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A mass MA 2kg slides down slippery ramp; reaching speed =10 m/$ at the botlom Aner reaching the level ground ma collides With mass m# 3k9 at botlom, sticking logeth...

Question

A mass MA 2kg slides down slippery ramp; reaching speed =10 m/$ at the botlom Aner reaching the level ground ma collides With mass m# 3k9 at botlom, sticking logether in a perfectly inelastic collision.What was the height of the ramp; assuming that mA started from rest?6 ) How fast were the combined masses mA and mB moving afer sticking together?The combined mass runs into spring: which brings It Io stop after being compressed cm, Whal is the spring constant ki

A mass MA 2kg slides down slippery ramp; reaching speed =10 m/$ at the botlom Aner reaching the level ground ma collides With mass m# 3k9 at botlom, sticking logether in a perfectly inelastic collision. What was the height of the ramp; assuming that mA started from rest? 6 ) How fast were the combined masses mA and mB moving afer sticking together? The combined mass runs into spring: which brings It Io stop after being compressed cm, Whal is the spring constant ki



Answers

A block of mass $m$, after sliding down a frictionless incline, strikes another block of mass $M$ that is attached to a spring of spring constant $k$ (see below). The blocks stick together upon impact and travel together. (a) Find the compression of the spring in terms of $m, M, h, g,$ and $k$ when the combination comes to rest. Hint: The speed of the combined blocks $m+M\left(v_{2}\right)$ is based on the speed of block $m$ just prior to the collision with the block $M\left(\mathrm{v}_{1}\right)$ based on the equation $v_{2}=(m / m)+M\left(v_{1}\right) .$ This will be discussed further in the chapter on Linear Momentum and Collisions. (b) The loss of kinetic energy as a result of the bonding of the two masses upon impact is stored in the so-called binding energy of the two masses. Calculate the binding energy.

Solving party of this problem. So here we have from conservation of energy. I can write one by two and we notice square is equal to one by two K a square. So the distant travel by observed before coming to race can be return. Urge is equal to we not root under em by K so simplifying it further and putting the value here I can write 1.24 m once again under 2.363 by 44.5. So on simplification, I get the value of A is equal to 0.11 to meet now for part B, we have given T is equal to one by four cap 30 according to the data. So T is equal to two by do tender and by K simplifying it further. I can write. The value of period of operation is equal to to buy a suit and a 0.363 by 44.5. So the final value of these coming edge 0.567 2nd. So as we all know that the time taken by spring to stop the mask is equal to T by four. So just putting the value 0.567 seconds by four. So t is coming edge zero point 142 seconds, which is our answer.

Solving party of this problem. So as we all know that the distance traveled by an object coming to rage can be calculated by the conservation of energy formula so half and he noticed square is equal to half k square. So the value of a can be return it if you're not root under AM by key. So just putting the value here so I can write 1.24 multiplication Underwood 0.363 by 44.5. And on simplification, I get the value of A is equal to 0.11 to meet it now solving part B. Yeah, so in part we we have t is equal to one by food. The according to the question also T is equal to two pilot tender m by K. So simplifying it further, I can't devalue Age T is equal to pi route and I'll 0.363 by 44.5. And on solving I get is equal to 0.567 seconds. So in this way, the time taken for the spring to stop the mass t is equal to t by four. So t is equal to just putting the value 0.567 2nd by four, so the final answer is 0.142 seconds.

That we have two blocks of mass one kitchen, three kids are on a frictionless san Francisco street spring. Right, So that means if this is a block one and block The 2nd 1 and it is corrupted by as being if this is one kg and this is three kg then it is easily stressed. And then when it is released then then velocity of the one kg block becomes what becomes what 1.7 m per second to ask this one. Okay, so what will be, what will the velocity of the other blog at that moment? Pretty simple how? Because uh you know, we can just assume that this is a system that means spring and to block a system and we know that over here the string is much less. So we are not going to talk about the masses and the city of Moment about the spring because these are these are best massless. That means the mass of the spring is just negotiable to this one so that you know that if net external force did you know that Mr spring is internal force the moment of contact. So if someone is in this direction, then let's say that means everyone the one Plus and equal to what initial moment of the zero. And when uh Hitler part of art minus of M one B one upon them to um 11 into 1.7 point three. So it would be like mhm you don't want 567 m parts again minus. So this will be this man is indicated that this will move to our stop one kg block.

Hello. So for this problem we've got this mass A. Which is two kg. Running into mass B. Which is five kg. S. A. Is moving at 10 m per second B. Is moving at three m per second. We know the K the spring constant of the spring and we need to figure out how far it will compress. We're told in the problem to note that the maximum compression at the point of maximum compression, they'll move as one unit or so we can treat this as a inelastic collision. So momentum will be conserved. So the total momentum before the collision is massive blockade times V. One blockade plus massive black B times V. One of Black B. And that is going to equal the combined mass as a black A. Plus massive black beat times their new speed as they move off together. We'll just call that be too. So if we solve this for me too, we're going to get the right side of this equation all over the combined mass And we'll get a V two five meters per second. So if you plug all this in all these numbers in um you should get five m per second, provided we used our parentheses correctly. So this is the final speed. Now that we know the final speed, we can we can start using conservation of energy. So we have uh an initial kinetic energy from each of these two things. We're going to have a final kinetic energy. And the difference between these two kinetic energies is how much energy gets stored in the spring. And so we're going to have um kinetic energy from black A plus kinetic energy from black B equals kinetic energy of block A. B. System plus the amount of energy lost to the spring. Mhm. So this Right side it's just 1/2 mm Eva one squared. So one half times two times 10 squared plus one half MBBB one Squared Arby's. So 1/2 times five times 3 square. And on this side if you plug in those numbers, you'll get 122.5 jules on the left side on the right side, we can figure this since we know all that stuff plus and be B two squared. So one half times seven is their combined mass times five squared For that, you'll get 87.5 jewels plus us or us. So plus us. So this you this potential energy here in the spring is going to be this minus that. Which ends up being 35 jules is the potential energy stored in the spring. And we were given the case so we can actually solve this as well. So this it's going to be one half K. X squared is the amount of energy stored in the spring where X. Is the compression of the spring. So this is what we're looking for right there. That X on this shift colors here for this part. So if we solve this for X, we're going to get X equals you times two divided by K, and then square root of the whole thing. So to you overcame Where you is 35 jules. Kay is 1120 newtons per meter. So we plug all that in. We're going to get any compression of 0.25 person.


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