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PHYS 207 Help

irishScott

Lifer
Oct 10, 2006
21,568
2
0
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 8.2 m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficeint of kinetic friction.

My prof wants us to use Conservation of Momentum/Impulse Momentum Theory to solve this. I've been crunching at it for 30 min now, and all I've got is that the final velocity of the smaller fragment is 7 times the velocity of the larger fragment.

Any help would be appreciated.
 

darthsidious

Senior member
Jul 13, 2005
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0
71
Assuming there is a constant force on the fragments (= mu*Normal force).

Your first step is correct: the initial velocity of the smaller fragment will be 7 times higher

use the formula v^2 = u^2 - 2*a*s , a = F/mass = (mu*mass*g)/mass = mu*g. s = distance travelled.

for the block that travels 8.2m and stops, final velocity (v1) is zero. use that to calculate u1. Once you know u1, u2 = 7*u1, and again final velocity is zero. Use this to calculate how long the smaller fragment travels before it comes to a stop.

 

irishScott

Lifer
Oct 10, 2006
21,568
2
0
Originally posted by: darthsidious
Assuming there is a constant force on the fragments (= mu*Normal force).

Your first step is correct: the initial velocity of the smaller fragment will be 7 times higher

use the formula v^2 = u^2 - 2*a*s , a = F/mass = (mu*mass*g)/mass = mu*g. s = distance travelled.

for the block that travels 8.2m and stops, final velocity (v1) is zero. use that to calculate u1. Once you know u1, u2 = 7*u1, and again final velocity is zero. Use this to calculate how long the smaller fragment travels before it comes to a stop.
Worked perfectly. Thanks. Knew there was something I was missing...
 

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