Yahoo Answers: Answers and Comments for Tough Physics Question  Grade 12  Kinematics/Forces/Energy all in one  Please show steps? [Physics]
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From Anonymous
enCA
Tue, 03 Dec 2013 06:17:14 +0000
3
Yahoo Answers: Answers and Comments for Tough Physics Question  Grade 12  Kinematics/Forces/Energy all in one  Please show steps? [Physics]
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https://ca.answers.yahoo.com/question/index?qid=20131203061714AA0ByJe
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From Jim: This question involves application of the two ...
https://ca.answers.yahoo.com/question/index?qid=20131203061714AA0ByJe
https://ca.answers.yahoo.com/question/index?qid=20131203061714AA0ByJe
Tue, 03 Dec 2013 06:42:51 +0000
This question involves application of the two conservation principles:
Conservation of Momentum is applicable when bullet collides into wood block.
Conservation of Energy is applicable when block/bullet compress (ideal) spring.
Since the question asks for the initial speed of bullet let that unknown = v
and
Let the speed of block/bullet combination = V
The momentum of block/bullet combination = (0.045 + 8.3)V = 8.345V
The KE of block/bullet combination = 1/2(8.345)V² = 4.1725V²
SPE = 1/2kx² = (0.5)(76)(0.28)² = 2.9792 J
by conservation of energy => 2.9792 = 4.1725V²
V² = 0.714008388
V = 0.84499 m/s
The momentum of block/bullet combination = (8.345)(0.84499) = 7.05144 kgm/s
The momentum of bullet prior to hitting block = mv = 0.045v
by conservation of momentum => 0.045v = 7.05144
v = 7.05144/0.045 = 157 m/s ANS

From Whome: To find the initial velocity of the bullet and...
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https://ca.answers.yahoo.com/question/index?qid=20131203061714AA0ByJe
Tue, 03 Dec 2013 08:26:16 +0000
To find the initial velocity of the bullet and block after impact we apply conservation of energy principles
The maximum spring potential plus the energy lost to friction will equal the initial kinetic energy
KE = PS + U
½mv² = ½kx² + Fd
as the spring is resting against the block before impact, then d = x
F will be the friction force or
F = μmg
½mv² = ½kx² + μmgx
v² = kx²/m + 2μgx
v = √(kx²/m + 2μgx)
v = √(76(0.28²) / (8.3 + 0.045) + 2(0.24)9.81(0.28))
v = 2.032 m/s
Now we use conservation of momentum to find the initial bullet velocity
mu1 + Mu2 = (m + M)v
as the block initial velocity, u2, is zero
u1 = (m + M)v / m
u1 = (0.045 + 8.3)2.032 / 0.045
u1 = 377 m/s