Discover the Angle of Swing: Solving for the Motion of a Bullet and Block System

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The discussion revolves around a physics problem involving a bullet embedded in a block, where participants analyze the energy transfer and motion of the system. Key points include the use of conservation of momentum to find the velocity of the bullet/block system after the collision, which is calculated to be approximately 8.67 m/s. Participants express confusion over the potential energy and height calculations, noting that the derived height of 3.8 m exceeds the string length of 1 m, indicating a flaw in the approach. There is a consensus that the problem may be incorrectly stated, as the energy calculations suggest the block would have enough energy to complete a loop indefinitely. The discussion highlights the complexities of energy conservation in inelastic collisions and the need for clarification from the instructor.
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Homework Statement


: A bullet of mass 25g is fired at 425m/s into a block of wood , mass 1.200kg,hanging at the end of a 1.0m long string which is attached to the ceiling .The bullet becomes embedded in the block causing it to swing back along an arc.Through what angle does the block swing back?


Homework Equations





The Attempt at a Solution


0.5*0.025*(425)^2=(0.025+1.2) * 9.81h +0.5(0.025+1.2)v^2
h=L-Lcosθ
you cannot use the conservation of momentum because its not a linear motion
the height needs the angle
i feel the many unknowns please help
 
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It looks like you took the initial kinetic energy of the bullet and set it equal to the potential energy of the bullet/block system plus that systems kinetic energy.

When I first saw this problem, I though that that was what you did, and then set the final velocity to 0; essentially saying the KE of the bullet goes into gravitational potential of the system. But you can't say this, as energy is lost in the collision.So you would have to take the kinetic energy of the bullet/block system immediately after the collision and then set that equal to the gravitational potential to find height.
 
So could i use conservation of momentum

(mv)1 =(m1 + m2)v
v= (0.025*425)/(0.025+1.2)= 8.67m/s

0.5mv^2 = (m1 +m2) gh
0.5*(0.025 +1.2) * (8.67) = (0.025+1.2)*(9.8)h
h=3.8m which cannot be true it cannot be greater than 1m
 
I got the same answer.

Maybe my approach is also flawed, but maybe the answer is just saying the block will have enough energy to complete a loop around the wire.
 
so the angle will be L-LCos(theta)=h
L=1 1-cos(theta)=3.8
theta is undefined
 
Because the system will have so much energy, the energy will never transfer completely to gravitational and some will remain kinetic.
 
so we have to include the kinetic energy at the top?
 
I'm not sure. When it gets to the top it will continue to rotate around in circles forever (ignoring any dampening effects).
Unless the block is moving too slowly at the top to keep the string taught, in which case you might have to find the angle at which the string slacks and that is what the question is asking for? But that seems a little too involved for what should be a simple problem.

Is it possible the question was written wrong?


The total energy here is 46.08J, and at the top the kinetic energy is 22.07J, which means the speed of the system is 6m/s.
Ac = v2/r, so g = v2, and .5v2 is ~36, meaning the block will complete the loop and swing indefinitely.
 
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Chances are yes,but thanks for helping I will consult my lecturer.
 
  • #10
I agree with all your calculations... I think there is something wrong in the original question.
 
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