[latex]Dane:\ v_1=3frac{m}{s}\ v_2=2frac{m}{s}\ m_2=25 kg\ Oblicz:\ m_1=?\ Rozwiazanie:\ p_1=p_2\ m_1*v_1=(m_1+m_2)*v_2\\ m_1*v_1=m_1*v_2+m_2*v_2\\ m_1*v_1-m_1*v_2=m_2*v_2\\ m_1(v_1-v_2)=m_2*v_2[/latex] [latex]m_1=frac{m_2*v_2}{v_1-v_2}\\ m_1=frac{25*2}{3-2}\\ m_1=frac{50}{1}\\ m_1=50kg[/latex] Masa Moniki wynosi 50kg. Pozdrawiam, Adam
Dane: [latex]V_{1}=3frac{m}{s}\ V_{2}=2frac{m}{s}\ m_{l}=25kg\ m_{m}=?\ \ Z zasady zachowania pedu:\ p_{1}=p_{2}\ p_{1}-ped Moniki\ p_{2}-ped Moni z lodka\ \ p=mV\ p-ped\ m-masa\ V-predkosc[/latex] [latex]\ \ p_{1}=p_{2}\ m_{m}V_{1}=(m_{m}+m_{l})V_{2}\ m_{m}-masa moni\ m_{l}-masa lodki\ \ m_{m}*3=(m_{m}+25)2\ 3m_{m}=2m_{m}+50\ \ m_{m}=50[kg][/latex]
