# Question 3ef6b

Nov 27, 2016

Let

${m}_{1} k g \to \text{1st mass having weight } 9.9 N$

${m}_{2} k g \to \text{2nd mass having weight } 22.2 N$

So ${m}_{1} = \frac{9.9}{g} = \frac{9.9}{10} = 0.99 k g$

and ${m}_{2} = \frac{20.2}{g} = \frac{20.2}{10} = 2.02 k g$

Since ${m}_{2} > {m}_{1}$
The combined system will have downward acceleration on ${m}_{2}$ and same upward acceleration on ${m}_{1}$. Let the acceleration of the combined system be a and tension on connecting rope be $T$

Now considering the force on ${m}_{1}$ we can write

$T - {m}_{1} g = {m}_{1} a \ldots \ldots . . \left(1\right)$

Agaun considering the force on ${m}_{2}$ we can write

${m}_{2} g - T = {m}_{2} a \ldots \ldots . . \left(2\right)$

Adding (1) and (2) we get

$\left({m}_{2} + {m}_{1}\right) a = \left({m}_{2} g - {m}_{1} g\right) = \left(20.2 - 9.9\right)$

$\implies a = \frac{10.3}{2.02 + 0.99} m {s}^{-} 2$

$\implies a = \frac{10.3}{3.01} m {s}^{-} 2 \approx 3.42 m {s}^{-} 2$