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Question Number 10620 by Saham last updated on 20/Feb/17

Commented by mrW1 last updated on 20/Feb/17

$p l e a s e c h e c k :$ $a f t e r c o l l i s i o n {\vec{V}}_{A} = - 5.0 i + 20 j (n o t - 5.0 j + 20 j)$ $a n s w e r (b) 500 J (n o t 525 J)$
	Answered by mrW1 last updated on 20/Feb/17

	$b e f o r e a f t e r$ $m_{A} {\vec{V}}_{A, 1} + m_{B} {\vec{V}}_{B, 1} = m_{A} {\vec{V}}_{A, 2} + m_{B} {\vec{V}}_{B, 2}$ $2 (15 i + 30 j) + 2 (- 10 i + 5 j) = 2 (- 5 i + 20 j) + 2 {\vec{V}}_{B, 2}$ $\Rightarrow {\vec{V}}_{B, 2} = (15 - 10 + 5) i + (30 + 5 - 20) j = 10 i + 15 j$ $V_{A, 1} =∣ {\vec{V}}_{A, 1} ∣= \sqrt{15^{2} + 30^{2}} = 15 \sqrt{5} m / s$ $V_{B, 1} =∣ {\vec{V}}_{B, 1} ∣= \sqrt{{(- 10)}^{2} + 5^{2}} = 5 \sqrt{5} m / s$ $V_{A, 2} =∣ {\vec{V}}_{A, 2} ∣= \sqrt{{(- 5)}^{2} + 20^{2}} = 5 \sqrt{17} m / s$ $V_{B, 2} =∣ {\vec{V}}_{B, 2} ∣= \sqrt{10^{2} + 15^{2}} = 5 \sqrt{13} m / s$ $K_{1} = K_{A, 1} + K_{B, 1} = \frac{1}{2} m_{A} V_{A, 1}^{2} + \frac{1}{2} m_{B} V_{B, 1}^{2}$ $= \frac{1}{2} \times 2 \times 1125 + \frac{1}{2} \times 2 \times 125 = 1250 J$ $K_{2} = K_{A, 2} + K_{B, 2} = \frac{1}{2} m_{A} V_{A, 2}^{2} + \frac{1}{2} m_{B} V_{B, 2}^{2}$ $= \frac{1}{2} \times 2 \times 425 + \frac{1}{2} \times 2 \times 325 = 750 J$ $Δ K = K_{2} - K_{1} = 750 - 1250 = - 500 J = L o s t$
	Commented by Saham last updated on 20/Feb/17

	$Thanks for everytime sir .$ $God bless you .$
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