Inelastic Collision Physics Problems In One Dimension - Conservation of Momentum
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- Опубликовано: 5 сен 2024
- This physics video tutorial Explains how to solve inelastic collision problems in one dimension using the law of conservation of linear momentum. In an inelastic collision, momentum is conserved but kinetic energy is not conserved. This video explains how to calculate the final speed of two objects sticking together and how to calculate the final speed of a bullet embedded in a wooden block as well as the maximum height attained by the bullet-block system. You need to use the law of conservation of energy to solve the 2nd part of the second problem where kinetic energy is converted to potential energy.
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I really like the way you break everything down. Sometimes, you just need to know the practical calculation first before really getting into the theory... at least for me. I get more confused if it's the other way around
SAME. I need an actual practical of the theory right away I can’t understand the theory by just it’s definition, linking it to reality to see the bigger picture helps me.
I am currently studying a course in mechanisc. These videos are near identical to the problems I've been trying to solve. We don't get much in the way of material so we got to teach ourselves. Thank you "The Organic Chemistry Tutor" you are awesome!
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In the last example, why was your m1 in the final momentum the mass of the block? While the m1 in your initial momentum is the mass of the bullet?
Same question😂
Question on the number 3
M1 is the mass of the bullet
V1 is the speed of the bullet
So why the m1v1' turn out into the
Mass and the speed of the block and the m2v2' turn out mass and speed of the bullet?
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Oh wow, he explained it pretty well
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How do you find the distance of two cars that collided together? You have the mass and the speeds of both cars
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Your videos have helped me a lot this semester, have you done a problem with an inelastic collision in one dimension on a incline. for example a box on a 7/12 pitch roof and a squirrel running jumps on it and it slides down the roof. How do you figure out where it will land on the ground.
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for question 3 isnt m1 for the bullet and m2 for the block ? ,, do we interchange it for after the collision ?
Thank you for this! I'm really having a hard time answering 1 of my homework and I need a reference to know how to solve this kind of problem and luckily it's in this video! again I wanna say thank you for this your videos are so helpful for someone who suck at Physics
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In the 2nd problem shouldn't it be the change in kinetic energy which is equal to the gain in gpe
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How do you find the initial velocity of the bullet before the collision when given the mass of the bullet and the mass of the block, the height of a table that the block is resting on and the distance the bullet and block travel after the collision?
4:30 but isn’t the height dependent on the length of the string? For example, if the length of the string was 1000m it will rise less than if it was 10cm.
I think the initial high would be assumed to be at level 0, so went it is hit; and rises, then the high is positive.
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8:04 are they not different masses? great video.
I thought the same thing. I thought the the ke was of the bullet
@@raynier250599it's the bullet-block system so both their masses, but he should've divided it by 9.81for gravity for a more accurate answer
Thanks, this really helped me
tysm you helped a lot
4:10 Who are you trying to kill with a 15kg bullet (But all jokes aside this man is a legend).
He said 15g.... whts wrong with you?
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These videos are great but I do have one small suggestion. If you edited out some of the gaps between speech it would make it a little easier to listen to. I tried playing the video at a higher speed but the problem is not the tempo of the actual speech itself the problem is the amount of time in between. I find it difficult to pay attention.
Thank you once again.
In the 1st part of 2nd question why he didnt put the value of M2 in the formula of inelastic collision but a 0 WHY?please tell me
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What if two objects moving in different directions collide and stick together, how do you calculate that?
plz make a video on physical optics & Coordination chemistry
thanks for the help appreciate it!
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please help me with this one
A bullet mass of 0.005kg is fired with horizontally into a wooden block of mass of 1.2kg which is resting on a horizontal surface.The cofficient of kinetic energy is between block and the surface is 0.20 .The bullet remains embedded in the block and the block is observed to have moved 0.230 m along the surface before stopping.Calculate intial velocity of the bullet
Thanks for the help :).
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Where does the lost kinetic energy go? Everyone says it becomes other kinds of energy such as thermal but I think that it is also lost in chaging the objects' momentum.
it goes in the form of friction, heat, sound and so on, I believe.
In elastic collisions KE and momentum are conserved so it doesn't go into changing momentum
@@manishniroula313 It also slightly deforms the objects involved
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Hey mn the masses are different not the same because kinetic energy is only related to the bullet and potential energy is related to both the bullet and the block
best teacher
This is a good video, however I would suggest an improvement: Solve equations algebraically before plugging in values.
thanks man
If a 10kg mass object with a speed 27.77m/s and an opposit object with the same mass and speed what will happen after collision ???
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how m2v2 is 0?
thx for the help my teacher is wack
Could someone please answer this for me I’ve tried solving tons of times but I keep getting confused
Q: A 5g bullet is fired horizontally and hits an 8kg block of wood (initially at rest) which can move freely. The wood and the bullet move with a velocity of 0.50 m/s after impact. What is the initial velocity of the bullet?
Assuming an inelastic collision, m1v1+m2v2 (momentum of bullet and momentum of block) = (m1+m2)v3 (momentum of the bullet block system).
Plugging in all values we get 0.005*v1 + 8*0 = (8.005)(0.5)
And get a final result of v1 = 800.5 m/s
We can use conservation of momentum here which states that m1u1+m2u2=m1v1+m2v2
So initial momentum of the block is 0 as it asrest as mentioned
Since they are moving together 0.05u1= 0.05+8*0.50 which gives 8.05m/s
thanks man 2021
so im a bit confused, why did you 1.5 as final velocity for mass 1 and 0.025 for mass 2, at 11:14
1.5 is the mass of the box and 0.025 is the mass of the bullet
@@meganrivera7653 Yes, but the masses are switched at the end. He uses M1 as 0.025 and then uses it for M2. Which is confusing because that would mean V1 initial equals V2 final.
Make the velocity of block 2 to be 20 m/s. Show the math that illustrates the problem with this.
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Legend
but why is organic chemistry teaching mechanics?
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2:49 Wrong! True equation must be (m1 + m2) * Vf.
That is incorrect!
A system is made up of three bodies with respective velocities as follows: body A of mass 1.5 kg moving at 2.0 m/s, north; body B of mass 2.0 kg… Can you please make a solution for this please
2 kg moving at 3 m/s; and C of mass 5.2 kg moving at 2.5 m/s.
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