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anonymous
 3 years ago
A 12 g bullet is fired into a 9.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.
What was the speed of the bullet?
(please help, I've tried setting this problem up multiple times and can't get it right)
anonymous
 3 years ago
A 12 g bullet is fired into a 9.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20. What was the speed of the bullet? (please help, I've tried setting this problem up multiple times and can't get it right)

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ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352789469767:dwdw:1352789574359:dw dw:1352789707768:dwdw:1352789857307:dw Using \( v^2=u^2+2as\) \({v_4}^2={v_3}^22as\) Here dw:1352790096724:dw Then using conservation of momentum principle \(mv+m_1v_1=(m+m_1)v_3\) \(0.012v=9.012*2as\) dw:1352790355574:dw

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0you might have also used work energy theorem simply : W(friction) = 1/2 (m1 v^2) where m1 = mass of bullet => U (m1+m2) g . d = 1/2 m1 v^2 m2 = mass of block U=coefficient of friction d= distance substituting, we have v^2 = (2 * 0.2 * 9.012 *10 * 0.05)/(0.012) please correct me if am wrong somewhere as i am not too strong with this topic..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A elegant set up is considering that: 1)The bullet transfers its momentum and energy completely to the block. 2)The blockbullet combo uses that energy to slide over a rough surface. Thus you might want to think of conservation of momentum (in the first case as no distance travelled is given to you.) and the workkinetic energy theorem(in the second case as the distance is given) Now, for the first case, where t is the bullet, k is the block, \(u_k=0,v_t=0\)is given. \(m_t u_t = (m_k+m_t) v_total\) so you get \(v_total=\frac{0.012 \times u_t}{0.012+9}\) Now, taking the workkinetic energy theorem, \( \Sigma E_{kinetic}=\Sigma W \) taking the fact that it slides to a stop, and the only force doing work on it is friction, \(F_f=\mu F_N\)and \(F_N=F_g={m_{total}}g\) Where it slides for S= 0.05m \(\frac{1}{2} m_{total}(0^2u^2_{total})=F_f(S)\) \(\frac{1}{2} m_{total}((\frac{0.012 \times u_t}{9.012})^2)=\mu(m_{total})(g)(0.05)\) Rearranging some of them, \(\frac{12u^2_t}{9012}=2(0.05)(0.2)(9.8)\) And you should get the answer that is approximately \(12ms^{1}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It should be noted that some of the energy of the bullet is used to drill into the block. Thus the kinetic energy of the bullet does not directly equals the work done by friction.

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0@Shadowys I dnt thnk the final answer is 12ms^1

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0I assumed g as 10ms^2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's a little more than that. I simply gave a general direction so that if the answer flies too far... yh it should differ by a few decimal points if you used g=10.

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352800019664:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're welcome :) @ajprincess your approach used the actionreaction force. However, in this case the action reaction is very complex as the forces acting on the bullet also includes the resistance of the block. Thus normally we don't include the analysis of such forces.

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0oh k. thanx:) dw:1352800258696:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lol You're welcome :)..Though is having \(12ms^{1}\) normal for a bullet? :)

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0but velocity comes as 332ms^1 @Shadowys That value is an approximated one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which velocity being 332? Yes, its an approximate. Don't want to leak too much of the answers to our poster, do we? :)

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0ya. velocity of the bullet:) I simplified ur answer.:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh right. LOL it's 332. I guess I forgot about the square.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol sorry about that.Eyes get blurry in a hurry.
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