[ /bananas/ / /boku/ / /dawa/ / /desu/ / /jum/ / /kashira/ / /md/ / /otousama/ / /ro/ / /unyuu/ / /yakult/ ] [ /a/ / /c/ / /h/ / /loli/ / /moonspeak/ / /nagato/ / /nij/ / /nipa/ / /touhou/ / /tr/ / /vn/ / /yan/ ] [ /do/ / /fi/ / /lit/ / /o/ / /pro/ / /tech/ / /tg/ / /v/ / /vic/ ] [ /arrrrr/ / /gar/ / /gif/ / /media/ / /ot/ / /r/ / /w/ ] [ /donate/ / /mod/ / /sandbox/ / /sugg/ ]

Hello there physicists, I have a question;A 0.50 kg wooden block is dropped from a height of 15.0 m. At what height must a 005.0 kg bullet traveling upward strike the block in order to momentarily stop its fall? Assume that the time required for the bullet to come to rest is in the block is negligibly short and that the bullet travels with a speed of 320 m/s immediately before striking the block.Does anybody know the solution?

Hello there physicists, I have a question;

A 0.50 kg wooden block is dropped from a height of 15.0 m. At what height must a 005.0 kg bullet traveling upward strike the block in order to momentarily stop its fall? Assume that the time required for the bullet to come to rest is in the block is negligibly short and that the bullet travels with a speed of 320 m/s immediately before striking the block.

Does anybody know the solution?

a 5kg bullet stopping a 0.5kg block?edit:I'm fudging your numbers here, but with a 5kg block and a 0.005 kg bullet the block would have to be 9.78 meters above the ground to momentarily come to a halt. Last edited 10/01/10(Sun)22:04.

a 5kg bullet stopping a 0.5kg block?

edit:I'm fudging your numbers here, but with a 5kg block and a 0.005 kg bullet the block would have to be 9.78 meters above the ground to momentarily come to a halt.

Last edited 10/01/10(Sun)22:04.

The bullet has a momentum of 1600 kg·m/s right before impact. Working under the assumption that the block's momentum needs to be the same in order for the bullet to stop its fall, it needs to be falling at 3200 m/s, which would occur after falling for ~326.53 s. So it would never happen if the block is dropped from 15 m.However, if the bullet is .5 kg and the block 5 kg, then you get a more realistic time of 3.26 s. But the block would have hit the ground in less than 2 seconds.It should be noted that I last took a physics class 2 years ago. So I may be a bit rusty.

The bullet has a momentum of 1600 kg·m/s right before impact. Working under the assumption that the block's momentum needs to be the same in order for the bullet to stop its fall, it needs to be falling at 3200 m/s, which would occur after falling for ~326.53 s. So it would never happen if the block is dropped from 15 m.

However, if the bullet is .5 kg and the block 5 kg, then you get a more realistic time of 3.26 s. But the block would have hit the ground in less than 2 seconds.

It should be noted that I last took a physics class 2 years ago. So I may be a bit rusty.

Now I realized I typed the numbers all wrong and yeah, so the wooden block is 5.0 kg and the bullet is 0.005 kg. What I'm confused about is how I set up the equation to get the answer because I tried using Einitial = Efinal. But I'm not sure what to do from there with finding the hfinal (h = height) of the bullet

>>10452looks like you got got bro

>>10452http://en.wikipedia.org/wiki/Equations_for_a_falling_bodyenjoyedit:tip. so you equated the kinetic energy and the variable you discovered was a number of seconds. now how much did it fall in those many seconds?would have liked the problem more if you only knew the exiting velocity of the bullet at ground level. Last edited 10/01/11(Mon)03:35.

>>10452http://en.wikipedia.org/wiki/Equations_for_a_falling_bodyenjoy

edit:tip. so you equated the kinetic energy and the variable you discovered was a number of seconds. now how much did it fall in those many seconds?

would have liked the problem more if you only knew the exiting velocity of the bullet at ground level.

Last edited 10/01/11(Mon)03:35.

It's been a while, but I know the kinematics involved. However, isn't this an inelastic collision, in which kinetic energy is not conserved?

also, http://www.brainycreatures.org/physics/collision.asp

>>10460Okay, but I'm still not seeing why you are apparently applying K_block = K_bullet. I am seeing this as an ideal inelastic collision, in whichm_bullet*v_bullet + m_block*v_block = (m_bullet + m_block) * 0(Both velocities are vectors.)This would mean thatm_bullet*mag(v_bullet) = m_block*mag(v_block)Thus, the speed of the block would be solved, divided by g for time (which would be very small), and placed into the equation h_initial - 0.5*g*t^2 for the height. I don't like the answer, though.Both bodies are instantaneously at rest after the collision, and the bullet is explicitly inside the block afterwards. Last edited 10/01/11(Mon)17:19.

>>10460

Okay, but I'm still not seeing why you are apparently applying K_block = K_bullet. I am seeing this as an ideal inelastic collision, in which

m_bullet*v_bullet + m_block*v_block = (m_bullet + m_block) * 0

(Both velocities are vectors.)

This would mean that

m_bullet*mag(v_bullet) = m_block*mag(v_block)

Thus, the speed of the block would be solved, divided by g for time (which would be very small), and placed into the equation h_initial - 0.5*g*t^2 for the height. I don't like the answer, though.

Both bodies are instantaneously at rest after the collision, and the bullet is explicitly inside the block afterwards.

