A ball of mass $4 \; kg$ moving with a velocity $10 \; m/s$ collides with another body of mass $16 \; kg$ moving with $4 \; m/s$ from the opposite direction and then coalesces into a single body. Compute the loss of energy on impact.

Given,

Mass of first ball $(m_1) = 4\;kg$

Initial velocity of first ball $(u_1) = 10\;m/s$

Mass of second ball $(m_2) = 16\;kg$

Initial velocity of second ball $(u_2) = 4\;m/s$

Loss of energy on impact $(∇E) = \;?$

Since ball are moving in opposite direction.

Then, from the principle of conservation of linear momentum,

$m_1\;.\;u_1 \;+\; m_2\;.\;(-\;u_2) = (m_1\; +\; m_2)\;v$               [$v$ is the common velocity]

⇒ $v = $$\frac{m_1\;u_1\; -\; m_2\;u_2}{m_1\; + \;m_2} = \frac{4\;*\;10 \;- \;16\;*\;4}{4\; +\; 16}$ $\;\;\; =\;-\;1.2\;m/s$

[i.e. is in the direction of $m_2$]

Kinetic Energy before collision $E_1 = $$\frac{1}{2}$$\;m_1\;u_1^2\; + \;$$\frac{1}{2}$$\;m_2\;u_2^2$

$= $$\frac{1}{2}$$\;*\;4\;*\;10^2 \;+ \;$$\frac{1}{2}$$\;*\;16\;*\;4^2 = 328\;J $

Kinetic Energy after collision $E_f = $$\frac{1}{2}$$\;(m_1\;+\;m_2)\;v^2$

$= $$\frac{1}{2}$$\;(4\;+\;16)\;*\;(1.2)^2 = 14.4\;J$

Loss of energy on impact (∇E) = Kinetic Energy before collision - Kinetic Energy after collision.

= 328 - 14.4 = 313.6\;J

∴ Loss of energy is 313.6 J.

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