Ashell that is initially at rest explodes into two fragments, one fragment 25 times heavier than the other. if any gas from the explosion has negligible mass, then:
a. the momentum change of the lighter fragment is exactly the same as the momentum change of the heavier fragment.
b. the kinetic energy change of the lighter fragment is 25 times as great as the kinetic energy change of the heavier fragment.
c. the momentum change of the lighter fragment is 25 times as great as the momentum change of the heavier fragment.
d. the kinetic energy change of the heavier fragment is 25 times as great as the kinetic energy change of the lighter fragment.
e. the momentum change of the heavier fragment is 25 times as great as the momentum change of the lighter fragment.
f. the kinetic energy change of the lighter fragment is exactly the same as the kinetic energy change of the heavier fragment.

(a) the momentum change is the same in magnitude but in opposite direction

(b) the kinetic energy change of the lighter fragment is 25 times as great as the kinetic energy change of the heavier fragment.

Explanation:

since there is no force applied to the shell , when it explodes

F= d(mv)/dt =0 therefore mv=constant

where F= force , m=mass , v=velocity , t=time

since the momentum is conserved , and denoting the heavier element as H and lighter as L

(mH+mL)* v initial  = mH*vH+ mL*vL

since the system is initially at rest v initial = 0, then

mH*vH+ mL*vL = 0 →   mL*vL = -mH*vH = M

therefore the momentum change if the lighter fragment is the same as the heavy one (a) in magnitude (but not in direction) , not (e) or (c)

since

- vL/vH= mH/mL = 25

the ratio of kinetic energy

KH/KL = 1/2 mH*vH² / (1/2 mL*vL)² = (mH/mL)*(vH/vL)² = 25* (1/25)² = 1/25

therefore

KL = 25 KH

thus (b) is also correct and not (d)

i think the answer would be b

mass and volume are the answers you are looking for