Question

A bullet of mass 10 g is shot from a gun of
mass 2 kg such that gun recoils with velocity 2
m/s. The bullet strikes a stationary block of
mass 490 g placed on a rough horizontal
surface and gets embedded in it. If frictional
force provided by the surface on block is 2 N
The distance travelled by block before coming
to rest is
10 g द्रव्यमान की एक गोली को 2 kg द्रव्यमान की एक
चन्दक से इस प्रकार दागा जाता है ताकि बन्दूक 2 m/s के​

Answer 1

Given:

A bullet of mass 10 g is shot from a gun of mass 2 kg such that gun recoils with velocity 2m/s. The bullet strikes a stationary block of mass 490 g placed on a rough horizontal surface and gets embedded in it. Frictional force provided by the surface on block is 2 N.

To find:

Distance travelled by block before coming

to rest is ?

Calculation:

First of all, let's calculate the velocity of the bullet using the principle of Conservation of Linear Momentum:

$\therefore \: P_{i} = P_{f}$

$\implies \: 0 + 0 = \dfrac{10}{1000} u + 2( - 2)$

$\implies \: 0 + 0 = \dfrac{10}{1000} u - 4$

$\implies \: \dfrac{u}{100} = 4$

$\implies \: u = 400 \: m {s}^{ - 1}$

Now, let's calculate the velocity of block-bullet system:

$\therefore \: P_{i} = P_{f}$

$\implies \: \dfrac{10}{1000} \times 400 = \dfrac{(490 + 10)}{1000} v$

$\implies \: 4= \dfrac{v}{2}$

$\implies \: v = 8 \: m {s}^{ - 1}$

Now , applying 3rd Equation of Kinematics:

$\therefore \: { (v_{f})}^{2} = {v}^{2} + 2as$

$\implies \: {(0)}^{2} = {(8)}^{2} + 2 (\frac{force}{mass} )s$

$\implies \: {(0)}^{2} = {(8)}^{2} + 2 ( - \dfrac{2}{0.5} )s$

$\implies \: 0 = 64 - 8s$

$\implies \: 8s = 64$

$\implies \: s = 8 \: metres$

So, the block-bullet system will move for 8 metres.