Question

An object of mass $40\,\,Kg$ having velocity $4\widehat i\,\,m{s^{ - 1}}$ collides with another objects of mass $40\,\,Kg$ having velocity $3\widehat i\,\,m{s^{ - 1}}$. If the collision is perfectly inelastic, then the loss of mechanical energy.
(A) $250\,\,J$
(B) $100\,\,J$
(C) $125\,\,J$
(D) $35\,\,J$

Answer 1

Hint:The given problem can be solved using the formula for the loss of mechanical energy in a perfectly inelastic collision. The formula is derived from the collision of inelastic objects and from the formula of mechanical energy.

Formulae Used:
The mechanical energy that is lost due to the inelastic collision is given by;
$\delta KE = \dfrac{1}{2}mv_{net}^2$
Where, $m$denotes the total mass of the ball and ${v_{net}}$denotes the net velocity of the ball.

Complete step-by-step solution:
The data give in the problem are;
Mass of the first ball, ${m_1} = 40\,\,Kg$.
Mass of the second ball, ${m_2} = 40\,\,Kg$.
velocity of the first ball, ${v_1} = 4\widehat i\,\,m{s^{ - 1}}$.
velocity of the second ball, ${v_2} = 3\widehat i\,\,m{s^{ - 1}}$.
The mechanical energy that is lost due to the inelastic collision is given by;
$\delta ME = \dfrac{1}{2}mv_{net}^2$
Where, $m = \left[ {\dfrac{{\left( {{m_1}{m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}} \right]$ and $v_{net}^2 = \left( {v_1^2 + v_2^2} \right)$
Substitute the values of $m$and $v_{net}^2$;
$\delta ME = \dfrac{1}{2} \times \left[ {\dfrac{{\left( {{m_1}{m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}} \right] \times \left( {v_1^2 + v_2^2} \right)$
Substitute the values of Mass of the first ball, ${m_1} = 40\,\,Kg$, Mass of the second ball, ${m_2} = 40\,\,Kg$, velocity of the first ball, ${v_1} = 4\widehat i\,\,m{s^{ - 1}}$, velocity of the second ball, ${v_2} = 3\widehat i\,\,m{s^{ - 1}}$.
$\delta ME = \dfrac{1}{2} \times \left[ {\dfrac{{\left( {40\,\,Kg \times 40\,\,Kg} \right)}}{{\left( {40\,\,Kg + 40\,\,Kg} \right)}}} \right] \times \left( {{4^2}\,\,m{s^{ - 1}} + {3^2}\,\,m{s^{ - 1}}} \right)$
Simplify the above equation;
$\delta ME = \dfrac{1}{2} \times \left[ {\dfrac{{1600}}{{80}}} \right] \times \left( {25} \right)$
$\delta ME = 250\,\,J$.
Therefore, the mechanical energy lost due to the inelastic collision is $\Delta ME = 250\,\,J$.
Hence the option (A) $250\,\,J$ is the correct answer.

Note:- In case of the inelastic collision the combined momentum of the two bodies continue the same, but some of the initial kinetic energy is changed into heat energy inside the bodies, which are used up in misshaping the bodies, or discharged away in some other method.