Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of –0.1 m/s. What is the resulting velocity of marble B after the collision?

Answer:The resulting velocity of marble B after the collision is 0.96 m/s.Explanation:Mass of marble A ,[tex]m_1= 0.08 kg[/tex]Initial velocity of the marble A,[tex]u_1 = 0.5 m/s[/tex]Mass of marble B,[tex]m_2 = 0.05 kg[/tex]Initial velocity of the marble B,[tex]u_2 = 0 m/s[/tex]final velocity of the marble A,[tex]v_1 = -0.1 m/s[/tex]final velocity of the marble B,[tex]v_2 = ?[/tex]Applying law of conservation of mass:[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex][tex]0.08 kg\times 0.5 m/s+0.05kg\times 0 m/s=0.08 kg\times (-0.1 m/s)+0.05kg \times v_2[/tex][tex] v_2=0.96 /s[/tex]The resulting velocity of marble B after the collision is 0.96 m/s.

Answer:the other ball will move with speed 0.96 m/sExplanation:Here in collision type of questions we can use momentum conservation because there is no force on the systemSo here we will have[tex]m_a v_{1i} + m_b v_{2i} = m_a v_{1f} + m_b v_{2f}[/tex]so here we have[tex]0.08(0.5) + 0.05(0) = 0.08(-0.1) + 0.05 v[/tex][tex]0.048 = 0.05 v[/tex][tex]v = 0.96 m/s[/tex]So the light ball will move with speed 0.96 m/s