PHysics
posted by Anne on .
a 15g bullet strikes and becomes embedded in a 1.1kg block of wood placed on a surface in front of the gun. if the coefficient of kinetic friction between the block and the surface is 0.25 and the impact drives the block a distance of 95m before it comes to rest, what was the muzzle speed of the bullet?
What i did was F = umg = 2.73 = ma. so a = 2.45.
then i used Vf^2 = Vo^2 + 2a(deltaX) so Vo = 6.8.
then M1V1 = (M1+M2)(V1)'
But my V1 is 507 m/s while the answer is 515 m/s.
Where did i go wrong?

m1v1=(m1+m2)v
but 1/2 (m1+m2)v^2=u(m1+m2)gd
v=sqrt2ugd
v1=(m1+m2)sqrt2ugd /m1
= (1.115)/.015 sqrt(2*.25*9.8*95)
I don't get either of those answers.
Your Vo is wrongly computed.
vo=sqrt(2ad)=sqrt(2*2.45*95)=21.6m/s