Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 5.10 m before stopping. How far does the lighter fragment slide?

Answer 1

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

`MV=mv`

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

`V = (mv)/(M)`

Also we have that
`m/M = 1/7 times`

On the other hand we have from law of conservation of energy that

`W_f = KE`

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

`F_f*S = (1)/(2)MV^2`

`\mu M*g*S = (1)/(2)MV^2`

`\mu g*S = (1)/(2)( (mv)/(M))^2`

`\mu = (1)/(2) ((1)/(7)v)^2`

`\mu = (1)/(98)v^2`

`\mu = (1)/(g(98)(5.1))v^2`

Here we can apply the law of conservation of energy for light mass, then

`\mu mgs = (1)/(2) mv^2`

Replacing the value of
`\mu`

`(1)/(g(98)(5.1))v^2  mgs = (1)/(2)mv^2`

Deleting constants,

`s= ((98*5.1))/(2)`

`s = 249.9m`