Could someone tell me whether this statement is true or false. This question appeared in a practise test and I think that the answer to it is incorrect.

Statement 1: If linear momentum of a system of particles is zero, then the kinetic energy of the system will also be zero.
Statement 2: Linear momentum of a particle is related with kinetic energy as K= P^2/2m

a) S-1 is True, S-2 is True; S-2 is a correct explanation for S-1
b) S-1 is True, S-2 is True; S-2 is a not correct explanation for S-1
c) S-1 is True, S-2 is False
d) S-1 is False, S-2 is True

I think the correct answer should be either a or b because I think that both the statements are at least correct, but the answer given is D, but it doesn't specify any reason.
It would be great if someone could help me out!

The answer is indeed D, because S-1 is false. The momentum of a system of particles is equal to the vector sum of each particle's momentun, so you need to take into account the direction of the particle's velocity. You can have two particules of equal mass (m) moving at the same speed but in opposite directions (at +v and -v respectively), and the total momentum of the system would be mv + m(-v) = 0. But the KE of a system is a scalar, so for this example you simply add the KE's of the two paticles together: KE = 1/2 mv^2 + 1/2 mv^2, which is not zero. Hence S-1 is false.

K=(PC)*(PC)/(E+E0)
K=Kinetic Energy
P=Momentom
E=Rilativistic Energy
E0=Rest Energy
C=Speed of the Light
in high speeds E0=0 soE=K & PC=K
but in other speeds
(PC)*(PC)=(K*(E+E0))
in law speeds E=E0 so
PC*PC=2KE PP=2KM

K=P²C²/(E+e)

K=Kinetic Energy

P=Momentom

E=Rilativistic Energy

e=Rest energy

C=Speed of the Light
in high speeds

e=0 so E=K & PC=K

but in other speeds

P²C²=K*(E+e)

in law speeds

E=e so

P²C²=2KE

P²=2KM
:o