Momentum and Projectile Motion

  • #1

Homework Statement


Can anyone help me with this problem?

A bullet of mass m is moving horizontally with speed Vo when it hits and embeds in a block of mass 100m that is at rest on a horizontal frictionless table (in the diagram it isn't placed at the edge of the table, i don't know if this matters). The surface of the table is a height h above the floor. After the impact the bullet and the block slide off the table and hit the floor a distance x from the edge of the table.

Derive expressions for the following quantities in terms of m, h, Vo and appropriate constants.
a.) the speed of the block as it leaves the table
b.) the change in kinetic energy of the bullet-block system during the impact
c.) the distance x

Suppose that the bullet passes through the block instead of remaining in it.
d.) state whether the time it takes the block to reach the floor from the edge of the table would now be greater, less, or the same
e.) state whether the distance x would now be greater, less, or the same


The Attempt at a Solution


i don't think i solved these right at all...

a.) mVo = (m+100m)Vf
Vf = (mVo)/(m+100m)

b.) KEf - KEi = change in KE
1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE

c.) x = Vit + 1/2at^2
x = Vft + 0

i don't know how to figure these out...
d.) would the amount of time increse becasue it has less mass??
e.) would x increase because it has more time to fall??

Thanks for any help!!
 

Answers and Replies

  • #2
Doc Al
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a.) mVo = (m+100m)Vf
Vf = (mVo)/(m+100m)
OK, but simplify that answer.

b.) KEf - KEi = change in KE
1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
That's the first step. Keep going. (Don't leave your answer in terms of Vf.)

c.) x = Vit + 1/2at^2
x = Vft + 0
Well? Solve for the horizontal distance. (Please use different letters to represent horizontal and vertical motion.)

You may as well solve these before worrying about d and e. But the hint for those two is: Does the block end up going faster or slower than in the earlier case.
 
  • #3
Is this right??
a.) Vf = (mVo)/(m+100m)
(mVo)/m(1+100)
Vo/101

b.) 1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
1/2(m+100m)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE
0 = change in KE
can that be right?
 
  • #4
arildno
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Is [itex]\frac{101}{(101)^{2}}=1??[/itex]
 
  • #6
arildno
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So why did you state that it was, when going from the 7th line to your 8th line in your previous post?

Aside from that silly little mistake, your work looks good! :smile:
 
  • #7
1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE

so you can't cancel the 101?
 
  • #8
Doc Al
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Is this right??
a.) Vf = (mVo)/(m+100m)
(mVo)/m(1+100)
Vo/101
Good.
b.) 1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
1/2(m+100m)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE
0 = change in KE
can that be right?
You made a mistake simplifying the term that I bolded in red. (As arildno has pointed out.)
 
  • #9
Doc Al
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so you can't cancel the 101?

You can cancel one of them. (Don't forget the square!)
 
  • #10
oh, i see.
so it would be 1/2m(Vo^2/101) - 1/2mVo^2
 
  • #11
Doc Al
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oh, i see.
so it would be 1/2m(Vo^2/101) - 1/2mVo^2

Good. Now simplify it.
 
  • #12
1/2mVo^2(-100/101) = change in KE
 
  • #13
for c.) can't i just use Vf=(Vo/101)
x=(Vf)t
x=(Vo/101)t
 
  • #14
Doc Al
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1/2mVo^2(-100/101) = change in KE
Simplify it further.

for c.) can't i just use Vf=(Vo/101)
x=(Vf)t
x=(Vo/101)t
Nothing wrong with that as a first step, but you can't stop there. (You can't just leave an unknown "t" in your answer.) Now write the equation for the vertical motion and solve both equations simultaneously.
 
  • #15
y = Vt + 1/2at^2
h = 0 + 1/2(-10)t^2
t = square root of (h/-5)

x = (Vo/101)(square root of (h/-5))
 
  • #16
Doc Al
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y = Vt + 1/2at^2
h = 0 + 1/2(-10)t^2
t = square root of (h/-5)

x = (Vo/101)(square root of (h/-5))

You realize that square roots of negative numbers presents a problem, right?

If you take the acceleration as a = -10 m/s^2; then the final height is y = -h (since it falls a distance h).
 
  • #17
Thanks for all the help!
I appriciate it!
 
  • #18
I have been thinking about parts d and e, and this is what I have decided:

d.) y = 1/2at^2
-h = 1/2(-10)t^2
t = square root of (h/5)
Since the height doesn't change, the time it takes for the block to fall would stay the same.

e.) Since momentum had to be conserved, the velocity of the block would be less if the bullet went through the block.
x = Vt
t = square root of (h/5)
That would mean that the distance would have to be smaller.

Is this correct?
 
  • #19
Hootenanny
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Please ignore sherlock, she is simply a secretary practising her typing and p#ssing everyone off :grumpy:
 
  • #20
Moonbear
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Thanks, Hootenanny, for pointing that out so pinkpolkadots would not be further confused.
 
  • #21
Hootenanny
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Thanks, Hootenanny, for pointing that out so pinkpolkadots would not be further confused.
No problem, she's been doing the same in a few other forums for an hour or so.

pinkpolkadots: I haven't had a detailed look at your problem, but both your answers look good to me (provided x is the distance the block lands from the table).
 
  • #22
yes, x was the distance the block falls from the table.

thank you!
 
  • #24
Doc Al
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d.) y = 1/2at^2
-h = 1/2(-10)t^2
t = square root of (h/5)
Since the height doesn't change, the time it takes for the block to fall would stay the same.
Good thinking!

e.) Since momentum had to be conserved, the velocity of the block would be less if the bullet went through the block.
You are correct, but can you explain your reasoning quantitatively? Hint: If the bullet passes through the block, what must be true about the bullet's final speed compared to the block's final speed?
 

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