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$\dfrac{v_{i_2}}{v_{i_1}} = 1.4$

Giancoli 6th Edition, Chapter 7, Problem 31

(1:55)

Chapter 7, Problem 31 is solved.

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Transcript for this Giancoli solution

In both cases for this ballistic pendulum we’ll have the gravitational potential at the end will equal the kinetic energy that it has initially. So we can say that ‘m’ times ‘g’ times ‘h’ equals one half times ‘m’ times ‘vi’ squared. This is the first scenario with projectile one, we’ll call it ‘h1’ and ‘vi1’, the ‘m’s cancel solving for ‘vi1’ equals square root of two times ‘g’ times ‘h1’. Using the same logic we can say the initial velocity of projectile two is the square root two times ‘g’ times ‘h2’. And we are told to find the second projectile was how many times faster than the first. So translating those words into algebra, the initial velocity of the second projectile equals what times the velocity of the first projectile. So we have to find ‘X’. So ‘X’ equals ‘vi2’ divided by ‘vi1’ which is; square root of two times ‘g’ times ‘h2’ divided by square root of two times ‘g’ times ‘h1’. The two times ‘g’ both cancels and we are left with: ‘X’ equals the square root of ‘h2’ over ‘h1’ which is the square root of five point two centimeters divided by two point six centimeters. No need to change that to meters because units are going to cancel anyway. So in the second case the projectile was going one point four times faster.