accession of capital M fine. So, accession of small m will be in this way let us call this as a1 and for capital M accession is backward direction we are calling it as a2 okay. Along the vertical direction we will get mg plus n1 minus n2 is equal to 0 right. So, we have four equations how many variables we have we have a1 a2 n1 and n2. So, we can solve these four equations to get the value of a1 and a2. So, now we can assume that we have the value of a1 and value of a2. Now the question says how much time it takes for this small m to topple from the capital M. Now if we see the motion of small m it goes from this end point number 1 and it must cross this point number 2 to topple over okay. So, the total length it covers is l now is this length an actual length or it is a relative length compared to or with respect to capital M. It is a relative displacement right what will happen the small m and capital M will move together with respect to capital M small m moves a distance of l fine and what is the acceleration along the length along the length it has acceleration a1 fine and capital M has acceleration backward direction which is a2. So, net acceleration of approach from 1 to 2 is what this one let us suppose is small m which is moving with a1 and this capital M is coming towards it with a2. So, acceleration of approach is a1 plus a2 let us call this as an we will say that this is a net acceleration okay. Now initial velocity of approach is how much we have seen that both have started moving together this point number 1 and point number 2 both are going in this direction with same velocity. So, their acceleration their velocity of approach is 0. So, initial velocity is 0 and displacement which is relative displacement since we are talking about relative variable is l. So, which equation comes in your mind to find the value of time s equals to ut plus half at square right. So, if I use s equal to ut plus half at square here and I substitute value of s as l u as 0 and a as a1 plus a2 I will get the value of t okay. So, this is how you solve the particular question I hope now you feel comfortable in applying the basics on problem like these okay. So, thanks for watching the video we will come back with few more problem solving videos which will be testing you on different type of concepts okay.