Question

a bullet of mass 20g moving with a velocity of 100m/s strikes a wooden block of mass 800 gram and gets embedded into it .calculate the velocity of the combined system​

Answer 1

Given ,

Mass of bullet = 20 g or 0.02 kg

Inital velocity of bullet = 100 m/s

Mass of wooden = 800 gram or 0.8

Initial velocity of wooden = 0 m/s

Let , the velocity of wooden block after collision be " x "

We know that ,

Law of conservation of momentum states that during any collision , if the external force is zero then the initial momentum will be same as final momentum after collision

Thus ,

$\mathtt{\large \fbox{m_{1} u_{1} + m_{2} u_{2} = ( m_{1} + m_{2})v}}$

0.02 × (100) + 0.8(0) = (0.02 + 0.8)v

2 + 0 = (0.82)v

v = 2/0.82

v = 2.44 m/s

Hence , the velocity of the combined system is 2.44 m/s

Answer 2

Mass of bullet, $\sf{m=20\ g=0.02\ kg}$

Initial velocity of bullet, $\sf{u=100\ m\,s^{-1}}$

Mass of wooden block, $\sf{M=800\ g=0.8\ kg}$

The wooden block was initially at rest, hence its initial velocity, $\sf{U=0\ m\,s^{-1}}$

Since the bullet gets embedded into the block, both move with same velocity.

Let the velocity of the combined mass after collision be $\sf{V.}$

By law of conservation of linear momentum,

$\longrightarrow\sf{mu+MU=(m+M)\,V}$

$\longrightarrow\sf{V=\dfrac{mu+MU}{m+M}}$

$\longrightarrow\sf{V=\dfrac{0.02\times100+0.8\times0}{0.02+0.8}}$

$\longrightarrow\sf{V=\dfrac{2}{0.82}}$

$\longrightarrow\sf{\underline{\underline{V=2.44\ m\,s^{-1}}}}$

Hence the combined system moves with a velocity of $\bf{2.44\ m\,s^{-1}.}$