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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)
ok here is the setup bu=bullet bl=Block B= Block and bullet (mbu)*(Vbu)o+(mbl)*(Vbl)o=mB*VB => yeilds VB= 12/21 (Vbu)o from vector diagram with positive x to the left uN/21=ax N=9*9.81 from kinematic equation Vf=VB+2*ax*(x-xo) Vf=0 & (x-xo)=5 (due to positive x to the left) and that should give you the variable chain that you solve for your answer. Let me know if it works
change that, N = 21*9.81
I made one more mistake though the mass of the block was 12 g but it is 12 kg , distribute this mass accourdingly
Consider the diagram attached. We know that the the block and bullet is being slowed by friction from some initial velocity from the impact (we will assume that the velocity imparted is instantaneous). Friction is force, and is equal to the product of the friction coefficient \(\mu\) and the normal reaction \(R=mg\) where m is the mass of the object, and g is the acceleration due to gravity. Upon impact the block will be given an initial velocity, \(u_{2}\), which will be slowed down over the distance of \(s=5\) cm by the acceleration \(a=\mu R\), acting in the opposite direction to the velocity, until it comes to rest \(v=0\). We can use the equation of motion, \[v^2=u_{2}^{2} +2as\] to find the initial velocity of the block just after impact. Having found the initial velocity we can then solve for the velocity of the bullet just before impact by either conservation of momentum, or by conservation of energy. Using momentum, we note that the momentum before must equal the momentum after and hence \[M_{(bullet)}U_{1} + M_{(block)}U_{(block)}=M_{(bullet+block)}U_{2}\], Then just solve for \(U_{1}\) since the velocity of the block before impact is zero. By conservation of energy, we know teh kinetic energy before is equal to kinetic energy after or \[\frac{1}{2}M_{(bullet)}U_{1}^{2} +\frac{1}{2}M_{(block)}U_{(block)}^{2}= \frac{1}{2}M_{(bullet+block)}U_{2}^{2}\]. Then just solve for \(U_{1}\) since the velocity of the block before impact is zero. Note be sure to have every quantity in SI units to make the calculation correct. i.e convert all masses to Kilograms and distance to metres.
Well I think I got the right answer. I'm just not sure. Is it 330.4 m/s?
i get 332.5 m/s using a g value of 9.8 m/s^2. If you used a g = 10, you will get 335.85 m/sec^2. There is a slight error in my post above. The acceleration term is given by \[a=\frac{\mu R}{m}=\frac{\mu mg}{m}=\mu g\]not \(\mu R\) as I erroneously stated. Also, I am being very silly in the using of conservation of energy. By friction being involved, energy will not be conserved, unless you could the energy loss from friction. So as the equation stands, it wont give you the correct answer. Apologies for confusing everyone. Use the conservation of momentum instead.
Vbullet=-(Mbullet/Mbullet+block)*sqrt(-2(u*g)(x-x0))