Question

A bullet hits a stationary wooden block with a velocity of $200\text{m/s}$. After the collision, the bullet lodges into it and the combined mass move together with a velocity of $10\text{m/s}$. If the mass of the block is $9\text{kg}$, then find the mass of the bullet.

• A. $473\text{g}$
• B. $380\text{g}$
• C. $527\text{g}$
• D. $428\text{g}$

Answer 1

The correct option is A $473\text{g}$

Initially block is at rest.Therefore $u_2=0$Let the mass of bullet is $m_1\text{kg}$.

After collision

Applying P.C.L.M,
$m_1{u}_1+m_2{u}_2=(m_1+m_2)v$
$m\times 200+9\times 0=(m+9)\times 10$
$200\times m=10m+90$
$m=\frac{90}{190}=0.473\text{kg}=473\text{g}$