Conservation of energy and momentum Problem

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SUMMARY

The discussion centers on a physics problem involving the conservation of energy and momentum, specifically applied to a 16 kg block that explodes into two fragments. The explosion adds 76.8 joules of kinetic energy, with one 10 kg fragment sliding 1.47 m before stopping. Using the equations of momentum and energy conservation, the solution determines that the 6 kg fragment slides 4.08 m before coming to rest. The equations used include momentum conservation (6va = 10vb) and energy conservation (76.8 = 0.5 * 6 * va^2 + 0.5 * 10 * vb^2).

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  • Understanding of conservation of momentum principles
  • Knowledge of kinetic energy equations
  • Familiarity with friction force calculations
  • Basic grasp of algebraic manipulation of equations
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  • Learn how to apply kinetic energy equations in collision problems
  • Research the effects of friction on motion and energy loss
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Homework Statement



A 16 kg block is initially at rest on a rough horizontal surface (uk= .2). The block has a small explosive charge inside it. The charge explodes, splinting the block into two fragments and adding 76.8 joules of kinetic energy. The other 10kg slides 1.47m along the rough surface before coming to Rest. How far does the other 6kg fragment slide along the rough surface before it comes to rest?
Must be solved using conservation laws.

Homework Equations





The Attempt at a Solution


Using conservation of momentum:
6va = 10vb
vb= .6 va
line 1: 76.8 = .5 * 6 *va^2 + .5* 10 * vb^2
va = 4
line 2: .5 * 6 * 4^2 + -.2 * 6 * g * da= 0
da= 4.08
I am not sure where the equation from line 1 and 2 came from.
Is line 1 using work = change in kinetic energy?
I am not sure what equation is used for line 2.
 
Physics news on Phys.org
line 1 computes total kinetic energy after explosion
line 2 uses the work of friction force and sets it equal to the change in kinetic energy
 

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