(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.
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]
Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]
since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]
Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]
since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]
