A body of mass 5,kg strikes another body of m | vedclass

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(A) Let the mass of the first body be $m_1 = 5\,kg$ and its initial velocity be $u$. Let the mass of the second body be $m_2 = 2.5\,kg$ with initial v

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(A) Let the mass of the first body be $m_1 = 5\,kg$ and its initial velocity be $u$. Let the mass of the second body be $m_2 = 2.5\,kg$ with initial velocity $u_2 = 0$.By the law of conservation of linear momentum: $m_1 u + m_2 u_2 = (m_1 + m_2) v$,where $v$ is the final common velocity.$5u + 0 = (5 + 2.5)v \Rightarrow 5u = 7.5v \Rightarrow v = \frac{5}{7.5}u = \frac{2}{3}u$.The final kinetic energy of the combined mass is $K_f = \frac{1}{2}(m_1 + m_2)v^2 = 5\,J$.Substituting the values: $\frac{1}{2} 	imes 7.5 	imes (\frac{2}{3}u)^2 = 5$.$3.75 	imes \frac{4}{9}u^2 = 5 \Rightarrow \frac{15}{9}u^2 = 5 \Rightarrow \frac{5}{3}u^2 = 5 \Rightarrow u^2 = 3$.The initial kinetic energy of the first body is $K_i = \frac{1}{2}m_1 u^2 = \frac{1}{2} 	imes 5 	imes 3 = 7.5\,J$.

$A$ body of mass $5\,kg$ strikes another body of mass $2.5\,kg$ initially at rest. The bodies after collision coalesce and begin to move as a whole with a kinetic energy of $5\,J$. The kinetic energy of the first body before collision is ............... $J$.

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