Question

Two unequal masses,(m1 and m2) are connected by a string which passes over a frictionless pulley (Fig. 3.1). If m1 and m2 and the table are frictionless, the acceleration of the masses
would be
(a) m1g/m1+m2
(b)m1+m2/m1+g
(c)m2g/m1+m2
(d)none of these

​

Answer 1

SoluTion :-

Given :

Two unequal masses M1 and M2 are connected by a string which passes over a frictionless pulley.

M1, M2 and table are frictionless.

To Find :

Acceleration of the whole system.

Concept :

We can calculate acceleration of the whole system by net force method.

Here, net force acts on the system is only gravitational force.

Calculation :

As per newton's second law of motion,

$\Rightarrow\bf\:F_{net}=(M_1+M_2)\times a\\ \\ \Rightarrow\sf\:M_1g=(M_1+M_2)\times a\\ \\ \Rightarrow\underline{\boxed{\bf{\gray{a=\dfrac{M_1g}{M_1+M_2}}}}}$

Additional information :

▪ Calculation of tension in string :

$\sf\:Net\:force\:on\:block\:of\:mass\:M_2\\ \\ \dashrightarrow\bf\:T=M_2\times a\\ \\ \dashrightarrow\sf\:T=M_2\times (\dfrac{M_1g}{M_1+M_2})\\ \\ \dashrightarrow\underline{\boxed{\purple{\bf{T=\dfrac{M_1M_2g}{M_1+M_2}}}}}$

Answer 2

$\Large{\sf{\green{\underline{Given}}}}$

• Two unequal masses (M1, M2) respectively

• All the surfaces are frictionless

$\Large{\sf{\green{\underline{To\:find}}}}$

• Acceleration of the masses (M1, M2)

_________________________________________________

$\boxed{\Large{\sf{\blue{Method\:1\: }}}}$

We can find the acceleration by simply using the Newtons second law of motion

where,

$\boxed{\large{\sf{\red{a_{\: net} = \frac{Force_{\:net}}{Mass_{\:net}}}}}}$

$\sf{\implies\: a = \frac{M_{1}g}{M_{1} + m{2}}}$

_________________________________________________

$\boxed{\Large{\sf{\blue{Method\:2}}}}$

We can also find the acceleration by using free body diagram of the masses

$\setlength{\unitlength}{1cm}</p><p>\begin{picture}(12,4) \thicklines</p><p>\put(3,1) {\line(1,0){1cm}}</p><p>\put(3,2){\line(1,0){1cm}}</p><p>\put(4,1){\line(0,1){1cm}}</p><p>\put(3,1){\line(0,1){1cm}}</p><p>\put(4,1.5){\vector(1,0){1.5cm}}</p><p>\put(3.5,2){\vector(0,1){1cm}}</p><p>\put(3.5,1){\vector(0,-2){1cm}}</p><p></p><p>\put(7,1){\line(1,0){1cm}}</p><p>\put(7,2){\line(1,0){1cm}}</p><p>\put(8,1){\line(0,1){1cm}}</p><p>\put(7,1){\line(0,1){1cm}}</p><p>\put(7.5,2){\vector(0,1){1cm}}</p><p>\put(7.5,1){\vector(0,-2){1cm}}</p><p></p><p>\put(2.8, 3.2){ N }}</p><p>\put(2.5, -0.2){ M_{2}g }</p><p>\put(5.7, 1.5){ T }}</p><p></p><p>\put(7.8, 3.2){ T }}</p><p>\put(8, -0.2){ M_{1}g }</p><p></p><p>\put(3.3, 1.45){ M_{2} }</p><p>\put(7.3, 1.45){ M_{1} }</p><p></p><p>\end{picture}$

In the above diagram, we have considered the acceleration to be in downward direction

$\sf{ M_{1}g - T = M_{1}a \: \: \rightarrow\: (1)}$

$\sf{T = M_{2}a \: \: \rightarrow\: (2)}$

Adding equation (1) and (2):

$\sf{\implies\: M_{1}g = M_{1}a + M_{2}a}$

$\sf{\implies\: M_{1}g = a (M_{1} + M_{2})}$

$\boxed{\sf{\green{\therefore\: a = \frac{M_{1}g}{M_{1} + M{2}}}}}$