# Please help me, this problem is killing me

A 20 g bullet is fired horizontally into a .5 kg wooden block, resting on a horizontal surface ( kinetic friction coefficient is .2), THe bullet goes through the block and comes out the other side with a speed of 200m/s... if the block travels 1 m before coming to rest, what was the Vi of the bullet.... (ignore work of gravity on the bullet....ps, this is a momentum problem i think, i just cant get it, i would GREATLY appriciate any help on this.... i have the answer but i need to work to get credit..... this is what i have so far

i tried using Wnet = change in KE

Fx - .2(.52)(1m)(9.8) = x - 1.02 J but i have no idea what to do next, i think it is a perfectly inelastic collision during the one meter, but than the bullet come out!!?!! how the heck do you account for that? again any help is appriaciated,,, thanks for y our time

## Answers and Replies

i'm not sure about this... (not really sure about wat kinetic coefficient is) but what i tink u can do is since there is kinetic friction .2 .. let's take as if u'r block is movin so use $$KE = 1/2 m v^2$$ to find u'r 'velocity' of u'r block/.. After doin that, u can use $$M(1) U(1) + M(2) U(2) = M(1) V(1) + M(2) V(2)$$ to find u'r initial speed.. i hope so.. note: u'r V(2)= 0... oso dunnoe y..

May not be rite.. though APOLOGIES! Juz try it.. hope u get u'r answer Last edited:
Friction force = coefficient of friction * normal force.
thus friction force is 0.5*9.8*0.2=0.98 N
this acts through 1 m, so the work done by friction is
work done = F*s = 0.98 * 1 = 0.98 J.
this work done by friction is absorbing the kinetic energy of the block initially,
so the initial kinetic energy of the block is 0.98 J
thus 0.98 = 0.5*m*v^2.
so the velocity of the block just after impact is 1.980 m/s.

now, apply momentum equation,
momentum initial = momentum final
= 0.5*1.980 + 0.020 * 200 = 4.99 kg m/s.

initial momentum of the bullet is 4.99 kg m/s also, so
the speed of the bullet is 4.99/0.02 = 249 m/s.

the BIG assumption we adopt here is the external friction is negligible during the interaction process which happens very-very fast, otherwise momentum is not conserved.

maybe that helps.

Sniffer (poor physicist)

thank a bunch, thats exactly the answer that was given, thanks again.....