Question

An 8.00-g bullet is fired into a 250-g block that is initially
at rest at the edge of a table of height 1.00 m (Fig. P6.30).
The bullet remains in the block, and after the impact the
block lands 2.00 m from the bottom of the table. Determine
the initial speed of the bullet.

Answer 1

I've been thinking

after colission we have:

\[8.0g*v _{0} +250g *0 =(8+250)v _{f} ?\]

Answer 2

@PsiSquared could you help me out ? please.

Answer 3

also I've been thinking in PE = KE to calculate when it lands or sort of

Answer 4

That's correct so far. You can also calculate the final velocity. How? First consider how you might calculate the block's time of flight. You know the block's initial vertical and horizontal speed, so how might you calculate that time of flight?

Answer 5

You can work the problem from an energy point of view, but you'll still need to find vf_horizontal. For that you need to find time.

Answer 6

but i have no magnitude in |v| i mean, I was thinking of Vx = |v|cos (a) for the horizontal . .

Answer 7

PE(block) = .25Kg * 9.8 m/s*s *1m

Answer 8

Try this: we know the block's initial speed vertically and horizontally is zero. We also know how far it has to fall, its initial height, and its acceleration when it falls. Those bits of information sound a lot like what is included in this equation:\[y=y _{0}+v _{0}t-0.5g t ^{2}\]where y0 is the initial height; v0 is the initial vertical velocity; t is time; y is the final height; and g is the acceleration of gravity.

Answer 9

Once you have time, you have a way of finding vf because you'll know the horizontal distance the block travels and the time it takes to travel that distance.

Answer 10

0= 1m + 0 - 4.4t*t --> \[t =\sqrt{\frac{ 1m }{ 4.4\frac{ m }{ s ^{2} } }}\]

Answer 11

I guess, now : v ( of the block) = 2m/t( some fraction)

Answer 12

Note that half of g is not 4.4 m/s^2. It's 4.9 m/s^2. You've got that part of the problem set up correctly, though.

Answer 13

Correct on v_block.

Answer 14

oh i see I'm sorry

Answer 15

so with that in mind, how is this related with the PE and KE OR I just have to replace by Vf ?

Answer 16

Now that you have vF horizontal, you can use either the conservation of momentum or the conservation of energy to find v0.

Answer 17

you mean : \[mgh = \frac{ 1 }{ 2 }mv ^{2}\]

\[v =\sqrt{2gh}\]

Answer 18

?

Answer 19

No, the very first equation you wrote:\[m _{b}v _{b}=\left( m _{b}+m _{B} \right)v _{Bb}\]where b stands for bullet; B stands for block; and Bb stands for combination of the bullet and block. vf_horizontal is vBb.

Answer 20

lol i just had to replace v

Answer 21

\[V _{0} = \frac{ 0.33Kg*Vh }{ 0,008Kg }\]

Answer 22

I guess that is the expression for the answer

Answer 23

Be careful with units. 8g+250g=258g=0.258Kg.

Answer 24

Otherwise your expression is correct.

Answer 25

ok i see it is cause I was thinking only in the variables, I did not remember well the values, thanks bro for your help !

Answer 26

I will set up well and then find the exact value.

Answer 27

but one last question, how would it be with energy conservation method ?

Answer 28

You're welcome.

Answer 29

could you show me

Answer 30

Using the conservation of energy it would be:\[\frac{ 1 }{2 }m _{b}v _{b}^{2}=\frac{ 1 }{ 2}\left( m _{b}+m _{B} \right)v _{Bb}^{2}\]The answer will ultimately be the same.

Answer 31

ok i get it, awesome, till the next one :)