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densk [106]
3 years ago
10

7. A 30.0-g rifle bullet traveling 185 m/s embeds itself in a 3.15-kg pendulum hanging on a 2.85-m-long string, which makes the

pendulum swing upward in an arc. Determine: a. The vertical component of the pendulum’s maximum displacement. (10pts) b. The horizontal component of the pendulum’s maximum displacement. (10pts) c. The angle of the pendulum’s maximum displacement with the vertical. (10pts)
Physics
1 answer:
EastWind [94]3 years ago
5 0

Answer:

Explanation:

c )

First of all we shall calculate the velocity of bullet just after the collision with the pendulum by applying conservation of momentum law.  

v₂ = mv₁ / ( m + M )

v₂ is velocity after the collision ,  m is mass of bullet v₁ is velocity of bullet and M  is mass of pendulum.

v₂ = .030 x 185 / 3.18

= 1.745 m /s

Let the angle of the pendulum’s maximum displacement with the vertical be θ

height attained by the pendulum  h = L ( 1 - cosθ)   ; L is the length  of the string.

Applying conservation of mechanical energy law

mgh = 1/2 m v₂²

m is mass of (bullet+ pendulum)  , v₂ is its velocity

g L ( 1 - cosθ) = v₂² / 2

9.8 x 2.85 ( 1 - cosθ) = 1.745² / 2

( 1 - cosθ) = .0545

cosθ = .9455

θ = 19 degree

a ) The vertical component of the pendulum’s maximum displacement.

L ( 1 - cosθ)

= 2.85 ( 1 - .9455

= .155 m

b ) Horizontal component :   L sin18

= 3.15 x .30

= .97 m .

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3 years ago
Tarzan, whose mass is 103 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets
Anettt [7]

Answer:

   v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

Explanation:

For this exercise let's start by using Newton's second law

Y axis

        N-W = 0

        N = W

X axis

         fr = m a

the expression for the friction force is

         fr = μ N

we substitute

        μ mg = m a

        μ g = a

calculate us

         a = 0.620  9.8

         a = 6.076 m / s²

now we can use the kinematics relations

          v² = v₀² - 2 a x

suppose v = 0

          v₀ = \sqrt{2ax}Ra 2ax

let's calculate

         v₀ = \sqrt{2 \ 6076 \ 60}

         v₀ = 27.00 m / s

let's slow down to the english system

          v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)

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With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

6 0
3 years ago
A pendulum consists of a 1.7-kg block hanging on a 1.6-m length string. A 0.01-kg bullet moving with a horizontal velocity of 82
Rzqust [24]

Answer:

0.42 m

Explanation:

mass of pendulum, M = 1.7 kg

Length of pendulum , l = 1.6 m

mass of bullet, m = 0.01 kg

initial velocity of bullet, u = 828 m/s

final velocity of bullet, v = 340 m/s

initial velocity of pendulum, U = 0

Let the final velocity of pendulum is V.

Use conservation of momentum for bullet and the pendulum

m x u + M x U = m x v + M x V

0.01 x 828 + 1.7 x 0 = 0.01 x 340 + 1.7 x V

8.28 + 0 = 3.4 = 1.7 V

V = 2.87 m/s

Now the kinetic energy of the pendulum is converted into potential energy of pendulum and let it raised to a height of h from the initial level.

Use energy conservation

Kinetic energy of the pendulum  = potential energy of the pendulum

0.5 x M x V² = M x g x h

0.5 x 2.87 x 2.87 = 9.8 x h

h = 0.42 m

6 0
3 years ago
1.
Ivanshal [37]
I think it’s 10 kg since 20/2=10
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