ok heres how i' did it, but i don't get your answer. but i still think i'm right :S
1) find friction of surface acting against movement of clock
2) use this to find out initial speed
3) apply conservation of momentum to find inital speed of bullet
F=mu R
R of clock= 1.2g
f=0.2 x 1.2g= 0.24g
f=ma
0.24g= 1.2a
a= 0.2g
use constant acceleration formula. you now know...
a= -0.2g (against movement)
v=0 (stops)
s= 0.23
and you want u
so use v^2=u^2+2as
0=u^2 -0.9016
u=0.9495
inital speed of clock slipping
now conservation of mometum states:
m(1)u(1)+m(2)u(2)=m(1)v(1)+m(2)v(2)
but the bullet coalesces with clock so...
0.05u+ 0= 1.25 x 0.9495
0.05u=1.1869
u=23.738m/s
inital speed of bullet.
If you think i'm wrong fair enough, but looking at the question... i think that the clock would slide more than 0.23m if it was shot.
and theres another way of doing it...which should give the same answer...
if you find the inital speed of the clock= 0.9495m/s
then you can say that the kinetic energy of the bullet must be equal to:
Ke bullet= k.e clock + workdone against friction
1/2 mv^2 so...
0.025v^2=( 0.625x0.9495^2) + (0.24g x 0.23)
which gives v=6.647
i guess these are different becuase you don't have the info on the recoil speed of the gun and shooter but surprised that they are sooo different. infers theres extra factors not shown in the question... or that my maths is not up to scratch. one of the two.
well i tried mate. maybe the question is just wrong?
