

Explanation
(a)(i) The momentum of a body is the product of its mass and its velocity.
(ii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.
(b) Mp Up + Mq Uq = Mp Vp + Mq Vq
0.25 Up + 0 = 0.25 x \(\frac{2}{3}\) Up + 0.25 x 2; 0.25
(Up - 2/3 Up) = 0.50; 1/3
Up = \(\frac{0.50}{0.25}\) = 2m/s
Up = 2 x 3 = 6m/s
(c)(i) Newton's second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force
(ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time
= \(\frac{mv - mu}{t}\)
From second law: \(\frac{mv - mu}{t}\) = KF
\(\frac{m(v - u)}{t}\) = KF
If m = 1kg, a = 1ms\(^{-2}\) and F = 1N, K = 1
ma = F
(iii) Resultant force on the vehicle = 9600 - 200 = 7400N.
But F = ma
a = \(\frac{F}{m} = \frac{7400}{3400}\) = 2.18ms\(^{-2}\) or 2.2ms\(^{-2}\)

