Question

A force of 10 N acts on 5 kg for 2 seconds. The same force acts on 2.5 kg for 2 seconds. Compare the momentum produced and compare the acceleration caused? [3 MARKS]

Answer 1

Solution:2.5 Marks
Reasoning: 0.5 Marks
Given, force on first body, $F_1=10N$. force on second body, $F_2=10N$, mass of first body, $m_1=5kg$, initial velocity of first body, $u_1=0$, initial velocity of second body, $u_2=0$, mass of second body, $m_2=2.5kg$ and time, $t=2s$.
$Step1$:
Let acceleration produced on first body be $a_1$.
Force on first body be $F_1$ and final velocity of first body be $v_1$.
According to Newton's second law,
$Force=mass\times acceleration$
So, $F_1=m_1{a}_1$
$10=5\times a_1$
$\Rightarrow$$a_1=\frac{10}{5}=2m{s}^{-2}$
From $1^{st}$equation of motion,
$a_1=\frac{v_1-u_1}{t}=\frac{v_1-0}{2}$
$\Rightarrow$$v_1=4m{s}^{-1}$
$Step2$:
Let acceleration produced on second body be $a_2$, force on second body be $F_2$ and final velocity of second body be $v_2$.
According to Newton's second law,
$Force=mass\times acceleration$
So, $F_2=m_2{a}_2$
$10=2.5\times a_2$
$\Rightarrow$$a_2=\frac{10}{2.5}=4m{s}^{-2}$
From $1^{st}$equation of motion
$a_2=\frac{v_2-u_2}{t}=\frac{v_2-0}{4}$
$\Rightarrow$$v_2=8m{s}^{-1}$
Let momentum of first body be $P_1=m_1{v}_1$ and second body be $P_2=m_2{v}_2$.
Comparing momentum=$\frac{P_1}{P_2}=\frac{m_1{v}_1}{m_2{v}_2}=\frac{5\times 4}{2.5\times 8}=\frac{20}{20}=1:1$.
Comparing acceleration $=\frac{a_1}{a_2}=\frac{2}{4}=1:2$.

Conclusion: In this equation, Since the force applied on two bodies is same we observe that the ratio of change in momentum remains same even though the ratio of acceleration is different.