91Ó°ÊÓ

in question the acceleration-versus-force graphs is given and this graph give the information about the rubber bands who pulled the two objects. so now observe the graph and calculate the ratio of masses of two rubber bands.

Here, use the Newton second law who say that rate of change of momentum is directly proportional to force.

$$\begin{gathered} F\mathrm{Î\pm }\frac{d\stackrel{→}{p}}{dt} \\ F=\frac{dmv}{dt} \\ F=m\frac{dv}{dt} \\ F=ma......\left(1\right) \end{gathered}$$

now used the equation 1 to get the mass of object 1

from the graph taken the value of force which is 3 and acceleration which is 5a₁

Newton second law

$$\begin{gathered} F_1=m_1{a}_1 \\ {m}_1=\frac{F_1}{a_1} \\ {m}_1=\frac{3}{5a_1}.......\left(2\right) \end{gathered}$$

again used the Newton second law to find the mass of the object 2

from the graph take the values of force which is 5 and the acceleration is 4a₁

$$\begin{gathered} F_2=m_2{a}_2 \\ {m}_2=\frac{F_2}{a_2} \\ {m}_2=\frac{5}{4a_1}.........\left(3\right) \end{gathered}$$

Now compare the equation (2) and (3) to get two masses ratio

$$\begin{gathered} \frac{m_1}{m_2}=\frac{{\displaystyle \frac{3}{5a_1}}}{{\displaystyle \frac{5}{4a_1}}} \\ \frac{m_1}{m_2}=\frac{3×4a_1}{5×5a_1} \\ \frac{m_1}{m_2}=\frac{12}{25} \end{gathered}$$

Therefor ratio of two masses are$\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$25$}\right.$