Respuesta :
Answer:
Explanation:
Given
Two masses M and 2 M with velocity v in opposite direction
After collision they stick together
Initial momentum
[tex]P_i=Mv-2Mv[/tex]
final momentum
[tex]P_f=3Mv'[/tex]
Conserving momentum
[tex]P_i=P_f[/tex]
[tex]-Mv=3Mv'[/tex]
[tex]v'=\frac{v}{3}[/tex]
i.e. system moves towards the direction of 2M mass
Initial kinetic Energy [tex]K_1=\frac{1}{2}Mv^2+\frac{1}{2}2Mv^2[/tex]
[tex]K_1=\frac{3}{2}Mv^2[/tex]
Final Kinetic Energy [tex]K_2=\frac{1}{2}\cdot (3M)\cdot (\frac{v}{3})^2=\frac{3}{18}Mv^2[/tex]
loss of Energy[tex]=K_1-K_2[/tex]
[tex]=\frac{3}{2}Mv^2-\frac{3}{18}Mv^2[/tex]
[tex]=\frac{4}{3}Mv^2[/tex]
The mechanical energy is lost to other forms of energy during the collision is [tex]\bold { \dfrac 43 mv^2}[/tex].
Given here,
Mass of the first object = M
Mass of the second object = 2M
Thy are moving at same speed = v,
The initial momentum,
[tex]\bold {Pi = Mv - 2mv}[/tex]
The Final momentum,
Pf = 3 Mv'
From the conservation of momentum,
Pi = pf
Mv - 2Mv = 3mv'
[tex]\bold {v' = \dfrac v3}[/tex]
The final velocity is one-third of the initial speed.
Since, the system moves towards the direction second object,
The loss of energy = Ki - Kf
Where,
Ki - Initial kinetic energy,
[tex]\bold {Ki = \dfrac 12 mv^2 + \dfrac 12 mv^2 = \dfrac 32 mv^2}[/tex]
Kf = final kinetic energy
[tex]\bold {Kf = \dfrac 12 (3m)(\dfrac v3)^2 = \dfrac {3}{18 } mv^2}[/tex]
Thus,
[tex]\bold {\Delta E = \dfrac 32 mv^2 - \dfrac 3{18} mv^2 }\\\\\bold {\Delta E = \dfrac 43 mv^2}[/tex]
Therefore, the mechanical energy is lost to other forms of energy during the collision is [tex]\bold { \dfrac 43 mv^2}[/tex].
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