Linear momentum - Bullet fired vertically

[Kmol6]

Homework Statement

A bullet is fired vertically into a 1.40 kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it?

Homework Equations

1. m1v1 +m2v2 = mfvf
2. x=xo +vot +1/2at²
3. Quadratic formula

The Attempt at a Solution

1.
(.029kg)(510m/s) + 0 = (1.42)(vf)
10.4 m/s = vf

2.
x=0 +10.4(t) + 1/2(9.8)t²
x= 10.4t + 4.9t²

3.
t= -10.4+ {(Sqrt) ( 10.4^2-4(1)(4.9)}/ 2(1)

t= -10.4+ 9.42/2 ( using -10.4- 9.42/2 = -9.91s which doesn't seem realistic and gives a final answer of +500m)
t= -.49 time cannot be negative.
t= .49 s

2.
.49 and plug it back into
x=10.4t+ 4.9t^2
x= 6.27meters.

I'm given the answer as 5.47 m so I know I did something wrong or I used the wrong equations, however this question is in the Linear momentum section and the momentum is vertical? I don't know, please help :)

 

[Physics-Tutor]

1st part (Conservation of momentum): OK
2nd part: Using conservation of energy would be a more straightforward approach...

However if you still wish to use kinematics, try another equation that does not involve time. The one I am thinking of will give you the same expression than if you used conservation of energy.

 

[Kmol6]

ok, If I use v^2 = vo^2 = 2a (x-x0) I think this is the one you're thinking of? ( a lot less work :P)
10.4^2=2(9.8)x
x=5.52m which is close, but still not exact even with rounding?
or I use
conservation of energy and I get
1/2(.029)(510)=1/2(1.42)(10.4)^2 + (1.42)(9.8)h
x= 4.98 m

 

[Physics-Tutor]

The differences you find are due to your choices of significant figures.
1st, be consistent. if you want 3 sf, use 9.81 m/s² for g and include the mass of the bullet in your first step (mass of block + mass of bullet = 1.429 Kg) .
2nd, use an extra significant figure for intermediary results (here your speed) to avoid any unwanted intermediary rounding. Here, You used 10.4 for your speed, i.e. a number starting with a 1 (the last digit carries an error of 1%!), You cannot expect to get the final result with less than 1% error! Use 4 sf for your speed.

Be careful when applying standard rules concerning sf: the third digit of a number does not carry the same weight when the first digit is a 1 or a 9, so take this into account in intermediary calculations.

 

[PeroK]

ok, If I use v^2 = vo^2 = 2a (x-x0) I think this is the one you're thinking of? ( a lot less work :P)
10.4^2=2(9.8)x
x=5.52m which is close, but still not exact even with rounding?
or I use
conservation of energy and I get
1/2(.029)(510)=1/2(1.42)(10.4)^2 + (1.42)(9.8)h
x= 4.98 m

You can't use conservation of energy for the whole problem, because you have an inelastic collision where energy is not conserved. You could, however, use conservation of energy for the upward motion after the bullet is embedded in the block.

One of the problems you have is that you follow the "plug the numbers in as soon as possible" method. And, by his method, you lose sight of what is happening. Also, you are then prone to exaggerated rounding errors.

You could try to solve the problem in two steps:

1) First show that $v = (\frac{m}{m+M})u$

Where $u$ is the speed of the bullet, $m$ the mass of the bullet and $M$ the mass of the block and $v$ is the speed of the block and bullet after impact.

2) Then you could show that $h = \frac{v^2}{2g}$. where $h$ is the height reached by the block and bullet (or by anything) shot upwards with speed $v$.

3) Then you could put the two together to get:

$h = (\frac{m}{m+M})^2 \frac{u^2}{2g}$

Now you have only one calculation to do and no intermediate rounding.

The advantages of this approach are many. Not least, you may begin to see that you are often solving the same physics problems over and over again, only with different numbers each time.

For example, $\frac{m}{m+M}$ comes up all the time in physics and you get to recognise it. But, if you just plug in the numbers, you get different numbers each time and you take very little learning from this problem into the next.

Finally, if the speed of the bullet was actually $550m/s$ then you'd have to do the whole problem over again from the start. Whereas, I could just repeat the last calculation using the formula for $h$, knowing that is valid for any bullet and block with any mass and any speed.

 

[Kmol6]

I definitely try and plug the numbers in as soon as I can, and I avoid combining equations. I don't trust my physics / math skills enough yet, it's been a decade since I've done any of this.

Thank you for the help and advice! I got the right number.