Respuesta :
Answer:
a) h = 0.0088 m
b) Kb = 794.2J
c) Kt = 0.88J
Explanation:
By conservation of the linear momentum:
[tex]m_b*V_b = (m_b+m_p)*Vt[/tex]
[tex]Vt = \frac{m_b*V_b}{m_b+m_p}[/tex]
[tex]Vt=0.42m/s[/tex]
By conservation of energy from the instant after the bullet is embedded until their maximum height:
[tex]1/2*(m_b+m_p)*Vt^2-(m_b+m_p)*g*h=0[/tex]
[tex]h =\frac{Vt^2}{2*g}[/tex]
h=0.0088m
The kinetic energy of the bullet is:
[tex]K_b=1/2*m_b*V_b^2[/tex]
[tex]K_b=794.2J[/tex]
The kinetic energy of the pendulum+bullet:
[tex]K_t=1/2*(m_b+m_p)*Vt^2[/tex]
[tex]K_t=0.88J[/tex]
a. The vertical height through which the pendulum rises is equal to 0.9 cm.
b. The initial kinetic energy of the bullet is equal to 794.2 Joules.
c. The kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum is equal to 0.883 Joules.
Given the following data:
- Mass of bullet = 11.0 g
- Speed = 380 m/s
- Mass of pendulum = 10.0 kg
- Length of cord = 70.0 cm
a. To determine the vertical height through which the pendulum rises:
First of all, we would find the final velocity by applying the law of conservation of momentum:
Momentum of bullet is equal to the sum of the momentum of bullet and pendulum.
[tex]M_bV_b = (M_b + M_p)V[/tex]
Where:
- [tex]M_b[/tex] is the mass of bullet.
- [tex]M_p[/tex] is the mass of pendulum.
- [tex]V_b[/tex] is the velocity of bullet.
- V is the final velocity.
Substituting the given parameters into the formula, we have;
[tex]0.011\times 380 = (0.011+10)V\\\\4.18 = 10.011V\\\\V = \frac{4.18}{10.011}[/tex]
Final speed, V = 0.42 m/s
Now, we would find the height by using this formula:
[tex]Height = \frac{v^2}{2g} \\\\Height = \frac{0.42^2}{2\times 9.8} \\\\Height = \frac{0.1764}{19.6}[/tex]
Height = 0.009 meters.
In centimeters:
Height = [tex]0.009 \times 100 = 0.9 \;cm[/tex]
b. To compute the initial kinetic energy of the bullet:
[tex]K.E_i = \frac{1}{2} M_bV_b^2\\\\K.E_i = \frac{1}{2} \times 0.011 \times 380^2\\\\K.E_i = 0.0055\times 144400\\\\K.E_i = 794.2 \; J[/tex]
Initial kinetic energy = 794.2 Joules
c. To compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum:
[tex]K.E = \frac{1}{2} (M_b + M_p)V^2\\\\K.E = \frac{1}{2} \times(0.011 + 10) \times 0.42^2\\\\K.E = \frac{1}{2} \times 10.011 \times 0.1764\\\\K.E = 5.0055 \times 0.1764[/tex]
Kinetic energy = 0.883 Joules.
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