10_P52InstructorSolution

10_P52InstructorSolution - 10.52. Model: Assume an ideal...

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10.52. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy gs K UU ++ is conserved. Visualize: We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m and the block’s mass is M . Solve: (a) The energy conservation equation 2s 2g 21s 1g 1 K UU KUU ++= becomes 22 ee 2 11 1 () mMv ky y mMgy y + + = + Noting 2 0 m/s, v = we can rewrite the above equation as 2 1 1 1 ()2 ( ) ( ) ( ) ky mMgy y mM v ky Δ+ + Δ + Δ = + + Δ Let us express 1 v in terms of the bullet’s initial speed B v by using the momentum conservation equation fi p p = which is 1 B block . mM v m v M v += + Since block 0 m/s, v = we have 1B m vv ⎛⎞ = ⎜⎟ + ⎝⎠ We can also find the magnitude of 1 y from the equilibrium condition 1e . M g −= 1 Mg y k Δ= With these substitutions for 1 v and 1 , y Δ the energy conservation equation simplifies to 2 2 B 21 2 2 2 2 B1 2 2 ( ) ( ) 2( ) ( ) mv Mg k k mMMg vg y y k y mm k m + Δ + Δ = + + + ⇒= Δ− + Δ We still need to include the spring’s maximum compression ( d ) into this equation. We assume that
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at University of Georgia Athens.

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10_P52InstructorSolution - 10.52. Model: Assume an ideal...

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