Two masses \[{{M}_{1}}=5\text{ }kg,\text{ }{{M}_{2}}=10\text{ }kg\] are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When masses are released, then acceleration of masses will be: [AIPMT 2000] |
A) g
B) \[\frac{g}{2}\]
C) \[\frac{g}{3}\]
D) \[\frac{g}{4}\]
Correct Answer: C
Solution :
In the case of masses hanging from a pulled a string, the tension in whole string is same say equal to T. |
As \[{{M}_{2}}>{{M}_{1}},\] so mass \[{{M}_{2}}\] moves down and mass \[{{M}_{1}}\] moves up with the same acceleration a (say). The arrangement of the motion is represented in the figure. |
Equation of motion of mass \[{{M}_{2}},\] is |
\[{{M}_{2}}g-T={{M}_{2}}a\] |
Equation of motion of mass M, is, |
\[T-{{M}_{1}}g={{M}_{1}}a\] |
Adding Eqs. (i) and (ii), we get |
\[\left( {{M}_{2}}g-T \right)+(T-\left( {{M}_{1}}g \right)=\left( {{M}_{1}}+{{M}_{2}} \right)a\] |
\[\left( {{M}_{2}}-{{M}_{1}}g \right)=\left( {{M}_{1}}+{{M}_{2}} \right)a\] |
\[\Rightarrow \] \[a=\left( \frac{{{M}_{2}}-{{M}_{1}}}{{{M}_{1}}+{{M}_{2}}} \right)g\] |
Given, \[{{M}_{1}}=5\,kg,\,{{M}_{2}}=10\,kg\] |
Hence, \[a=\left( \frac{10-5}{5+10} \right)g=\frac{5}{15}g=\frac{g}{3}\] |
Note: In a mass-pulley system, the tension in the string is always towards the pulley. |
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