Question

A bullet whose mass is 50g moves horizontally with velocity $100m/\sec $, hits and gets embedded in a wooden block having mass $450g$ placed on a vertical height of $19.6m$. Calculate how far from the wall does the block fall on the ground, if the bullet gets embedded in the wall? It is given that, $g = 9.8m/{\sec ^2}.$

Answer 1

Hint: Use the formula of momentum and find the initial momentum and final momentum and then use the conservation of momentum to find the velocity. The system will fall as the horizontal projectile.

Complete step by step solution:
Let the initial momentum be ${P_i}$ and final momentum be ${P_f}$. We know that momentum can be defined as the product of mass and velocity. Mathematically, it can be represented as –
$P = mv$
where, $m$ is the mass and $v$ is the velocity.
According to the question, it is given that –
Mass of bullet, $m = 50g = \dfrac{{50}}{{1000}}kg$
Mass of wooden block, $M = 450g = \dfrac{{450}}{{1000}}kg$
$\therefore {P_i} = m \times 100$
Putting the value of mass of bullet in the above equation, we get –
$
   \Rightarrow {P_i} = \dfrac{{50}}{{1000}} \times 100 \\
  \therefore {P_i} = 5kg - m/\sec \\
 $
Now, when it gets embedded in the wooden block let the velocity of the combined system be $Vm/\sec $.
So, the momentum of the system after the bullet gets embedded inside the block will be –
$
   \Rightarrow {P_f} = \left( {\dfrac{{450 + 50}}{{1000}}} \right) \times V \\
   \Rightarrow {P_f} = \left( {\dfrac{V}{2}} \right)kg - m/\sec \\
 $
Now, using the conservation of momentum which states that in the system the momentum remains constant or momentum can neither be created nor be destroyed. Hence, the initial momentum and final momentum will be equal.
$\therefore {P_i} = {P_f}$
$
  \therefore 5 = \dfrac{V}{2} \\
   \Rightarrow V = 10m/\sec \\
 $
Therefore, the combined system moves with the velocity of $10m/\sec $ after the bullet gets embedded inside the block.
So, now the system falls as a horizontal projectile. In the question it is given that height, $H = 19.6m$.
We know that, time taken by the projectile can be calculated by –
$T = \sqrt {\dfrac{{2H}}{g}} $
Putting the values of height and $g$ in the above equation, we get –
$
   \Rightarrow T = \sqrt {\dfrac{{2 \times 19.6}}{{9.8}}} \\
   \Rightarrow T = \sqrt 4 \\
  \therefore T = 2\sec \\
 $
Now, we have to get the range covered by the projectile before it touches the ground so, it can be calculated as –
$
  R = V \times T \\
   \Rightarrow R = 10 \times 2 \\
  \therefore R = 20m \\
 $
Hence, the distance from the wall to the falling block on the ground is $20m$.

Note: As it is given that bullet is embedded inside the block then, the total mass of the system will be the sum of mass of bullet and mass of block which is used during the finding of final momentum.