Answer:
The kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block
Explanation:
Kinetic energy of smaller object
[tex]K=(1/2)mv^2[/tex]
where m= mass of smaller object and v= velocity of smaller object
Also, it is given that heavier object is four times the mass of lighter object and consider its velocity as V
kinetic energy of heavier block [tex]K '= (1/2) (4m) V^2[/tex]
Now, For smaller block , [tex]v^2 - u^2=2aS[/tex]
[by Newtons laws of motion]
Also, [tex]v^2 = 2(F/m)S[/tex]
Where S= displacement, F= force, u= initial velocity
So, [tex]K=(1/2)m[2(F/m)]S[/tex]
[tex]\Rightarrow K = FS[/tex]
For heavier block ,
[tex]V^2 - u^2 = 2a'S[/tex]
or, [tex]V^2=2(F /4m)S[/tex]
So,[tex]K '= (1/2)(4m)[2(F/4m)S[/tex]
[tex]\Rightarrow K'= FS[/tex]
Therefore the kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block