1.

A block of mass m can slide freely in a slot made in a bigger of mass M as shown in fig. There is no friction anywhere in the system. The block m is connected to one end of a string whose other end is fixed at point P. system is released from rest when the string at P makes an angle theta with horizontal. Find the acceleration of m and M after release.

Answer» <html><body><p></p>Solution :Method 1: Just after the release of <a href="https://interviewquestions.tuteehub.com/tag/system-1237255" style="font-weight:bold;" target="_blank" title="Click to know more about SYSTEM">SYSTEM</a>, block M will move towards left (say with acceleration A) and block m will move down with respect ot M (say with acceleration a). As there is no motion of M is vertical direction, hence the acceleration of m in verticle direction will be same as that of M, i.e. A, because horizontal both move together. <br/> Finding constraint Method 1: Let just after release, during a small time dt, the displacement of m(w.r.t. M) is dy downwards and that of M is dx horizontally asshown in fig. <br/> Decreases in the length of segment I`=underset(bar)d xcos theta` <br/> Increases in the length of segment II' `underset(bar)d y` <br/> As the total length of the string is constant , so <br/> `underset(bar)d = underset(bar)d x cos theta` ..(i) <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S01.png" width="80%"/> <br/> Divinding both sides by dt, we get <br/> `(dy)/(dt)=(dx)/(dt) cos theta implies v = V cos theta` ..(ii) <br/> differenting w.r.t. `t=(dv)/(dt)=V(- sin theta)(d theta)/(dt)+(dV)/(dt) cos theta` <br/> But initially `V=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`, as `a=A cos theta` ...(iii) <br/> (`:. (dv)/(dt) = a , (dV)/(dt)=A`) <br/> Method 2: we can obtain (i) in the following way. <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S02.png" width="80%"/><br/> The displacment of M towards left is dx, then its component along AP is `dx cos theta`should be decreased in the length of segment I and this should be equal to increases in the length of segment II which is dy. Hence, `dy=dx cos theta`. <br/> `rArr a = A cos theta`<br/> Method 3: <br/>Total length of string : `l=sqrt(h^(2)+y^(2))+x` <br/> differentiating w.r.t. `t=(dl)/(dt) =(d(h^(2)+x^(2))^(1//2))/(dt) +(dx)/(dt)` <br/> `rArr 0=(2y(dy)/(dt))/(2sqrt(h^(2)+y^(2)))+(dx)/(dt)rArr0= cos theta (-V)+r` <br/> (`because(dx)/(dt)=v, (dy)/(dt)=-V`(as y decreases with time)<br/> and `cos theta=(y)/(sqrt(h^(2)+y^(2)))` <br/> `implies v=V cos theta` which is same as (ii) <br/> Now finding acceleration of blocks <br/> We can use any of the following method to find the value of A and a . <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S03.png" width="80%"/> <br/> Method 1: From FBD of M: <br/> `T cos theta-N = MA` ...(<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>) <br/> From FBD of m: `N=mA` ..(v) <br/> `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>-t=ma` ..(vi) <br/> Combining (iv) and (v) we get <br/> `T cos theta=(m+M)A`...(vii)<br/> Putting `a=A cos theta` in (iv) we get <br/> `mg-T=mA cos theta`..(viii) <br/> Put the value of T form (viii) into (vii) we get <br/> `(mg-mA cos theta)cos theta=(m+M)A` <br/> `implies A=(mg cos theta)/(M+m(1+cos^(2)theta))` <br/> and `a=A cos theta=(mg cos^(2)theta)/(M+m(1+cos^(2) theta))` <br/> The net acceleration of m is <br/> `sqrt(A^(2)+a^(2))=A sqrt(1+cos^(2) theta)=(mg cos thetasqrt(1+cos^(2) theta))/(M+m(1+cos^(2) theta))` <br/> Method 2: Analyzing the block with w.r.t. wedge <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S04.png" width="80%"/> <br/> W.r.t M, m has only vertcal acceleration which is a downwards. Here we have to apply pseudo force because M is a non inertial frame. <br/> FBD of m w.r.t. M: <br/> `N=mA, mg-T=ma` <br/> These equation are same as (v) and (vi). Now solve to get answer. <br/> Method 3: Analyzing the block and wedge together <br/> Analyzing the system of `(M+m)`, i.e. , block and wedge together in horizontal direction, in horizontal direction both masses have same acceleration which is A. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S05.png" width="80%"/> <br/> `T cos theta=(M+m)A` ...(ix) <br/> Analyzing the motion of blcok in vertical direction <br/> `mg-T=ma` ..(x) <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_S01_054_S06.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> (ix) and (x) , we get <br/> `A=(mg cos theta)/(M+m(1+cos^(2) theta))` <br/> and `a=(mg cos^(2)theta)/(M+m(1+cos^(2) theta))`</body></html>


Discussion

No Comment Found

Related InterviewSolutions