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dorkymichelle
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Homework Statement
In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.18 kg) and embeds itself in block 2 (mass 1.85 kg). The blocks end up with speeds v1 = 0.540 m/s and v2 = 1.47 m/s (Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.
Homework Equations
P(momentum)=mv
Variables:
mass of bullet = m
mass of block 1 = m1
mass of block 2 = m2
initial speed of bullet before entering = Vi
speed of block 1= v1
speed of block 2 = v2
speed bullet has after leaving block 1 = vf
The Attempt at a Solution
momentum is always conserved in a system so
mVi = mvf+m1v1 < --even after the bullet leaves the system, both the block and the bullet's momentum would be changed but still should add up to the starting momentum
Vi=(mvf+m1v1)/m
mVf = (m+m2)V2 <--because the block and bullet stick together
vf=(m+m2)V2/m
putting in numbers,
bullet's mass is 3.50g=0.0035kg
Vf = (0.0035+1.85/0.0035)*0.0035 = 778.47m/s
Vi=0.0035*778.47 + 1.18*0.540 /0.0035 = 960.52m/s
what did i miss here...