An object of mass 40 kg having velocity $4\hat{\mathrm{i}}$ m/s collides with another object of mass 40 kg having velocity $3\hat{\mathrm{j}}$ m/s. If the collision is perfectly inelastic, then the loss of mechanical energy in (in SI unit)

The momentum is conserved in both the directions.

So after collision the objects stick together and move with a combined velocity

By applying conservation of linear momentum in x-direction

$40\mathrm{x}4\hat{\mathrm{i}}+40\mathrm{x}0=80{\mathrm{v}}_{\mathrm{x}}\Rightarrow {\mathrm{v}}_{\mathrm{x}}=2\hat{\mathrm{i}}$

$40\mathrm{x}0+40\mathrm{x}3\hat{\mathrm{j}}=80{\mathrm{v}}_{\mathrm{y}}\Rightarrow {\mathrm{v}}_{\mathrm{y}}=1.5\hat{\mathrm{j}}$

So the final kinetic energy of the combined mass is given by

$\mathrm{K}.\mathrm{E}=\frac{1}{2}.80.(2^2+1.5^2)=250\mathrm{J}$

Initial kinetic energy of the objects is given by $\mathrm{K}.{\mathrm{E}}_1+\mathrm{K}.{\mathrm{E}}_2=\frac{1}{2}40.{\left(4\right)}^2+\frac{1}{2}40{\left(3\right)}^2=500\mathrm{J}$

So the loss in Mechanical energy= 500J -250J