What is a free body diagram in Class 11 Physics Chapter 4 Laws of Motion and why is it useful?

= 0.92 * 1260 = 1161 m/s

$v_1=\sqrt[]{2g10}$

$v_2=\sqrt[]{2g5}$

$\overrightarrow{l}=\Delta \overrightarrow{p}$

$l=0.1\left\{\sqrt[]{2g10}+\sqrt[]{2g5}\right\}$

$=0.1\left\{10\sqrt[]{2}+10\right\}$

$=\left(\sqrt[]{2}+1\right)Ns$

For 2 kg block

T – 2g sin37 = 2a        . (i)

For 4 kg block

4g – 2T =  $\frac{4a}{2}$

2g – T = a                    . (ii)

T = (2g – a)

2g – a – 2g ×  $\frac{3}{5}$  = 2a

3a = 2g ×  $\frac{2}{5}$

$a=\frac{4g}{15}$

$a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}=\frac{g}{8}$

8m₁ – 8m₂ = m₁ + m₂

7m₁ = 9m₂

$\frac{m_1}{m_2}=\frac{9}{7}$          

f_K = μmg cosq

= 0.1 × $\frac{50\times \sqrt[]{3}}{2}$

$=2.5\sqrt[]{3}N$

 F₁    = mg sinq + f_K

= 25 + 2.5 $\sqrt[]{3}$  

F₂    = mg sinq – f_K

= 25 – 2.5 

F₁ – F₂ = 5 $\sqrt[]{3}$  N