Question

A 50 gram bullet moving with a velocity of 10 m/s gets embedded into a 950 gm stationary body. The loss in kinetic energy of the system will be:
A. 5%
B. 50%
C. 100%
D. 95%

Answer 1

In this question since the mass and the initial velocity of the bullet is given so by law of conservation of momentum we will find the final velocity of the bullet and then we will find the change in kinetic energy of the system. By the law of conservation of the momentum of a body, initial momentum of the body is equal to its final momentum.

Complete step by step answer:
The mass of the bullet $m = 50gm = 50 \times {10^{ - 3}}kg$
Initial velocity of the bullet ${v_i} = 10\dfrac{m}{s}$
Mass of the stationary body $M = 950gm = 950 \times {10^{ - 3}}kg$
We know the momentum of the body of mass m and moving with a velocity v is given by the formula $P = mv$.
So the initial momentum of the bullet of mass $m = 50gm$and moving with a velocity ${v_i} = 10\dfrac{m}{s}$ will be equal to ${P_i} = m{v_i} = \left( {50 \times {{10}^{ - 3}}} \right) \times 10 = 0.5kg\dfrac{m}{s} - - (i)$
After the bullet is fired it embed a stationary body of $M = 950gm$ so the final momentum of the bullet becomes

$${P_f} = m{v_f} \\\ = \left( {50 + 950} \right) \times {10^{ - 3}} \times {V_f} \\\ = {V_f}kg\dfrac{m}{s} - - (ii) \\\$$

Where, m is the total mass of the system and ${V_f}$ is the final velocity.
Now by conservation of momentum we can write${P_i} = {P_f}$, hence by substituting the values of momentum from the equation (i) and (ii) we can write

$${P_i} = {P_f} - - (iii) \\\ {V_f} = 0.5\dfrac{m}{s} \\\$$

Therefore the final velocity of the bullet becomes${V_f} = 0.5\dfrac{m}{s}$
Now since the velocity of the bullet is changes when the bullet embed the stationary body hence so the change in kinetic energy will be

$$\Delta K.E = K.{E_i} - K.{E_f} \\\ = \dfrac{1}{2} \times m \times V_i^2 - \dfrac{1}{2} \times m \times V_f^2 - - (iv) \\\$$

Hence by substituting the values we get the change in kinetic energy

$$\Delta K.E = \dfrac{1}{2} \times m \times V_i^2 - \dfrac{1}{2} \times m \times V_f^2 \\\ = \dfrac{1}{2}m\left( {V_i^2 - V_f^2} \right) \\\ = \dfrac{1}{2} \times 50 \times {10^{ - 3}} \times \left( {{{10}^2} - {{0.5}^2}} \right) \\\ = 25 \times {10^{ - 3}} \times \left( {100 - 0.25} \right) \\\ = 25 \times {10^{ - 3}} \times 99.75 \\\ = 2.5J \\\$$

Therefore the change in kinetic energy $= 2.5J$
So the percentage change in kinetic energy will be

$$K.E\% = \dfrac{{\Delta K.E}}{{K.{E_i}}} \times 100 \\\ = \dfrac{{2.5J}}{{\dfrac{1}{2} \times \left( {50 \times {{10}^{ - 3}}} \right) \times {{10}^2}}} \times 100 \\\ = \dfrac{{2.5J}}{{\left( {25 \times {{10}^{ - 3}}} \right)}} \\\ = 100\% \\\$$

Hence the loss in kinetic energy of the system $= 100\%$

So, the correct answer is “Option C”.

Note:
Students must note that for a collision occurring between two objects in an isolated system the total momentum of two objects before the collision is equal to the total momentum of two objects after collision. While in a non-isolated system, the momentum does not hold any specific property.