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8.9 [-/2 Polnts] oBook 1 1 (a) What 1 object undcrgocs DETAILS m/s? 1 Is thc magnltude applled 1 acccleration SERCP1O 4.P.003 resultant force 0t 2.3 kg object; ms? ...

Question

8.9 [-/2 Polnts] oBook 1 1 (a) What 1 object undcrgocs DETAILS m/s? 1 Is thc magnltude applled 1 acccleration SERCP1O 4.P.003 resultant force 0t 2.3 kg object; ms? . acting on It? acceleratlon 1Your W 8

8.9 [-/2 Polnts] oBook 1 1 (a) What 1 object undcrgocs DETAILS m/s? 1 Is thc magnltude applled 1 acccleration SERCP1O 4.P.003 resultant force 0t 2.3 kg object; ms? . acting on It? acceleratlon 1 Your W 8



Answers

A force of $1 \mathrm{~N}$ acts on a body of mass $1 \mathrm{~kg}$, which is able to move freely. Then the body acquires (a) a speed of $1 \mathrm{~m} \mathrm{~s}^{-1}$ (b) an acceleration of $1 \mathrm{~m} \mathrm{~s}^{-2}$ (c) an acceleration of $9.8 \mathrm{~m} \mathrm{~s}^{-2}$ (d) an acceleration of $1.02 \mathrm{~m} \mathrm{~s}^{-2}$

Okay. In this problem you have a massive an object of mass m colliding with a object of mass. To him, they combine and slide off together. Since this isn't any lasted collision, I know that the momentum is conserved, but the kinetic energy is not conserved. Since the moment um is conserved. I could take the mass times the original velocity, which is the entire momentum of the system because the two m block is not moving in. That equals the final minimum of the system, which is three m times the final velocity. Now I can express these two velocities in terms of kinetic energy. The original kinetic energy is one half times m b, not squared. The final kinetic energy is one half times three m times. BF squared so should be squared. I'm having trouble well, right. And that's why I'll write it again. K f is equal to one half times three m comes VF squared. So my velocity is the square root of two ko over em. That's my original velocity In my final velocity is the square root of Tuesday over three m. So I could substitute my two velocities into my mo mentum conservation equation and get mm e I'm just screwed of two k o. It's the original kinetic energy divided by mass eyes equal to three m times the square root to K f divided by three m. So let's square everything to get rid of the radicals and cancel out that M I don't wanna cancel this one out yet just because I like to keep my calculations nice and clean. So I've got to k o over em is equal to nine times to k f over three m now my EMS will cancel two's canceled, some left with Ko eyes equal to 9/3. I'm SKF okay. F is equal to three nights Yo. So a f you could have one third of a Oh, so the final kinetic energy is one third of the original kinetic energy. Yeah,

So in this problem, we have two objects. Subject one has mass m an object to has missed two. So object the winds moving toward object to with velocity. And after the Clyde together, the ah, they combined together. And ah, we want to know what fraction of the energy of the collision is, is it? Ah, compared to the original connector Kinetic energy of the object one. So, uh so before the collision, we know that never meant him off. The system is that only comes from the object. What that is and puffy. And after the clich is the mo mentum comes from both and it becomes m times V prime plus two m times every Friday. Right? So this is a three impound three. Prime obtained that Toby Prime equal 1/3 will be right so we can compare the energy, the kinetic energy before and after the cliche. So before the collision, the Connecticut the kinetic energy is Ikea, which is half em be squid, right? And after the closure of the kind of energy is Ikea Prime, which is half and sorry. 3 a.m. Times of the prime squared, which is 1/3 time's okay says we can see that


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