Question #247344

The bumper cars A and B in Fig. Q5 below each have a mass of 150 kg and are coasting with the velocities shown before they freely collide head on. If no energy is lost during the collision, determine their velocities after collision. [10]


1
Expert's answer
2021-10-08T09:23:57-0400

ma=mb=150  kg(vA)1=3  m/s(vB)1=2  m/sm_a=m_b = 150 \; kg \\ (v_A)_1=3 \;m/s \\ (v_B)_1 = 2 \;m/s

No energy lost

El=0e=1e=[(VB)2(VA)2](VB)1(VA)1E_l=0 \\ e=1 \\ e = \frac{-[(V_B)_2 -(V_A)_2]}{(V_B)_1 -(V_A)_1}

A and B are in opposite direction

(VA)1=3  m/s(VB)1=2  m/s1=[(VB)2(VA)2]23(VB)2(VA)2=5(V_A)_1 = 3 \;m/s \\ (V_B)_1 = -2\;m/s \\ 1 = \frac{-[(V_B)_2 -(V_A)_2]}{-2-3} \\ (V_B)_2 -(V_A)_2 = 5

Conservation of momentum

ma(VA)1+mb(VB)1=ma(VA)2+mb(VB)2ma=mb(VA)1(VB)1=(VA)2+(VB)232=(VA)2+(VB)21=(VA)2+(VB)25=(VB)2(VA)2(VA)2=2  m/s(VB)2=3  m/sm_a(V_A)_1 +m_b(V_B)_1 =m_a(V_A)_2 +m_b(V_B)_2 \\ m_a=m_b \\ (V_A)_1 -(V_B)_1 = (V_A)_2 +(V_B)_2 \\ 3-2 = (V_A)_2 +(V_B)_2 \\ 1 = (V_A)_2 +(V_B)_2 \\ 5 = (V_B)_2 -(V_A)_2 \\ (V_A)_2 = -2 \;m/s \\ (V_B)_2 = 3 \;m/s


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