Question

A body of mass30×10−3kg, when acted upon by a force for 5s, attains a velocity of 100 m/s. If the same force is applied for 2 minutes on a body of mass 10 kg at rest, what will be its velocity?

Answer 1

$\bigstar$ Answer :

$\bigstar$ Given :

$mass,m_1=30*10^{-3}\;kg\\\\Force,F_1=F\;N\\\\Time,t_1=5s\\\\velocity,v_1=100\;m/s\\\\Force,F_2=F\;N\\\\Time,t_2=2\;min=120\;s\\\\mass,m_2=10\;kg\\\\velocity,v_2=?\;m/s\;[Initial\;velocity\;given\;is\;zero]$

$\bigstar$ Here given that two same forces are acted. $\bigstar$

$\implies \;\bold{F_1=F_2}\\\\\implies\;m_1.a_1=m_2.a_2\\\\\implies\;m_1.\frac{v_1}{t_1}=m_2.\frac{v_2}{t_2}\\\\ \implies\;30*10^{-3}.\frac{100}{5}=10.\frac{v_2}{120}\\\\ \implies\;0.6=\frac{v_2}{12}\\\\ \implies\bold{\bf{\blue{ \;v_2=7.2\;m/s}}}$

Answer 2

Given:

For First body

Mass of body, m₁= 30×10⁻³ kg

Time for which force is applied, t₁= 5s

Initial velocity, u₁= 0

Final velocity, v₁= 100 m/s

For Second body

Mass of body, m₂= 10 kg

Time for which force is applied, t₂= 2min= 120s

Initial velocity, u₂= 0

To Find:

Final velocity of second body

Solution:

We know that,

• For a body moving in uniform motion or having constant acceleration, acceleration 'a' of body is given by

$\pink{\boxed{\bf{a=\dfrac{v-u}{t}}}}$

where,

v is final velocity

u is initial velocity

t is time taken

• Force=mass×acceleration

$\rule{190}{1}$

Let the acceleration produced in first body after applying force be a

So,

$\longrightarrow\rm{a=\dfrac{v_{1}-u_{1}}{t_{1}}}$

$\longrightarrow\rm{a=\dfrac{100-0}{5}}$

$\longrightarrow\rm{a=\dfrac{100}{5}}$

$\longrightarrow\rm{a=20\:m/s^{2}}$

Let the force applied on first body be F

So,

$\longrightarrow\rm{F=m_{1}\times a}$

$\longrightarrow\rm{F=30\times10^{-3}\times20}$

$\longrightarrow\rm{F=600\times10^{-3}}$

$\longrightarrow\rm{F=0.6\:N}$

$\rule{190}{1}$

Let the acceleration of second body after applying same force be a'

So,

$\longrightarrow\rm{F=m_{2}\times a'}$

$\longrightarrow\rm{0.6=10\times a'}$

$\longrightarrow\rm{10a'=0.6}$

$\longrightarrow\rm{a'=0.06\:m/s^{2}}$

Also, let the final velocity of second body be v₂

So,

$\longrightarrow\rm{a'=\dfrac{v_{2}-u_{2}}{t_{2}}}$

$\longrightarrow\rm{0.06=\dfrac{v_{2}-0}{120}}$

$\longrightarrow\rm{0.06=\dfrac{v_{2}}{120}}$

$\longrightarrow\rm{\dfrac{v_{2}}{120}=0.06}$

$\longrightarrow\rm{v_{2}=120\times0.06}$

$\longrightarrow\rm\green{v_{2}=7.2\:m/s}$

Hence, the final velocity of second body is 7.2 m/s.