Question

A bullet of 5gm mass is fired horizontally with a speed of 1100m/s into a wooden block of mass 12 kg that is initially at rest the speed it will acquire after absorbing bullet is_m/s.

Answer 1

Mass of bullet, $\displaystyle\sf{m=5\times10^{-3}\ kg}$

Initial speed of the bullet, $\displaystyle\sf{u=1100\ m\ s^{-1}}$

Mass of the block, $\displaystyle\sf{M=12\ kg}$

Given that the mass is initially at rest.

Let the bullet move along with the block after penetration so that both move with same velocity, say $\displaystyle\sf{v.}$ We have to find this.

By linear momentum conservation,

$\displaystyle\longrightarrow\sf{mu+M(0)=(m+M)v}$

$\displaystyle\longrightarrow\sf{v=\dfrac{mu}{m+M}}$

$\displaystyle\longrightarrow\sf{v=\dfrac{5\times10^{-3}\times1100}{5\times10^{-3}+12}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{v=0.458\ m\ s^{-1}}}}$

Answer 2

The speed it acquire after absorbing bullet is 0.458 m/s

• We can solve this types of problems by using law of conservation of momentum,

Initial momentum = final momentum

• Given,

Mass of bullet , m = 0.005 kg

Mass of block , M = 12 kg

Initial speed of bullet, u = 1100 m/s

Initial speed of block, v = 0

After collision,

Total mass = 12+0.005 kg

                   = 12.005 kg

• Let common speed be s. Then,

mu + Mv = ( m + M ) * s

0.005* 1100 = 12.005 * s

s = 0.458 m/s