Inelastic Collision Physics Problems In One Dimension - Conservation of Momentum

[@TheOrganicChemistryTutor]

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[@thelonewolf9866]

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[@kenaldrich4575]

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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[@undergraduate6330]

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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I know this is a late comment.. but I find your videos helpful! however, my homework usually asks these questions by giving only values for mass or for velocity.. They don't usually have both numerical values in the question. Is there a video you can make on how to solve for problems like these? Without numerical values? Thanks!!

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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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[@jigglejogglers2606]

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?

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Same question😂

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[@tessasinsen6730]

for question 3 isnt m1 for the bullet and m2 for the block ? ,, do we interchange it for after the collision ?

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Thanks for the help :).

[@DrSpooglemon]

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[@diyajain3070]

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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[@brianadeleye2448]

4:10 Who are you trying to kill with a 15kg bullet (But all jokes aside this man is a legend).

[@kivenraj4922]

He said 15g.... whts wrong with you?

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[@DeanE351]

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?

[@BIGTTSNORLAX]

This is a good video, however I would suggest an improvement: Solve equations algebraically before plugging in values.

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[@jesutekevweboye]

What if two objects moving in different directions collide and stick together, how do you calculate that?

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[@AyeshaZeeshan819]

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

[@yassinewaterlaw6597]

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 ???

[@narenathnaelkhalanyane2681]

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

[@rye-bread5236]

8:04 are they not different masses? great video.

[@raynier250599]

I thought the same thing. I thought the the ke was of the bullet

[@Just_Deacon]

@@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

[@szlep8095]

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.

[@antoniavelez5150]

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.

[@kasaggajovanbenedict7059]

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

[@A_Wise_Pleb]

Legend

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[@MrBananazombie]

ily

[@dianneleroux1563]

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[@osvaldo1496]

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.

[@manishniroula313]

it goes in the form of friction, heat, sound and so on, I believe.

[@raynier250599]

In elastic collisions KE and momentum are conserved so it doesn't go into changing momentum

[@IdanKashani]

@@manishniroula313 It also slightly deforms the objects involved

[@fizixx]

Make the velocity of block 2 to be 20 m/s. Show the math that illustrates the problem with this.

[@angadsingh5042]

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

[@meganrivera7653]

1.5 is the mass of the box and 0.025 is the mass of the bullet

[@studiouscarpet34]

@@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.

[@elu5062]

Peoples who disliked this video are narrow minded😤

[@Business829]

how m2v2 is 0?

[@perennialwealth]

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?

[@annastein1128]

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

[@KRATOS-dq6sd]

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

[@nip-rx6em]

but why is organic chemistry teaching mechanics?

[@dmurray2978]

Because he is a generous god of science

[@borlasapaulexyt1.0]

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

[@borlasapaulexyt1.0]

2 kg moving at 3 m/s; and C of mass 5.2 kg moving at 2.5 m/s.

[@cain372]

2:49 Wrong! True equation must be (m1 + m2) * Vf.

[@ICantSeeYourRepliesDickhead]

That is incorrect!

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