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9point8
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)
|dw:1352789469767:dw||dw:1352789574359:dw| |dw:1352789707768:dw||dw:1352789857307:dw| Using \( v^2=u^2+2as\) \({v_4}^2={v_3}^2-2as\) 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|
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..
A elegant set up is considering that: 1)The bullet transfers its momentum and energy completely to the block. 2)The block-bullet 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 work-kinetic 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 work-kinetic 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 \(\frac{1}{2} m_{total}(0^2-u^2_{total})=-F_f(S)\) \(\frac{1}{2} m_{total}(-(\frac{0.012 \times u_t}{9.012})^2)=-\mu(m_{total})(g)(0.05)\) 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 \(12ms^{-1}\)
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.
@Shadowys I dnt thnk the final answer is 12ms^-1
I assumed g as 10ms^-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.
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You're welcome :) @ajprincess your approach used the action-reaction 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.
oh k. thanx:) |dw:1352800258696:dw|
Lol You're welcome :)..Though is having \(12ms^{-1}\) normal for a bullet? :)
but velocity comes as 332ms^-1 @Shadowys That value is an approximated one.
Which velocity being 332? Yes, its an approximate. Don't want to leak too much of the answers to our poster, do we? :)
ya. velocity of the bullet:) I simplified ur answer.:)
Oh right. LOL it's 332. I guess I forgot about the square.
lol sorry about that.Eyes get blurry in a hurry.