Answer: (b)
Explanation:
Given
mass of Student M=40 kg
mass of box m=3 kg
velocity of box v=8 m/s
suppose u is the velocity of the student after the throw
As there is no external force is applied so momentum is conserved
Initial momentum=final momentum=0
0=Mu+mv
Mu=-mv
[tex]u=-\frac{3}{40}\times 8=-\frac{3}{5}\\u=-0.6\ m/s[/tex]
Here negative sign indicates the velocity is opposite in direction i.e. opposite to the direction of the box