(a) Define (i) Linear momentum; (ii) Impulse (b) State the principle of conservation of linear momentum. (c) A tractor of mass 5.0 x 10\(^{3}\)kg is used a ...

(a) Definitions

(i) Linear momentum is the product of the mass of a body and its velocity: \(p = mv\). Its S.I. unit is kg m s^-1 (N s).

(ii) Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces: \(J = Ft = \Delta(mv)\). Its S.I. unit is N s.

(b) Principle of conservation of linear momentum

In a system of colliding bodies on which no external resultant force acts, the total linear momentum before impact is equal to the total linear momentum after impact.

(c) Calculations

Tractor: \(m_1 = 5.0\times10^{3}\) kg at \(u_1 = 3.0\) m s^-1; car: \(m_2 = 2.5\times10^{3}\) kg at rest. When the rope becomes taut they move together with common velocity v.

(i) Common speed after the rope is taut

\[ m_1 u_1 = (m_1 + m_2)v \] \[ v = \frac{5.0\times10^{3}\times3.0}{(5.0\times10^{3}+2.5\times10^{3})} = \frac{15\,000}{7\,500} = 2.0\ \text{m s}^{-1} \]

(ii) Loss in kinetic energy

\[ KE_i = \tfrac{1}{2}m_1 u_1^2 = \tfrac{1}{2}\times5.0\times10^{3}\times3.0^2 = 22\,500\ \text{J} \] \[ KE_f = \tfrac{1}{2}(m_1+m_2)v^2 = \tfrac{1}{2}\times7.5\times10^{3}\times2.0^2 = 15\,000\ \text{J} \] \[ \text{Loss} = 22\,500 - 15\,000 = 7\,500\ \text{J} \]

(iii) Impulse in the rope

Impulse equals the change in momentum of the car: \[ J = m_2 v - 0 = 2.5\times10^{3}\times2.0 = 5.0\times10^{3}\ \text{N s} \]