A 2 kg lab cart is sliding across across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?

Question

Update 2: what i don't get is i did force=(m*change in velocity)/change in time, so 6=[2(v final-4)/2 and got 10.
Update 3: but its not right
Update 4: my teacher used m(v final- v initial)=F*change in time. why is that?

Answer 1

-->J = Impulse = F * t = change in momentum
--> Units for Impulse can be N*s or kg*m/s
--> p = momentum = m * v units are kg*m/s
--> East is +, West is -
The change in momentum is = m(v final- v initial)
Impulse = F * t so...
m(v final- v initial) = F * t
(2kg * v final) - (2kg * 4 m/s) = -[6N * 2s]
(2kg*v) - 8 kg m/s = -12 kg m/s
+8 kg m/s +8 kg m/s
(2kg*v) =-4 kg m/s
v= -2 m/s (the negative signifies West)
Answer=2 m/s West