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…

Question

asked Mar 10, 2020 196k views

1 Answer

Waleed Muaz · Answer 1 · 2020-03-16T19:40:55+0000

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