Question

A bullet of mass $20g$ moving with a velocity of $100{ms}^{-1}$​ strikes a wooden block of mass $80g$ and gets embedded into it. Calculate velocity of the combined system.

• A. $100{ms}^{-1}$​
• B. $30{ms}^{-1}$​
• C. $20{ms}^{-1}$​
• D. $5{ms}^{-1}$​

Answer 1

The correct option is C

$20{ms}^{-1}$​

Let mass of bullet be $m_1=20g$
Let mass of the wooden block be $m_2=80g$
Let initial velocity of the bullet be $u_1$ = $100m{s}^{-1}$ and final velocity of the bullet be $v_1$.
Let the initial velocity of wooden block be $u_2$, since it was at rest, the initial velocity is zero, i.e $u_2=0m{s}^{-1}$
Let final velocity of the wooden block be $v_2$.
By applying law of conservation of momentum,
intial momentum of the system = final momentum of the system
i.e, $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
After the bullet is fired, the bullet gets embedded on the block.
Therefore, it acts as a system
Hence , the final velocity of the bullet and wooden block after collision will be equal as it works as a system
i.e $v_1=v_2=v$ where v is the velocity of the system after collision.
Substituting on the equation, we get
$(0.020\times 100)+(0.080\times 0)=v\times (0.020+0.080)$
$\Rightarrow$ $2=0.1\times v$
$\Rightarrow$ $v=20{ms}^{-1}$​

$\therefore$Velocity of system after collision = $20{ms}^{-1}$​