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9point8
Group Title
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)
 one year ago
 one year ago
9point8 Group Title
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)
 one year ago
 one year ago

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ajprincess Group TitleBest ResponseYou've already chosen the best response.0
dw: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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you 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..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
in this topic*
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
A 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 \(12ms^{1}\)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
It 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.
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
@Shadowys I dnt thnk the final answer is 12ms^1
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
I assumed g as 10ms^2.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
It'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.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i see..thanks..
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
dw:1352800019664:dw
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
You'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.
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
oh k. thanx:) dw:1352800258696:dw
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
Lol You're welcome :)..Though is having \(12ms^{1}\) normal for a bullet? :)
 one year ago

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

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
Which velocity being 332? Yes, its an approximate. Don't want to leak too much of the answers to our poster, do we? :)
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
ya. velocity of the bullet:) I simplified ur answer.:)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
Oh right. LOL it's 332. I guess I forgot about the square.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.2
lol sorry about that.Eyes get blurry in a hurry.
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
that's k:)
 one year ago
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