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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 armourpiercing 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 armourpiercing 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 × 102 m/s. Physics 30 – June 1999 xii...
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This note was uploaded on 01/14/2014 for the course PHYSICS Physics 30 taught by Professor Quinlan during the Fall '09 term at Centennial High School.
 Fall '09
 Quinlan
 Physics

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