Object X of mass 4 kg travels with a speed of 3 ms toward object Y of mass 2 kg that is initially at rest. Object X then collides with and sticks to object Y . After the collision, object X and object Y remain stuck together. How much mechanical energy is converted into nonmechanical energy during the collision?

The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.

It is given in the question that

Mass of the body X is [tex]M_X=4\ kg[/tex]

The velocity of the body X is =[tex]V_x=3 \ \frac{m}{s}[/tex]

Mass of the body Y is [tex]M_Y= 2\ kg[/tex]

The velocity of body Y is [tex]V_y= 0[/tex]

Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.

Now from the conservation of the momentum

[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]

[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]

[tex]V_F= \dfrac{12}{6}[/tex]

[tex]V_F = 2 \frac{m}{s}[/tex]

Now to find the Change in the energy of the body

[tex]\Delta E=E_i-E_f[/tex]

[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]

[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]

[tex]\Delta E= 6\ J[/tex]

Thus the nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]

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