
#1
Nov2110, 09:09 AM

P: 5

1. The problem statement, all variables and given/known data
A 50g bullet traveling horizontally at 200 m/s embeds itself in a wooden block, initially at rest, on a horizontal surface (μ_{k}=0.1). The block slides 1.2m toward a spring and collides with it. The block compresses the spring (k=600 N/m) by 20 cm. Find the mass of the wood. m_{bullet}=0.05kg v_{1bullet}=200 v_{1wood}=0 m_{wood}= ? m_{bullet}+m_{wood}= t d_{without spring}=1.2 d_{with spring}=1.4 k=600 x=0.02m 2. Relevant equations Ff=μ_{k}mg W_{friction}=Ffxd (where d is 1.4m) m_{1}v_{1}+....=m_{2}v_{2}....... E_{k}=mv^{2}/2 E_{e}=kx^{2}/2 3. The attempt at a solution W_{f}=0.98t*1.4 W_{f}= 1.372t m_{b}v_{1}+m_{w}v_{1}=m_{b+w}v .05*200=tv v=10/t E_{kbullet}=E_{ksystem}+E_{e}+W_{f} 1/2*.05*200^{2}=1/2*t*(100/t)^{2}+1/2*600*.2^{2}+1.372t 1000= 50/t + 12 +1.372t t=720 t=m_{w}+m_{b} 7200.05=m_{w} m_{w}=719.95 I stated that the kinetic energy of the bullet, becomes the kinetic energy for the block and the bullet to move, plus the elastic energy and work done by friction. Am I right in assuming this? Thank you for the help! 



#2
Nov2110, 10:58 AM

P: 607

This is an inelastic collision between the bullet and the block so KE is NOT conserved. Use
conservation of momentum to find the velocity immediately after the collision. And thereafter use the conservation of energy. 



#3
Nov2110, 11:21 AM

P: 5

m_{bv1}+m_{wv1}=m_{b+w}v
.05*200=tv v=100/t So would it only be: E_{k(bullet)}=E_{e} + energy lost ? what about when the block slides before hitting the spring? 



#4
Nov2110, 11:42 AM

P: 607

Bullet Hitting Wood the Spring, check of solution
ok now you got the velocity of the bullet+block system. Now find the expression for the kinetic energy of this system , with velocity, [tex]v=\frac{10}{m_b +m_w}[/tex]
Now use the conservation of energy. some of this energy will be turned into internal energy due to friction and some of it will be converted into the elastic potential energy of the spring. and 0.05 times 200 is not 100, its 10. careful about arithmetic in physics class 



#5
Nov2110, 11:49 AM

P: 5

Thank you for pointing out the error.
I have gotten that the Ek system= 50/t (aka. 50/(mw+mb) And now I set the total kinetic energy to equal elastic energy plus work done by friction? Ek=energy lost + Ee The one thing that I don't understand about that is wouldn't the block also be using kinetic energy to slide the 1.2m before hitting the spring? 



#6
Nov2110, 12:04 PM

P: 607

the block and bullet together slide. the kinetic energy lost during the inelastic collision is already gone ( in the form of sound etc)Now we have the kinetic energy of the system at our disposal. Think of energy as the nature's currency. Now we have to spend this remaining currency. Part of it will be spent in work related to the friction and other part will be stored in the spring. While using energy approach to do physics problems, always think of 'energy transfer'. I don't like to use the language of "work done" etc. Energy is an universal thing in the world. Always think of 'energy transfer' while doing such problems. Always ask, where did my energy go (like bankers always check the money in their account) ? And one more piece of advice while doing physics problems. ALWAYS use vector notation wherever possible. Then your calculations will never be wrong. Intro physics textbooks have the habit of converting a vector equation into a scalar equation involving the magnitudes. But if you stick with the vector notation, then the signs arising out of the vectors will be taken care of automatically since you specify vectors with the directions. This is very important in Workenergy related problems because , the
[itex]\pm[/itex] signs have deep meaning. + means energy is added into the system and  means energy is taken out of the system (and converted to some other form). Now where I am located, in India, its late night. So your further questions, I will answer tomorrow. 



#7
Nov2110, 12:08 PM

P: 5

Thank you very much for the great explanation! :D
Have a good night! 


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