Question

A girl exerts a force of 6 N on a
body of mass m1, to produce an
acceleration of 12 m s -2? When the
same force is applied on a body
of mass m2, it is accelerated to
24 ms-2. If both the bodies are
joined together, and the same
force is applied on them, what is the
acceleration produced in the
combination.

Answer 1

Given :-

• Force = 6N
• Acceleration of $\bold{{m}_{1}(a_{1})}$= 12m/s²
• Acceleration of $\bold{{m}_{2}(a_{2})}$=24m/s²

According to Question,

if we sum $\bold{{m}_{1}}$and $\bold{{m}_{2}}$ bodies then new body form let $\bold{{m}_{3}}$ i.e,

$\: \: \: \: \: \: \: \: \boxed{ \large{ \rm{{m}_{1} + {m}_{2} = {m}_{3}}}} \: - - (1)$

Find :-

• Value of $\rm{{m}_{1},{m}_{2},{m}_{3}}$
• Acceleration of $\rm{{m}_{3}} \:\rm{{a}_{3}}$

Solution :-

We know that, F = m × a

$\bold{F = 6\: (\red{\rm{Given}})}$

Now, We have to find the mass of $\rm{{m}_{1},{m}_{2}}$

Mass of 1st body

$\: \: \: \: \: \: \: \: \: \rm{F = m_{1 } \times \: a_{1 } } \\ \: \: \: \: \: \: \: \ \rm{6 = m_{1 } \times 12} \\ \: \: \rm{m_{1 } = \frac{6}{12} } \\ \: \: \rightarrow \boxed{ \rm \underline{{m_{1 } = 0.5 \: kg}}}$

Mass of 2nd body

$\: \: \: \: \: \: \: \: \: \rm{F = m_{2 } \times \: a_{1 } } \\ \: \: \: \: \: \: \: \ \rm{6 = m_{2 } \times 24} \\ \: \: \rm{m_{2 } = \frac{6}{24} } \\ \: \: \rightarrow \boxed{ \rm \underline{{m_{2 } = 0.25 \: kg}}}$

Sum of $\underline{\bold{m_{1},m_{2}}}$ /Mass of 3rd body

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: • \: \bold{m_{1} = 0.5 \: kg} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: • \: \bold{m_{2} = 0.25 \: kg}$

Now, putting these values in equation (1)

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{{m}_{1} + {m}_{2} = {m}_{3}} \\ \: \: \: \: \: \: \rightarrow \: \boxed{ \underline{\rm{ {m}_{3}} = 0.75 \: kg}}$

Acceleration of $\underline{\bold{\rm{{m}_{3}}}}(\rm{a_{3}})$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{F = m_{3} \times a _{3} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{6 = 0.75 \times a_{3}} \\ \: \: \: \: \: \: \: \: \rm{ a_{3} = \frac{6}{0.75 } } \\ \: \: \: \: \: \: \: \rightarrow \boxed{ \underline{ \rm{ \green{a_{3} = 8 }}}}$

#Gunjan