Question-10620

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

Commented by mrW1 last updated on 20/Feb/17

please check: after collision V_A ^→ =−5.0i+20j (not −5.0j+20j) answer (b) 500 J (not 525 J)

$${please}\:{check}: \\ $$$${after}\:{collision}\:\overset{\rightarrow} {{V}}_{{A}} =−\mathrm{5}.\mathrm{0}{i}+\mathrm{20}{j}\:\:\:\left({not}\:−\mathrm{5}.\mathrm{0}{j}+\mathrm{20}{j}\right) \\ $$$${answer}\:\left({b}\right)\:\mathrm{500}\:{J}\:\:\:\left({not}\:\mathrm{525}\:{J}\right) \\ $$

Answered by mrW1 last updated on 20/Feb/17

before after m_A V_(A,1) ^→ +m_B V_(B,1) ^→ =m_A V_(A,2) ^→ +m_B V_(B,2) ^→ 2(15i+30j)+2(−10i+5j)=2(−5i+20j)+2V_(B,2) ^→ ⇒V_(B,2) ^→ =(15−10+5)i+(30+5−20)j=10i+15j V_(A,1) =∣V_(A,1) ^→ ∣=(√(15^2 +30^2 ))=15(√5) m/s V_(B,1) =∣V_(B,1) ^→ ∣=(√((−10)^2 +5^2 ))=5(√5) m/s V_(A,2) =∣V_(A,2) ^→ ∣=(√((−5)^2 +20^2 ))=5(√(17)) m/s V_(B,2) =∣V_(B,2) ^→ ∣=(√(10^2 +15^2 ))=5(√(13)) m/s K_1 =K_(A,1) +K_(B,1) =(1/2)m_A V_(A,1) ^2 +(1/2)m_B V_(B,1) ^2 =(1/2)×2×1125+(1/2)×2×125=1250 J K_2 =K_(A,2) +K_(B,2) =(1/2)m_A V_(A,2) ^2 +(1/2)m_B V_(B,2) ^2 =(1/2)×2×425+(1/2)×2×325=750 J ΔK=K_2 −K_1 =750−1250=−500 J=Lost

$$\:\:\:\:\:{before}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{after} \\ $$$${m}_{{A}} \overset{\rightarrow} {{V}}_{{A},\mathrm{1}} +{m}_{{B}} \overset{\rightarrow} {{V}}_{{B},\mathrm{1}} ={m}_{{A}} \overset{\rightarrow} {{V}}_{{A},\mathrm{2}} +{m}_{{B}} \overset{\rightarrow} {{V}}_{{B},\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{15}{i}+\mathrm{30}{j}\right)+\mathrm{2}\left(−\mathrm{10}{i}+\mathrm{5}{j}\right)=\mathrm{2}\left(−\mathrm{5}{i}+\mathrm{20}{j}\right)+\mathrm{2}\overset{\rightarrow} {{V}}_{{B},\mathrm{2}} \\ $$$$\Rightarrow\overset{\rightarrow} {{V}}_{{B},\mathrm{2}} =\left(\mathrm{15}−\mathrm{10}+\mathrm{5}\right){i}+\left(\mathrm{30}+\mathrm{5}−\mathrm{20}\right){j}=\mathrm{10}{i}+\mathrm{15}{j} \\ $$$$ \\ $$$${V}_{{A},\mathrm{1}} =\mid\overset{\rightarrow} {{V}}_{{A},\mathrm{1}} \mid=\sqrt{\mathrm{15}^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} }=\mathrm{15}\sqrt{\mathrm{5}}\:{m}/{s} \\ $$$${V}_{{B},\mathrm{1}} =\mid\overset{\rightarrow} {{V}}_{{B},\mathrm{1}} \mid=\sqrt{\left(−\mathrm{10}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{5}}\:{m}/{s} \\ $$$${V}_{{A},\mathrm{2}} =\mid\overset{\rightarrow} {{V}}_{{A},\mathrm{2}} \mid=\sqrt{\left(−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{17}}\:{m}/{s} \\ $$$${V}_{{B},\mathrm{2}} =\mid\overset{\rightarrow} {{V}}_{{B},\mathrm{2}} \mid=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{13}}\:{m}/{s} \\ $$$$ \\ $$$${K}_{\mathrm{1}} ={K}_{{A},\mathrm{1}} +{K}_{{B},\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{{A}} {V}_{{A},\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{{B}} {V}_{{B},\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{1125}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{125}=\mathrm{1250}\:{J} \\ $$$$ \\ $$$${K}_{\mathrm{2}} ={K}_{{A},\mathrm{2}} +{K}_{{B},\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{{A}} {V}_{{A},\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{{B}} {V}_{{B},\mathrm{2}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{425}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{325}=\mathrm{750}\:{J} \\ $$$$ \\ $$$$\Delta{K}={K}_{\mathrm{2}} −{K}_{\mathrm{1}} =\mathrm{750}−\mathrm{1250}=−\mathrm{500}\:{J}={Lost} \\ $$

Commented by Saham last updated on 20/Feb/17

Thanks for everytime sir. God bless you.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{everytime}\:\mathrm{sir}.\: \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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