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

$V = \frac{mv}{M}$

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.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

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

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$