Question

A bullet of mass $10g$ fired with an initial velocity of $500m{s^{ - 1}}$ hits a $20kg$ wooden block at rest and gets embedded into it. (i) Calculate the velocity of the block after the impact. (ii) How much energy is lost in the collision?

Answer 1

Hint: In order to solve this question, we will first calculate the momentum of bullet and then final momentum of block and bullet and using law of conservation of momentum we will find velocity of the block as well as loss in kinetic energies of the system.

Formula used: If P and P’ be the initial and final momentum of a system before and after collision then, according to law of conservation of linear momentum $P = P'$
Kinetic energy of a body of mass m and velocity v is $K.E = \dfrac{1}{2}m{v^2}$

Complete step by step answer:
According to the question, we have given that
${m_{bullet}} = 10g = 0.01kg$ mass of bullet
${v_{bullet}} = 500m{s^{ - 1}}$ velocity of bullet before collision then, initial momentum of system before collision can be written as
$P = {m_{bullet}}{v_{bullet}}$ on putting the value of parameters we get,
$P = 0.01 \times 500$
$P = 5kgm{s^{ - 1}} \to (i)$
Now, after collision bullet embedded into block so, let’s combined mass of the system is
$m = 20 + 0.01 = 20.01kg$ mass of bullet plus mass of block
let v be the final velocity of block with bullet embedded in it, then final momentum P’ can be written as
$P' = mv$ on putting the value of parameters we get,
$P' = 20.01 \times v \to (ii)$
now, according law of conservation of momentum $P = P'$ so equation equations (i) and (ii) we have,
$5 = 20.01 \times v$
$ \Rightarrow v = 0.25m{s^{ - 1}}$
(i) Hence, the velocity of the block after the impact is $0.25m{s^{ - 1}}$
Now, Initial Kinetic energy of the system is $K.{E_{initial}} = \dfrac{1}{2}{m_{bullet}}{v^2}_{bullet}$ on putting the value of parameters we get,
$K.{E_{initial}} = 0.5 \times 0.01 \times 250000$
$K.{E_{initial}} = 1250J$
After collision, let Kinetic energy of the system is $K.{E_{final}} = \dfrac{1}{2}m{v^2}$ on putting the value of parameters we get,
$K.{E_{final}} = 0.5 \times 20.01 \times 0.0625$
$K.{E_{final}} = 0.6253J$
so, loss in energy can be calculated as
$Loss = K.{E_{initial}} - K.{E_{final}}$ on putting the values we get,
$Loss = 1250 - 0.6253$
$Loss = 1249.37J$
(ii) Hence, the loss in energy in collision is $1249.37J$

Note: It should be remembered that, when velocity is in metre per second always convert mass from grams in to Kilograms using basic conversion as $1g = 0.001kg$ and here bullet embedded in to block and moved with it is an example of perfectly inelastic collisions.