Velocity of 350kg driver
m g h = 1 2 m v 1 2 ⟹ v 1 = 2 g h = 2 ∗ 9.8 ∗ 2.8 = 7.41 m / s mgh=\frac{1}{2}mv^2_1 \implies v_1=\sqrt{2gh}=\sqrt{2*9.8*2.8}=7.41m/s m g h = 2 1 m v 1 2 ⟹ v 1 = 2 g h = 2 ∗ 9.8 ∗ 2.8 = 7.41 m / s
i. common velocity of the pile and driver after the impact
( m 1 v 1 + m 2 v 2 ) = ( m 1 + m 2 ) v c o m m o n v e l o c i t y (m_1v_1+m_2v_2)=(m_1+m_2)v_{common \space velocity} ( m 1 v 1 + m 2 v 2 ) = ( m 1 + m 2 ) v co mm o n v e l oc i t y
v c o m m o n v e l o c i t y = ( m 1 v 1 + m 2 v 2 ) ( m 1 + m 2 ) v_{common \space velocity}= \frac{(m_1v_1+m_2v_2)}{(m_1+m_2)} v co mm o n v e l oc i t y = ( m 1 + m 2 ) ( m 1 v 1 + m 2 v 2 )
v c o m m o n v e l o c i t y = ( 350 ∗ 7.41 ) ( 350 + 80 ) = 6.03 m / s v_{common \space velocity}= \frac{(350*7.41)}{(350+80)}=6.03 m/s v co mm o n v e l oc i t y = ( 350 + 80 ) ( 350 ∗ 7.41 ) = 6.03 m / s
ii percentage loss of kinetic energy after the impact
% L o s s = K E 1 − K E 2 K E 1 = 0.5 m 1 v 1 2 − 0.5 m 12 v 1 2 0.5 m 1 v 1 2 = m 1 v 1 2 − m 12 v 1 2 m 1 v 1 2 \% 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} % L oss = K E 1 K E 1 − K E 2 = 0.5 m 1 v 1 2 0.5 m 1 v 1 2 − 0.5 m 12 v 1 2 = m 1 v 1 2 m 1 v 1 2 − m 12 v 1 2
= 350 ∗ 7.4 1 2 − 430 ∗ 6.0 3 2 350 ∗ 7.4 1 2 ∗ 100 = 18.64 % = \frac{350*7.41^2-430*6.03^2}{350*7.41^2}*100=18.64\% = 350 ∗ 7.4 1 2 350 ∗ 7.4 1 2 − 430 ∗ 6.0 3 2 ∗ 100 = 18.64%
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