Physics 30 Dec 06 Diploma Exam

Significant digits or rounding physics 30 june 1999

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Unformatted text preview: he kinetic energy of the regular bullet. • How much energy is released by the explosion of the gunpowder if the transfer of energy from the explosion to the regular bullet is 90.0% efficient? • The regular bullet is in the rifle barrel for 1.42 × 10–3 s. What is the average force exerted on the regular bullet by the expanding gases? Physics 30 – June 1999 viii Use this additional information to answer the next part of the question. A second test performed by the company has the regular bullet strike the vest at a glancing angle. The mass of the vest and the dummy is 56.0 kg. The bullet–vest collision is inelastic. vBullet = 98.0 m/s mBullet = 8.00 g Vest 120° Path of bullet Dummy vBullet = 650 m/s • Determine the resultant speed of the vest and the dummy following the glancing collision shown above. Clearly communicate your understanding of the physics principles that you are using to solve this question. You may communicate this understanding mathematically, graphically, and/or with written statements. Sample Solution • Energy of the armour-piercing bullet compared to energy of the regular bullet. Method 1: EK AP EK 1 2 B 2 (8.00 × 10−3 kg • 1.2)(1.2 • 650 m/s) 1 2 (8.00 × 10 − 3 kg)(650 m/s) 2 EK AP EK = = 1 2 mAPvAP2 1 2 mBvB2 1.2(1.2)2 1 = 1.73 B The armour-piercing bullet has 1.73 times as much kinetic energy as the regular bullet. Physics 30 – June 1999 ix Method 2: EK B = = 1m v 2 2 BB 1 (8.00 × 2 10–3 kg)(650 m/s)2 = 1.69 × 103 J EK AP = 1 [(1.2)(8.00 2 3 × 10–3 kg)][(650 m/s)(1.2)] 2 = 2.92 × 10 J The EK or The EK AP AP is greater by 1.23 × 103 J is 1.73 times greater • The energy released by the explosion of the gunpowder at 90% efficiency. Ek = = 1 2 1 2 mv2 (8.00 × 10 −3 kg)(650 m/s) 2 Ek = 1.69 × 103 J 90% of Ep = Ek Ek = energy of gunpowder 0.9 1.69 × 103 J = 1.88 × 103 J 0.9 The energy released by the gunpowder is 1.88 × 103 J. • The average force exerted on the regular bullet by the expanding gases. F∆t = m∆v F = m∆v t (8.00 × 10 −3 kg)(650 m/s − 0) F= (1.42 × 10 − 3 s) F = 3.66 × 103 N The force of the expanding gas on the bullet is 3.66 × 103 N. Physics 30 – June 1999 x • The resultant velocity of the vest and dummy. Conservation of momentum in 2D implies two separate sets of calculations. Momentum is conserved in X direction and momentum is conserved in Y direction. ′ pB = mv = (8.00 × 10–3 kg)(98.0 m/s) ′ pB = 0.784 N.s ′ ′ pBy = pB sin 60$ ′ pB y = 0.784 sin 60° = 0.67896 kg.m/s ′ pB x = p′ cos 60$ B ′ pB x = 0.784 cos 60° = 0.392 kg.m/s ′ ′ pB x + pD x = pB x + pD x ′ (8.00 × 10–3 kg)(650 m/s) + 0 = (0.392 kg.m/s) + pD x ′ (5.20 kg.m/s) – (0.392 kg.m/s) = pD x ′ 4.808 kg.m/s = pD x ′ ′ pB y + pD y = pB y + pD y ′ ′ 0 + 0 = pB y + pD y Therefore: ′ ′ pD y = – pBy ′ pD y = –0.67896 kg.m/s Physics 30 – June 1999 xi To find the resultant vector from x and y components, use Pythagoras: pD′ = = ′ ( pD x ) 2 + ( p′ y ) 2 D (4.808)2 + (0.67896)2 pD′ = 4.85 kg.m/s pD′ = mv′ 4.85 kg⋅m/s 56.0 kg v′ = 0.0867 m/s v′ = The vest and dummy will have a speed of 8.67 × 10-2 m/s. Physics 30 – June 1999 xii...
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