Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
7
Linear Momentum
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7-1 and 7-2: Momentum and Its Conservation
7-3: Collisions and Impulse
7-4 and 7-5: Elastic Collisions
7-6: Inelastic Collisions
7-7: Collisions in Two Dimensions
7-8: Center of Mass
7-9: CM for the Human Body
7-10: CM and Translational Motion

Problem 33
A
a) Mm+M\dfrac{-M}{m+M}
b) 0.96-0.96
Giancoli 6th Edition, Chapter 7, Problem 33 solution video poster
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VIDEO TRANSCRIPT

We’re going to figure out what fraction of kinetic energy is lost in the inelastic collision with this ballistic pendulum. So it’s the amount of energy that disappeared plus the change in kinetic energy divided by the amount we started with. So that’s kinetic energy final minus kinetic energy initial divided by the initial kinetic energy. So the example in the textbook uses a few formulas to start with. So we can say that the kinetic energy final is one half times ‘m’ plus ‘M’, ‘m’ being the mass of the bullet and ‘M’ being the mass of the pendulum, times ‘v`’ squared. Where the book really helps actually is in this kinetic energy initial where that’s one half of ‘m’ because there is only the bullet there is no kinetic energy in the pendulum and initially because it’s not moving at the beginning until it’s hit by the bullet. So we have one half ‘m’ times ‘v’ squared where ‘v’ is the initial velocity of the bullet before it hits the pendulum and this is one half ‘m’ times ‘m’ plus ‘M’ divided by ‘m’ times ‘v`’ squared and we can express this instead as one half ‘m’ plus ‘M’ squared over ‘m’ times ‘v`’ squared. This ‘m’ is going to get squared because of that exponent and one of them is going to cancel and so that’s why there is no squared in the bottom there. One half ‘m’ plus ‘M’ times ’v`’ squared minus one half ‘m’ plus ‘M’ squared divided by ‘m’ times ’v`’ squared, and we can factor out a whole bunch of common factors here, so that all gets multiplied by one minus ‘m’ plus ‘M’ over ‘m’. So we have one half times ‘m’ plus ‘M’ times ’v`’ squared times ‘m’ minus ‘m’ minus ‘M’ all over ‘m’, so we have is negative ‘M’ over two times ‘m’ times ‘m’ plus ‘M’ times ‘v`’ squared. So that’s what the difference in the kinetic energy is, that’s how much kinetic energy is lost. So now we’ll have to divide that by the initial kinetic energy and that will answer this question. So the change in kinetic energy divided by the initial kinetic energy is going to be this expression we just figured out ‘M’ over two times ‘m’ times ‘m’ plus ‘M’ times ’v`’ squared divided by initial kinetic energy, that’s one half ‘m’ plus ‘M’ squared over little ‘m’ times ‘v`’ squared. Simplifying what we are left with in the end is negative ‘M’ over ‘m’ plus ‘M’. So that’s the answer for part a. And in part b we actually have to evaluate using numbers and this is going to be: negative three hundred and eighty grams divided by fourteen grams plus three hundred and eighty grams. We don’t need to convert to kilograms since these grams are going to cancel anyway and we are left with the dimensionless number or in other words a number with no units of zero point nine six. So this says that ninety six percent of the energy is lost which is not that much of a surprise because it’s an inelastic collision when the objects stick together and that inelastic collision will have the maximum amount of kinetic energy lost while still being consistent with conservation of momentum.

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