Answer
c) is correct.
Work Step by Step
According to the conservation of linear momentum, $$m_1v_{01}+0=m_1v_{f1}+m_2v_{f2}$$ $$m_1v_{01}=m_1v_{f1}+m_2v_{f2}$$
Since $m_1=m_2$, we can eliminate all the masses: $$v_{01}=v_{f1}+v_{f2}(1)$$
Similarly, according to the conservation of kinetic energy, $$v_{01}^2=v_{f1}^2+v_{f2}^2(2)$$
Take (1) to the power of 2 and set it equal to (2); we have $$v_{f1}^2+v_{f2}^2+2v_{f1}v_{f2}=v_{f1}^2+v_{f2}^2$$ $$2v_{f1}v_{f2}=0$$ $$v_{f1}v_{f2}=0$$
This means either object 1's or object 2's final speed has to be zero.
Only c) has one of the objects stopping completely, so c) is correct.
