Respuesta :
Given that the mass of clown 1 is m1 = 55.5 kg
Mass of clown 2 is m2 = 72.5 kg
The initial velocity of clown 1 is
[tex]v_i1\text{ = 2.5 m/s}[/tex]The initial velocity of clown 2 is zero as it is at rest.
[tex]v_i2\text{ = 0 m/s}[/tex]After the collision, the mass of the clown will be added up as they are stuck together.
So, the mass after the collision will be
[tex]m1+m2\text{ = 128 }kg[/tex]According to conservation of momentum,
[tex]m1v_i1+m2v_i2=(m1+m2)v_f[/tex]Here, v_f is the final velocity that is after the collision.
Substituting the values, the final velocity will be
[tex]\begin{gathered} v_f=\frac{m1v_i1+m2v_i2}{\mleft(m1+m2\mright)} \\ =\frac{55.5\times2.5+72.5\times0}{128} \\ =1.08\text{ m/s} \end{gathered}[/tex]Kinetic energy lost during this process will be
[tex]\begin{gathered} \Delta K.E.\text{ = }K.E._{f\text{inal}}-K.E._{\text{initial}} \\ =\frac{1}{2}(m1+m2)(v_f)^2-\frac{1}{2}m1(v_i1)^2 \\ =\frac{1}{2}\times128\times(1.08)^2-\frac{1}{2}\times55.5\times(2.5)^2 \\ =\text{ 74.65-172.5} \\ =-97.85\text{ J} \end{gathered}[/tex]Here, the negative sign indicates the loss in kinetic energy.
Thus, 97.85 J kinetic energy is lost during this process.
