A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg mass into it from directly above, and the cart continues moving.

Which equation best represents the horizontal momentum in this situation?

15vi + 2vi = 15vf + 2vf

15vi + 2 (0) = 15vf + 2vf

15vi + 2 (0) = (15 + 2) vf

15vi + 2vi = (15 + 2) vf

C. 15 vi + 2 (0) = (15 + 2) v f

The answer is 15vi + 2 (0) = 15vf + 2vfThe explanation behind this answer is:Pcart = Mcart x VcartPcart = 15viPstudent = 0 the meaning of this is the student doesn’t have momentum horizontal direction

Total horizontal momentum before the event is denoted as = 15vi + 0 = 15vi

The momentum after the event is: Pcart + student = (Mcart + Mstudent) vf

Knowing that the cart and student move together, we can denote this as:Pb = Pcart + student15vi = 2 (0) = 15vf + 2vf