9.5

9.5 Figure 9.13 shows two blocks of masses $M$ and $m$ that slide on smooth planes inclined at angles $\alpha$ and $\beta$ to the horizontal.The blocks are connected by a light inextensible string that passes over a light frictionaless pulley. Find the acceleration of the block of mass $m$ up the plane, and deduce the tension in the string.

Sol^-n Let the system consists of two blocks of mass $M$ and $m$ .
Initially,The system is at rest.
At time $t = t$ , two blocks have
i)the same distance ( which is parallel to the incline) $S$
ii)the same velocity $v = \dot{S}$
iii)the same acceleration $a = \dot{v}$
because they are fixed with the same string,and this string is light and inextensible.Thus tension $T$ is constant.

1.Find the acceleration of mass $m$

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Method 1 Newton 's law of motion

From the newton 's second law (in the direction which is parallel to the incline) of mass

$m$ ; $T\;-\;mg\sin\beta$ = $ma$ ........(1)

$M$ ; $Mg\sin\alpha\;-T\;$ = $Ma$ ........(2)

(1)+(2); $(Mg\sin \alpha\;-\;mg\sin \beta)$ = $(m+M)a$

$a$ = $\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}$

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Method 2 Energy conservation
From energy conservation principle which can simplify as

the decreasing of potential energy = the increasing of kinetic energy.

$-\Delta V$ = $+\Delta K$

$-(mgh_{1} - Mgh_{2})$ = $+(\frac{1}{2}mv^{2} + \frac{1}{2}Mv^{2})$

$-(mgS\sin \beta - MgS\sin \alpha)$ = $\frac{1}{2}(m+M)v^{2}$

$(M\sin \alpha\;-\;m\sin \beta)gS$ = $\frac{1}{2}(m+M)v^{2}$

differentiate with respect to t all the equation,and use $v\;=\;\dot{S}=\frac{dS}{dt}$ and $a\;=\;\dot{v}=\frac{dv}{dt}$ .then

$\frac{d}{dt}(M\sin \alpha\;-\;m\sin \beta)gS$ = $\frac{d}{dt}\frac{1}{2}(m+M)v^{2}$

$(M\sin \alpha\;-\;m\sin \beta)g\frac{dS}{dt}$ = $\frac{1}{2}(m+M)\frac{d}{dt}v^{2}$

$(M\sin \alpha\;-\;m\sin \beta)gv$ = $\frac{1}{2}(m+M)2va$

and divided by $v$ ,thus

$(M\sin \alpha\;-\;m\sin \beta)g$ = $(m+M)a$

$\therefore a\;=\;\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}$

2.Find the tension $T$ .
From the newton 's second law of mass $m$ .

$T\;-\;mg\sin\beta$ = $ma$

$T$ = $m(a\;+\;g\sin\beta)$

substitute $a\;=\;\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}$ then

$T$    = $m(\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}\;+\;g\sin\beta)$

      = $mg(\dfrac{M\sin \alpha\;-\;m\sin \beta+\;m\sin \beta+\;M\sin \beta}{m+M})$

   = $mMg\dfrac{(sin \alpha\;+\sin \beta)}{m+M}$

$\therefore T\;=\;mMg\dfrac{(sin \alpha\;+\sin \beta)}{m+M}$

7.19 A spacecraft travelling with speed V approaches a planet of mass M along straight lin whos perpendicular distance from the center of the planet is p. When the spacecraft is at a distance c from the planet, it fires its engines so as to multiply its current speed by a factor k( 0 < k < 1 ), its direction of motion being unaffected. [ You may neglect the time taken for this operation.] Find the condition that the spacecraft should go into orbit around the planet.

$\oint_{C}Mdx + Ndy$ = $\iint \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$

\frac{\partial N}{\partial x}

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