(a)(i) Define the term linear momentum.
(ii) State the law of conservation of linear momentum.
(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\(^{-1}\) in the original direction of P. Calculate the initial velocity of R
(c)(i) State Newton’s second law of motion.
(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.
(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.
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\(^{-2}\)