A bullet is fired into a fixed target loses half of its velocity after penetrating 3 cm. how much further will it penetrate before coming to rest assuming that it faces constant retardation?

Asked by Ravikant | 18th Nov, 2015, 09:14: PM

Expert Answer:

Let u be initial velocity of the bullet, after penetrating 3 cm into the target the velocity reduces to u/2 .
i.e After travelling s=3 cm with initial velocity u, the final velocity v = u /2. 
begin mathsize 14px style From space the space equation space of space motion comma straight v squared minus straight u squared equals 2 as straight a equals fraction numerator straight v squared minus straight u squared over denominator 2 straight s end fraction equals fraction numerator open parentheses begin display style straight u over 2 end style close parentheses squared minus straight u squared over denominator 2 cross times 3 space cm end fraction straight i. straight e. comma space straight a equals fraction numerator straight u squared minus 4 straight u squared over denominator 4 cross times 2 cross times 3 end fraction 24 space straight a equals negative 3 straight u squared straight a equals fraction numerator negative 3 straight u squared over denominator 24 end fraction equals fraction numerator negative straight u squared over denominator 8 end fraction Retardation equals straight u squared over 8 Let space apostrophe straight x apostrophe space be space the space distance space through space which space it space will space penetrate space further space and space stop. Then space straight v equals 0 comma space straight s equals straight x space and space straight u equals straight u over 2 therefore straight v squared minus straight u squared equals 2 as 0 minus open parentheses straight u over 2 close parentheses squared equals 2 cross times fraction numerator negative straight u squared over denominator 8 end fraction cross times straight x rightwards double arrow straight x equals 1 space cm end style

Answered by Faiza Lambe | 19th Nov, 2015, 07:49: AM