Answer:
The 43kg student will be sliding at 1.79m/s opposite the direction the 34kg student is going.
Explanation:
Conservation of linear momentum!
The law of conservation of momentum says that in an isolated system, the momentum before must equal the momentum after:
[tex]mv_1+m_1v_2=m_1_v_{1f}+m_2v_{2f}[/tex].
For our two students
[tex](32kg)(v_1)+(43kg)(v_2)= (32kg)+(43kg)(-2.4m/s)+(43kg)(v_{2f})[/tex] (notice the - sign in -2.4m/s, this means going to the left)
since the students were not moving at first, [tex]v_1=v_2= 0[/tex], therefore we have
[tex]0= (32kg)(-2.4m/s)+(43kg)(v_{2f})[/tex]
solving for [tex]v_{2f}[/tex] gives
[tex]76.8=(43kg)(v_{2f})[/tex]
[tex]\boxed{v_{2f} = 1.79m/s}[/tex]
Hence the 43kg student will be sliding at 1.79m/s to the right.
