Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com