Answer to Question #200648 in Mechanics | Relativity for Master j

Question

A pile driver of mass 350 kg falls freely from a height of 2.8 m and strikes an 80 kg pile. Determine the:

i. common velocity of the pile and driver after the impact

ii percentage loss of kinetic energy after the impact

Answer 1

Velocity of 350kg driver

$mgh=\frac{1}{2}mv^2_1 \implies v_1=\sqrt{2gh}=\sqrt{2*9.8*2.8}=7.41m/s$

i. common velocity of the pile and driver after the impact

$(m_1v_1+m_2v_2)=(m_1+m_2)v_{common \space velocity}$

$v_{common \space velocity}= \frac{(m_1v_1+m_2v_2)}{(m_1+m_2)}$

$v_{common \space velocity}= \frac{(350*7.41)}{(350+80)}=6.03 m/s$

ii percentage loss of kinetic energy after the impact

$\% Loss = \frac{KE_1-KE_2}{KE_1} = \frac{0.5m_1v^2_1-0.5m_{12}v^2_1}{0.5m_1v^2_1}= \frac{m_1v^2_1-m_{12}v^2_1}{m_1v^2_1}$

$= \frac{350*7.41^2-430*6.03^2}{350*7.41^2}*100=18.64\%$