Question

A bomb of mass $12\text{kg}$ is dropped by a fighter plane moving horizontally with a speed of $100{\text{ms}}^{-1}$ from a height of $1\text{km}$ from the ground. The bomb exploded after $10\text{s}$ into two pieces of masses in the ratio $1:5$. If the small part started moving horizontally with a speed of $600{\text{ms}}^{-1},$ the speed of bigger part will be $(g=10{\text{ms}}^{-2})$

• A. $100{\text{ms}}^{-1}$
• B. $10\sqrt{65}{\text{ms}}^{-1}$
• C. $120{\text{ms}}^{-1}$
• D. $100\sqrt{2}{\text{ms}}^{-1}$

Answer 1

The correct option is C $120{\text{ms}}^{-1}$

Given that,
Initial horizontal velocity of bomb $u_x=100\text{m/s}$
Initial vertical velocity of the bomb $u_y=0\text{m/s}$

Vertical velocity of bomb after $10s$ will be
$v_y=u_y+gt$
$v_y=gt=100\text{m/s}$

$m_1:m_2=1:5$
$m_1+m_2=12\text{kg}\therefore m_1=2\text{kg}\text{and}{m}_2=10\text{kg}$

Applying conservation of linear momentum in $X-$direction
$m{u}_x=m_1{v}_{x_1}+m_2{v}_{x_2}$
$12\text{kg}\times 100\text{m/s}=2\text{kg}\times 600\text{m/s}+10\text{kg}{v}_{x_2}$
$\Rightarrow v_{x_2}=0$

Applying conservation of linear momentum in $Y-$ direction,
$m{v}_y=m_1{v}_{y_2}$
$12\text{kg}\times 100\text{m/s}=2\text{kg}\times 0+10v_{y_2}$
$v_{y_2}=120\text{m/s},v_2=\sqrt{v_{x_2}^2+v_{y_2}^2}=120\text{m/s}$
$\therefore \left(c\right)$