anonymous
  • anonymous
A 6 g bullet is fired horizontally into a 5 kg block of wood suspended by a long cord. The bullet sticks in the block. Compute the velocity of the bullet if the impact causes the block to swing 10 cm above its initial level.
Mathematics
schrodinger
  • schrodinger
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sohailiftikhar
  • sohailiftikhar
let the mass of bullet is m1=6/1000kg and mass of block m2=5kg
anonymous
  • anonymous
116806.67 cm/s^2 ?
anonymous
  • anonymous

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sohailiftikhar
  • sohailiftikhar
the initial velocity for both bullit and block is zero you have to find their final velocity which you can take as v because it is same for both as the got combined
anonymous
  • anonymous
can't i use the formula \[ (m1 + m2)\sqrt{2gy} \over (m1) \]
anonymous
  • anonymous
?
sohailiftikhar
  • sohailiftikhar
lol you got two madel nice answer
anonymous
  • anonymous
Use the Principle of Conservation of Energy Before impact, The bullet moved with suppose velocity v, therefore it's kinetic energy was \[E_{1}=\frac{1}{2}mv^2\] Where m is the mass of the bullet After impact the system of block+bullet is raised to a height of 10cm Therefore increase in the potential energy would be \[E_{2}=(m+M)gh\] Here M is the mass of the block and h is the height to which it rises (10cm) Since m is quite small compared to M, we can approximate the potential energy as \[E_{2} \approx Mgh\] By the principle of conservation of energy, the kinetic energy before should be equal to the increase in the potential energy of the block+bullet system assuming no loss in energy due to sound heat etc. We are not considering initial potential energies of the block and bullet as they would just cancel out anyway \[E_{1}=E_{2}\]
sohailiftikhar
  • sohailiftikhar
ok you can use the law of conservation of momentum here
IrishBoy123
  • IrishBoy123
OK, so just how can you use momentum here?? we know from energy conservation assumption that \(\frac{1}{2}m_1v_1^2 = (m_1 + m_2)gh\) \(v_1 = \sqrt{ \frac{2(m_1 + m_2)gh}{m_1} }\) [which solves the OP's question ] "if" this is a closed system [and momentum is conserved just between m1 and m2] we can also say that \(m_1 v_1 = (m_1 + m_2)v_2\) \(v_2 = \frac{m_1 }{m_1 + m_2} \sqrt{ \frac{2(m_1 + m_2)gh}{m_1}} = \sqrt{ \frac{2m_1gh}{m_1+m_2}}\) so the total energy of the system, again conserved, gives us \(\frac{1}{2}(m_1+m_2)v_2^2 = \frac{1}{2}(m_1 + m_2) \frac{2m_1gh}{m_1+m_2} = m_1gh\) which is plainly wrong IOW m1 and m2 is not a closed system and you cannot use momentum conservation, that is why the solution requires energy conservation to see this we can say \(\frac{1}{2}(m_1+m_2)v_2^2 = \frac{1}{2}m_1v_1^2\) \(v_2 = \sqrt{ \frac{m_1}{m_1+m_2} } \ v_1\) \(= \sqrt{ \frac{m_1}{m_1+m_2} } \sqrt{ \frac{2(m_1 + m_2)gh}{m_1} } \\ = \sqrt{2gh}\) which is the answer we should be getting
anonymous
  • anonymous
So am I right?:P
thomas5267
  • thomas5267
Conservation of momentum does not apply in here as gravity (external force) is acting on the system.
thomas5267
  • thomas5267
I guess it would apply if you include Earth in the system as well?

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