Finding Distance with Mechanical Energy Problem

[raisatantuico]

Homework Statement

A bullet of mass 0.0800 kg is fired onto an empty box of mass 0.920 kg initially at rest. Upon hitting the box, the bullet has a speed of V= 900.0 m/s and is oriented horizontally. After hitting the box, the bullet passes through it and emerges with a speed of 1/2V. The box also moves but due to the rough surface it is on, it stops after moving a distance of s. If the coefficient of kinetic friction is 0.500, find s.

Homework Equations

W other= ∆E (mechl energy) + ∆E therm

∆E = (K2+U2) - (K1+U1) + f∆x

The Attempt at a Solution

what i did is i used the Wother formula to solve for ∆x.. my answer was .10m.
this is wrong because the choices to the multiple choice question are: A. 3.99m B. 39.1 C. 72.0m D. 99.7 E156m

I think there is more to this problem then using the Wother formula.. do i need to use the conservation of momentum??

 

[Filip Larsen]

do i need to use the conservation of momentum??

Yes.

 

[raisatantuico]

i am confused on how to use the conservation of momentum. so initial momentum is when the bullet hasnt hit the box yet, and the final momentum is when the bullet has hit the box?
so m(bullet)xV(bullet) + m(box)xV(box)= (m bullet + m box)Vfinal

then, in solving for the s displacement of the box,we use the Wother formula, ∆E mech+∆E therm. kf+uf - (ki+ui) + F∆x=0 ?

all the potential energies are equal to zero, so that leaves:
kf- ki + F∆x = 0

what is kinetic energy final and kinetic energy initial? is kinetic energy final equal to the 2 masses times the final velocity of the bullet once it is out of the box (1/2v = 450 m/s) ? initial kinetic energy is when the bullet hasnt hit the box yet?

i am so confused.

 

[Filip Larsen]

You haven't quite got the correct velocities for your momentum equation. The velocity of the box before collision has a special value (hint: its not moving) and afterwards it may have a different velocity than the bullet. You know three of the velocities (and all the masses) so you should be able to solve your momentum equation to get the velocity of the box after collision.

 

[raisatantuico]

Thanks! ended up with 156m = ∆s. i hope this is right. from what i understand, using the work other formula, we just consider the box and surface system. we just use the intial and final velocities of the box.

 

[Filip Larsen]

I also get 156 m.

Note, that the box is under constant deceleration so it is possible to find distance s from velocity v and deceleration a using the standard kinematic relationship 2as = v². This equation will also emerge directly if you consider the mechanical energy of the box and set work done by the deceleration force equal to kinetic energy.