Question

A bullet of mass ‘m', moving with a speed ‘v’ strikes a wooden block of mass M and gets embedded in it. The speed of this embedded block will be?

• A. $\left(\sqrt{\frac{M}{M+m}}\right)v$
• B. $\left(\sqrt{\frac{m}{M+m}}\right)v$
• C. $\left(\frac{m}{M+m}\right)v$
• D. $\left(\frac{m+M}{Mm}\right)v$

Answer 1

The correct option is C

$\left(\frac{m}{M+m}\right)v$

Let the velocity of the embedded block be V.
Total momentum before impact = mv
Total momentum after impact = (M+m)V
Now, applying conservation of momentum
mv = (M+m)V
$\therefore V=\left(\frac{m}{M+m}\right)v$