A wooden block of mass M rests on a table over a large hole as in Figure P6.84. A bullet of mass m with an initial velocity v i , is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of h . (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this topic. (b) Find an expression for the initial velocity of the bullet. Figure P6.84 Problems 84 and 85.
A wooden block of mass M rests on a table over a large hole as in Figure P6.84. A bullet of mass m with an initial velocity v i , is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of h . (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this topic. (b) Find an expression for the initial velocity of the bullet. Figure P6.84 Problems 84 and 85.
Solution Summary: The author explains how the initial velocity of a bullet can be found using the conservation of momentum and mechanical energy.
A wooden block of mass M rests on a table over a large hole as in Figure P6.84. A bullet of mass m with an initial velocity vi, is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of h. (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this topic. (b) Find an expression for the initial velocity of the bullet.
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12. A stone is dropped from the top of a cliff. It is seen to hit the ground below
after 3.55 s. How high is the cliff?
13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming
no air resistance, what is the speed of the ball just before it strikes the ground?
14. Estimate (a) how long it took King Kong to fall straight down from the top
of the Empire State Building (280m high), and (b) his velocity just before
"landing".
Useful equations
For Constant Velocity:
V =>
D
X = V₁t + Xo
For Constant Acceleration:
Vr = V + at
X = Xo+Vot +
v=V+2a(X-Xo)
\prom = V +V
V velocity
t = time
D Distance
X = Final Position
Xo Initial Position
V = Final Velocity
Vo Initial Velocity
a = acceleration
For free fall
Yf
= Final Position
Yo Initial Position
g = 9.80
m
$2
For free fall:
V = V + gt
Y=Yo+Vo t +
+gt
V,² = V₁²+2g (Y-Yo)
V+Vo
Vprom=
2
6
Solve the problems
A 11 kg weight is attached to a spring with constant k = 99 N/m and subjected to an external force
F(t) =-704 sin(5t). The weight is initially displaced 4 meters above equilibrium and given an
upward velocity of 5 m/s. Find its displacement for t> 0.
y(t)
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Applications and Investigations in Earth Science (9th Edition)
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