Answer :
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/s
Explanation:
Here in collision type of questions we can use momentum conservation because there is no force on the system
So 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
