Respuesta :
(M1xVi1)+(M2xVi2)=(M1xVf1)+(M2xVf2)
(.08x.5)+(.05x0)=(.08x-.1)+(.05xVf2)
Marble B's resulting velocity is .64
(.08x.5)+(.05x0)=(.08x-.1)+(.05xVf2)
Marble B's resulting velocity is .64
Explanation:
Given that,
Mass of marble A, [tex]m_A=0.08\ kg[/tex]
Mass of marble B, [tex]m_B=0.05\ kg[/tex]
Marble B is still, [tex]u_B=0[/tex]
Initial speed of marble A, [tex]u_A=0.5\ m/s[/tex]
Final speed of marble A after the collision, [tex]v_A=-0.1\ m/s[/tex]
We need to find the resulting velocity of marble B after the collision. It can be calculated using conservation of momentum as :
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bv_B[/tex]
[tex]0.08\times 0.5+0.05\times 0=0.08\times (-0.1)+0.05v_B[/tex]
[tex]v_B=\dfrac{0.032}{0.05}=0.64\ m/s[/tex]
So, the resulting velocity of marble B after the collision is 0.64 m/s. Hence, this is the required solution.