Question 95: Laws of Motion

Question

Two masses $$M_{1}$$ and $$M_{2}$$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $$M_{2}$$ is twice that of $$M_{1}$$, the acceleration of the system is $$a_{1}$$. When the mass $$M_{2}$$ is thrice that of $$M_{1}$$, the acceleration of the system is $$a_{2}$$. The ratio $$\frac{a_{1}}{a_{2}}$$ will be :

Accepted Answer

$${a_1} = {{{M_2} - {M_1}} \over {{M_2} + {M_1}}} 	imes g = {{2{M_1} - {M_1}} \over {3{M_1}}} 	imes g$$
$$ = {g \over 3}$$
And, $${a_2} = {{3{M_1} - {M_1}} \over {4{M_1}}} 	imes g = {g \over 2}$$
$$	herefore$$ $${{{a_1}} \over {{a_2}}} = {{g/3} \over {g/2}} = {2 \over 3}$$

Answer

$$\frac{1}{3}$$

Answer

$$\frac{2}{3}$$

Answer

$$\frac{3}{2}$$

Answer

$$\frac{1}{2}$$

Laws of Motion

Choose the correct answer: