Answer:
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping
Step-by-step explanation:
As we know that both mass have same horizontal force opposite to their motion
So we will have



now the stopping distance of an object moving with initial speed v is given as


so here we have


for other object we have


So correct answer will be
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping