Question

a body of mass 10kg collides with a stationary body of mass 4kg elastically find the fraction of connecting kinetic energy loss by the first is​

Answer 1

Change in kinetic energy of the first body is,

$\longrightarrow\sf{\Delta K=\dfrac{1}{2}\cdot10[(v_1)^2-(u_1)^2]}$

$\longrightarrow\sf{\Delta K=5[(v_1)^2-(u_1)^2]}$

By momentum conservation,

$\longrightarrow\sf{10u_1+4(0)=10v_1+4v_2}$

$\longrightarrow\sf{v_2=\dfrac{5(u_1-v_1)}{2}}$

Since collision is elastic, kinetic energy is conserved.

$\longrightarrow\sf{\dfrac{1}{2}\cdot10(u_1)^2+\dfrac{1}{2}\cdot4(0)^2=\dfrac{1}{2}\cdot10(v_1)^2+\dfrac{1}{2}\cdot4(v_2)^2}$

$\longrightarrow\sf{10(u_1)^2=10(v_1)^2+4\left(\dfrac{5(u_1-v_1)}{2}\right)^2}$

$\longrightarrow\sf{10(u_1)^2=10(v_1)^2+25(u_1-v_1)^2}$

From this,

$\longrightarrow\sf{\Delta K=\dfrac{25}{2}(u_1-v_1)^2}$

But since it is an elastic collision,

$\longrightarrow\sf{v_1=\left(\dfrac{10-4}{10+4}\right)u_1+\left(\dfrac{2\times4}{10+4}\right)0}$

$\longrightarrow\sf{v_1=\dfrac{3}{7}\ u_1}$

Therefore,

$\longrightarrow\sf{\underline{\underline{\Delta K=\dfrac{200}{49}(u_1)^2}}}$