(a) State the principle of conservation of momentum. (b) Ball A moves with speed v along a horizontal frictionless surface towards a stationary ball B, as shown in Fig. 3.1. [Figure 3.1] Ball A has mass 4.0 kg and ball B has mass 12 kg. The balls collide and then move apart as shown in Fig. 3.2. [Figure 3.2 (not to scale)] Ball A has velocity 6.0 m s⁻¹ at an angle of θ to the direction of its initial path. Ball B has velocity 3.5 m s⁻¹ at an angle of 30° to the direction of the initial path of ball A.
- (a)State the principle of conservation of momentum.[2]
- (b)(i)By considering the components of momentum at right-angles to the direction of the initial path of ball A, calculate θ.[3]
- (b)(ii)Use your answer in (i) to show that the initial speed v of ball A is 12 m s⁻¹. Explain your working.[2]
- (b)(iii)By calculation of kinetic energies, state and explain whether the collision is elastic or inelastic.[3]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- 3(a) sum/total momentum (of system of bodies) is constant M1 or sum/total momentum before = sum/total momentum after for an isolated system/no (resultant) external force A1
- 3(b)(i) p = mv C1 (4.0 x 6.0 x sin θ) – (12 × 3.5 × sin 30°) = 0 M1 or (m_A v_A × sinθ) – (m_B v_B × sin 30°) = 0 θ = 61° A1
- 3(b)(ii) shows the horizontal momentum component of ball A or of ball B as (4.0 × 6.0 × cos θ) or (12 × 3.5 × cos 30°) C1 (4.0 × 6.0 × cos 61°) + (12 × 3.5 × cos 30°) = 4.0v so v = 12 (ms⁻¹) A1
- 3(b)(iii) initial E_k (= ½ × 4.0 × 12²) = 290 (288) (J) M1 final E_k (= ½ × 4.0 × 6.0² + ½ × 12 × 3.5²) = 150 (145.5) (J) M1 (initial E_k > final E_k) so inelastic [both M1 marks required to award this mark] A1
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