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Object 1 travels with an initial speed of $v_{0}$ toward Object $2,$ which is traveling in the same direction as Object $1 .$ Object 1 collides perfectly inelastically into Object 2 and the velocity after impact is 0.5$v_{0} .$ If Object 2 has twice the mass of Object $1,$ what must have been the initial speed of Object 2 (assuming no external forces)?(A) 0.25$v_{0}$(B) 0.5$v_{0}$(C) 0.75$v_{0}$(D) $v_{0}$

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Chapter 13

Practice Test 3

Section 1

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high in the given problem. Initially speed of object one is we I one is going to be not and initially speed of object to that is missing the I do. We have to find it out. No. As the collision is perfectly in the last week, so both the objects moved together with finally speed V. F. It's about 2 0.5 of the note. It is given that if mass of the first object is M, then that of the second object is twice To that of the 1st 1. So using conservation of linear momentum as we know, linear momentum remains conserved in a perfectly inelastic collision. Also, so using the conservation of leading momentum, the initial linear momentum means linear momentum before the coalition for the first of all, this is anyone Until we are one plus for the second one, this is M two in two. G I two is equal to M one plus M two. As the objects stick together after coalition and their combined speed is we're so plugging in all known values for everyone. This is M for we I won, this is we not plus for him to this is twice of him and this we I do that is missing is equal to M one M M two To MVF. And that we have is given a 0.5 times off. We not which comes out to be three M in 20.5 means this is 1.5 am into me not Hence you can stay to em into a VI two Is equal to 1.5. I mean do we not minus Absing to be not? Which will remain as 0.5 am we not then canceling this. M VI two means initially speed of the second object here will come out to be 0.5 were not divided by two means this is 0.25 times of me. Not. Hence, here, we can say our option e is correct here. Thank you.

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