Forensic science . Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.50.) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 cm long, causing the block to swing upward to a maximum angle of 34.7° from the vertical. What was the original speed of this bullet? Figure 8.50 Problem 65. 8.65. Set Up: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. Let + x be to the right. Solve: Collision : P i, x = P f, x gives (4.20 × 10 −3 kg) v = (4.20 × 10 −3 kg + 2.50 kg) v ′ Motion after the collision : The kinetic energy of the block immediately after the collision is converted entirely to gravitational potential energy at the maximum angle of swing. The figure below shows that the maximum height h the block swings up is given by h = (0.750 m)(l − cos34.7°) = 0.133 m. Conservation of energy gives 1 2 m tot v ′ 2 = m tot g h and v ′ = 2 g h = 1.61 m / s . Then the conservation of momentum equation gives v = ( 4.20 × 10 − 3 kg + 2.50 kg 4.20 × 10 − 3 kg ) ( 1.61 m/s ) = 960 m/s Reflect: The original speed of the bullet is nearly three times the speed of sound in air. Kinetic energy is not conserved in the collision. Our analysis assumes that the block moves very little during the collision, while the bullet is coming to rest relative to the block. This is a good approximation since the velocity the block gets in the collision is much less than the initial velocity of the bullet.
Forensic science . Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.50.) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 cm long, causing the block to swing upward to a maximum angle of 34.7° from the vertical. What was the original speed of this bullet? Figure 8.50 Problem 65. 8.65. Set Up: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. Let + x be to the right. Solve: Collision : P i, x = P f, x gives (4.20 × 10 −3 kg) v = (4.20 × 10 −3 kg + 2.50 kg) v ′ Motion after the collision : The kinetic energy of the block immediately after the collision is converted entirely to gravitational potential energy at the maximum angle of swing. The figure below shows that the maximum height h the block swings up is given by h = (0.750 m)(l − cos34.7°) = 0.133 m. Conservation of energy gives 1 2 m tot v ′ 2 = m tot g h and v ′ = 2 g h = 1.61 m / s . Then the conservation of momentum equation gives v = ( 4.20 × 10 − 3 kg + 2.50 kg 4.20 × 10 − 3 kg ) ( 1.61 m/s ) = 960 m/s Reflect: The original speed of the bullet is nearly three times the speed of sound in air. Kinetic energy is not conserved in the collision. Our analysis assumes that the block moves very little during the collision, while the bullet is coming to rest relative to the block. This is a good approximation since the velocity the block gets in the collision is much less than the initial velocity of the bullet.
Forensic science. Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.50.) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 cm long, causing the block to swing upward to a maximum angle of 34.7° from the vertical. What was the original speed of this bullet?
Figure 8.50
Problem 65.
8.65. Set Up: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. Let +x be to the right.
Motion after the collision: The kinetic energy of the block immediately after the collision is converted entirely to gravitational potential energy at the maximum angle of swing. The figure below shows that the maximum height h the block swings up is given by h = (0.750 m)(l − cos34.7°) = 0.133 m.
Conservation of energy gives
1
2
m
tot
v
′
2
=
m
tot
g
h
and
v
′
=
2
g
h
=
1.61
m
/
s
.
Then the conservation of momentum equation gives
v
=
(
4.20
×
10
−
3
kg
+
2.50
kg
4.20
×
10
−
3
kg
)
(
1.61
m/s
)
=
960
m/s
Reflect: The original speed of the bullet is nearly three times the speed of sound in air. Kinetic energy is not conserved in the collision. Our analysis assumes that the block moves very little during the collision, while the bullet is coming to rest relative to the block. This is a good approximation since the velocity the block gets in the collision is much less than the initial velocity of the bullet.
Scientific analysis of crime scene evidence to determine the events of a crime. Forensic science includes analyzing biological evidence, such as DNA fingerprinting, and non-biological evidence, such as computer forensics.
Part C
Find the height yi
from which the rock was launched.
Express your answer in meters to three significant figures.
Learning Goal:
To practice Problem-Solving Strategy 4.1 for projectile motion problems.
A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m before hitting the ground. From what height was the rock thrown? Use the value g = 9.800 m/s2 for the free-fall acceleration.
PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems
MODEL: Is it reasonable to ignore air resistance? If so, use the projectile motion model.
VISUALIZE: Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Define symbols and identify what the problem is trying to find. For a launch at angle θ, the initial velocity components are vix=v0cosθ and viy=v0sinθ.
SOLVE: The acceleration is known: ax=0 and ay=−g. Thus, the problem becomes one of…
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