(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 tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\(^{-1}\) just before the towing rope becomes taut. Calculate the:
(i) Speed of the tractor immediately the rope becomes taut
(ii) Loss in kinetic energy of the system just after the car has started Moving;
(iii) Impulse in the rope when it jerks the car into motion.
(a)(i) Linear momentum is the product of the mass of a body and its velocity.
(ii) Impulse is the product of a large force and short time for which the force acts.
(b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system.
(c) (i) \(m_1u_1 + m_2U_2 = v(m_1 + m_2)\) 5 x 10\(^{3}\) x 3 + 0
= v(5 x 10\(^{3}\) + 2.5 x 10\(^{3}\)) v = \(\frac{15 \times 10^3}{7.5 \times 10^3}\) = 2m/s
(ii) Initial K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} MV^2 = \frac{1}{2} times 5 \times 10^{3} \times 3^2\)
= 2.25 x 10\(^4\)
Final K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} (5 + 2.5) \times 10^3 \times 2^2\)
= 1.5 x 10\(^{4}\)J
Loss in K.E of the couple = (2.25 - 1.5) x 10\(^4\)
= 75 x 10\(^3\)J
(iii) Impulse = (\(mu_1U_1 + m_2U_2\)) - v
(\(mu_1 +m_2) (0 + 5 \times 10^3) - 2 (2.25 + 5) 10^3\) = 0Ns
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