Calculate the energy loss when a 5 kg object moving at 5 ms^(−1) collides head −on with and sticks to a stationary 3 kg object. [Answer: 3.75 J] I need help. How do I arrive at this answer?


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Question Number 130066 by Don08q last updated on 22/Jan/21

$Calculate the energy loss when a 5 k g$ $object moving at 5 {m s}^{- 1} collides head$ $- on with and sticks to a stationary$ $3 k g object .$ $[A n s w e r : 3.75 J]$ $I n e e d h e l p . How do I arrive at this$ $answer ?$
Commented by mr W last updated on 22/Jan/21

$a n s w e r i s w r o n g!$
	Answered by ajfour last updated on 22/Jan/21

	$M u = (M + m) v$ $e n e r g y l o s s = \frac{1}{2} {M u}^{2} - \frac{1}{2} (M + m) {(\frac{M}{M + m})}^{2} u^{2}$ $= \frac{{M u}^{2}}{2} (1 - \frac{M}{M + m})$ $= \frac{{m M u}^{2}}{2 (M + m)} = \frac{(3 k g) (5 k g) {(5 {m s}^{- 1})}^{2}}{2 (5 k g + 3 k g)}$ $= \frac{3 \times 125}{16} J = \frac{375}{16} J$ $= 23 ∙ 437 J$
	Commented by Don08q last updated on 23/Jan/21

	$T h a n k y o u S i r$
	Answered by mr W last updated on 22/Jan/21

	$m_{1} = 5 k g$ $m_{2} = 3 k g$ $v_{1 i} = 5 m / s$ $v_{2 i} = 0 m / s$ $v_{1 f} = v_{2 f} = v_{f}$ $m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}$ $\Rightarrow m_{1} v_{1 i} = (m_{1} + m_{2}) v_{f}$ $\Rightarrow v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1 i} = \frac{5 \times 5}{5 + 3} = \frac{25}{8} m / s$ $l o s s o f e n e r g y :$ $Δ E = \frac{1}{2} m_{1} v_{1 i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}$ $= \frac{5 \times 5^{2}}{2} - \frac{(5 + 3) \times 25^{2}}{2 \times 8^{2}} = 23.4375 J$
	Commented by Don08q last updated on 22/Jan/21

	$T h a n k y o u S i r$
	Answered by Don08q last updated on 22/Jan/21

	$someone sent me this solution$ $K E = \frac{p^{2}}{2 m}$ $since momentum is conserved,$ $m_{1} u_{1} = (m_{1} + m_{2}) v = p$ $\Rightarrow p = 5 k g \times 5 {m s}^{- 1} = 25 {k g m s}^{- 1}$ ${K E}_{i} = \frac{p^{2}}{2 m_{1}}$ $Missing \left or extra \right$ $Energy loss = {K E}_{i} - {K E}_{f}$ $= \frac{p^{2}}{2} (\frac{1}{m_{1}} - \frac{1}{m_{1} + m_{2}})$ $= \frac{{(25)}^{2}}{2} (\frac{1}{5} - \frac{1}{5 + 3})$ $= 23.4375 J$
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