Question #110655
Q15.
The ram of a pile driver has mass 800 kg and is released from rest at a
height of 2 m above the top of a 2400 kg pile. If the ram rebounds to a
height of 0.1 m after a direct central impact with the pile, determine the
following:
(a) the velocity of the pile immediately after impact;
(b) the coefficient of restitution;
(c) the percentage of the energy lost in the impact.
1
2020-04-19T14:06:49-0400

Mass of Ram= 800 Kg , and mass of pile= 2400 , ram strikes to pile at height of 2m , here

initial velocity of ram at time of striking = "\\sqrt {2gh}= \\sqrt{2\\times9.8\\times2}=6.26 m\/s" and

and initial velocity of pile = 0 m/s and velocity of ram after impact=

"v_1=\\sqrt {2gh}= \\sqrt{2\\times 9.8\\times .1}=1.4 m\/s"

Now the velocity of pile after imapct,

(a)

"m_1u_1+m_2u_2=m_1v_1+m_2v_2"

"800\\times6.26+2400\\times0=800\\times1.4+2400v_2"

"v2=1.62" m/s

(b) for coefficient of restitution

e= "\\frac{velocity after impact}{velocity after impact}= \\frac{1.62+1.4}{6.26}=0.482"

(c) Lost in energy = Change in kinetic energy = final kinetic energy- initail kinetic energy

"\\Delta KE= {(\\frac{1}{2}\\times m_1 (u_1)^2)+(\\frac{1}{2}\\times m_2 (u_2)^2)}- {(\\frac{1}{2}\\times m_1 (v_1)^2)+(\\frac{1}{2}\\times m_2 (v_2)^2)}"

"\\Delta KE= {(\\frac{1}{2}\\times 800(6.26)^2)+(\\frac{1}{2}\\times 2400 (0)^2)}- {(\\frac{1}{2}\\times 800 (1.4)^2)+(\\frac{1}{2}\\times 2400 (1.62)^2)}"

"\\Delta KE=18040.32 J"

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