Answer to Question #110655 in Mechanical Engineering for ahmad alahesh

Question

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.

Answer 1

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

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

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

$\Delta KE=18040.32 J$