Last edited 10/01/11(Mon)17:19.

Also note that I am not the OP.

>>10472what is your answer then?the reasoning behind using kinetic energy is simply because the vector direction are directly opposed and during the collision all kinetic energy is instantly transformed into 2 forces, both slowing down 1 object to a momentary halt. since they are not moving at that one point, both their kinetic energy must be 0.using the formulas for inelastic collisions and all that are, I believe, useful in any other colision where after the collision the 2 objects are not stopped but move in different directions and likely different speeds. Last edited 10/01/11(Mon)19:15.

>>10472what is your answer then?

the reasoning behind using kinetic energy is simply because the vector direction are directly opposed and during the collision all kinetic energy is instantly transformed into 2 forces, both slowing down 1 object to a momentary halt. since they are not moving at that one point, both their kinetic energy must be 0.

using the formulas for inelastic collisions and all that are, I believe, useful in any other colision where after the collision the 2 objects are not stopped but move in different directions and likely different speeds.

Last edited 10/01/11(Mon)19:15.

I don't like my answer because it does not seem right. I have 14.995 m.It is still mathematically possible, at least, for the formula associated with inelastic collisions for the final velocity of the two objects to be a zero vector. It is clear that we started out with more kinetic energy prior to impact than after, since before, we had 0.5*m_bullet*|v_bullet|^2 + 0.5*m_block*|m_block|^2 and afterwards, we have, as you said, 0. However, kinetic energies cannot "cancel out" the way forces and fields do. What occurred was a dissipation of kinetic energy as work deforming the block and heat emitted due to friction. I submit that this does not tell us what the relationship between the kinetic energy of each object is, but assuming a completely elastic collision (which this problem describes), we can set the right side of the respective momentum equation to a zero vector and proceed from there.But then, like I said, I think I went wrong somewhere with my solution. Last edited 10/01/12(Tue)02:30.

I don't like my answer because it does not seem right. I have 14.995 m.

It is still mathematically possible, at least, for the formula associated with inelastic collisions for the final velocity of the two objects to be a zero vector. It is clear that we started out with more kinetic energy prior to impact than after, since before, we had 0.5*m_bullet*|v_bullet|^2 + 0.5*m_block*|m_block|^2 and afterwards, we have, as you said, 0. However, kinetic energies cannot "cancel out" the way forces and fields do. What occurred was a dissipation of kinetic energy as work deforming the block and heat emitted due to friction. I submit that this does not tell us what the relationship between the kinetic energy of each object is, but assuming a completely elastic collision (which this problem describes), we can set the right side of the respective momentum equation to a zero vector and proceed from there.

But then, like I said, I think I went wrong somewhere with my solution.

Last edited 10/01/12(Tue)02:30.

>>10487I argue that the kinetic energy of the brick and the bullet have to be equal to one another or otherwise they wouldn't cancel each others motion at the collision.this leads me to believe that just before the moment of impact the bullet would have Ek=0.5*0.005*320^2=256J of energy. the brick needs to pick up speed while falling until it's fast enough to have equal kinetic energy.v=sqrt(256/(0.5*5))=10.12 m/s(rounded)v=gt so 10.12/9.81=1.03 s (rounded)after 1.03 seconds the brick goes fast enough.d=0.5gt^2 so d=0.5*9.81*1.03^3=5.20 meter (rounded)so, the height is around 15-5.20=9.80 meters Last edited 10/01/13(Wed)04:20.

>>10487I argue that the kinetic energy of the brick and the bullet have to be equal to one another or otherwise they wouldn't cancel each others motion at the collision.

this leads me to believe that just before the moment of impact the bullet would have Ek=0.5*0.005*320^2=256J of energy. the brick needs to pick up speed while falling until it's fast enough to have equal kinetic energy.

v=sqrt(256/(0.5*5))=10.12 m/s(rounded)v=gt so 10.12/9.81=1.03 s (rounded)

after 1.03 seconds the brick goes fast enough.d=0.5gt^2 so d=0.5*9.81*1.03^3=5.20 meter (rounded)

so, the height is around 15-5.20=9.80 meters

Last edited 10/01/13(Wed)04:20.

sup all

Forgot to submit my comment earlier.Momentum is always conserved in a system with no outside work done to it; kinetic energy, however, is not conserved in an inelastic collision.Why exactly, then, would the inelastic collision equation not apply?

Forgot to submit my comment earlier.

Momentum is always conserved in a system with no outside work done to it; kinetic energy, however, is not conserved in an inelastic collision.

Why exactly, then, would the inelastic collision equation not apply?

if it were an inelastic collision the bullet wouldn't dig inside the brick and get stuck there, but the 2 would bounce right off like 2 approaching snooker balls. consider newtons cradle: http://upload.wikimedia.org/wikipedia/commons/d/d3/Newtons_cradle_animation_book_2.gif)only in this case would momentum be preserved. a true completely inelastic collision however does not exist in real life.>>10512yo. Last edited 10/01/15(Fri)20:21.

if it were an inelastic collision the bullet wouldn't dig inside the brick and get stuck there, but the 2 would bounce right off like 2 approaching snooker balls. consider newtons cradle: http://upload.wikimedia.org/wikipedia/commons/d/d3/Newtons_cradle_animation_book_2.gif)only in this case would momentum be preserved. a true completely inelastic collision however does not exist in real life.

>>10512yo.

Last edited 10/01/15(Fri)20:21.

- wakarimasen beta + futaba + futallaby -