Momentum problem of blocks and bullet

[dorkymichelle]

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...

 

[ehild]

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

that is correct, but what do you mean by the following lines?

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

ehild

 

[dorkymichelle]

The numbers is where I did the calculations, i submitted them for hw, it's online and it said they were wrong

 

[ehild]

The numerical expressions you wrote for vi and vf are entirely wrong. The results are correct, but you need to round them to three significant figures.

ehild

 

[dorkymichelle]

I'm sorry, I'm a bit confused, did I plug in the numbers into the equation wrong? When I entered the answer in, they were rounded to 778m/s and 960 m/s

 

[ehild]

"Vf = (0.0035+1.85/0.0035)*0.0035 = 778.47m/s" This is wrong.

Correctly: ((0.0035+1.85)/0.0035)*1.47 = 778.47m/s"Vi=0.0035*778.47 + 1.18*0.540 /0.0035 = 960.52m/s" It is wrong. Write
Vi=(0.0035*778.47 + 1.18*0.540) /0.0035 = 960.52m/s.

The results seem correct, I got the same. Are you sure you input them at
at the proper place? 960 m/s for the speed the bullet enters the first block and 778 m/s for the speed it leaves?

 

[dorkymichelle]

OOPS.. i did put it in the wrong place = /
thanks for your time!