A 11.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 10.0 kg, suspended from a cord 70.0 cm long.a) Compute the vertical height through which the pendulum rises.(cm)
b) Compute the initial kinetic energy of the bullet;(j)
c) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.(j)

Answers

Answer 1

Answer:

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.

$M_bV_b = (M_b + M_p)V$

Where:

$M_b$ is the mass of bullet.

$M_p$ is the mass of pendulum.

$V_b$ is the velocity of bullet.

V is the final velocity.

Substituting the given parameters into the formula, we have;

$0.011* 380 = (0.011+10)V\n\n4.18 = 10.011V\n\nV = (4.18)/(10.011)$

Final speed, V = 0.42 m/s

Now, we would find the height by using this formula:

$Height = (v^2)/(2g) \n\nHeight = (0.42^2)/(2* 9.8) \n\nHeight = (0.1764)/(19.6)$

Height = 0.009 meters.

In centimeters:

Height = $0.009 * 100 = 0.9 \;cm$

b. To compute the initial kinetic energy of the bullet:

$K.E_i = (1)/(2) M_bV_b^2\n\nK.E_i = (1)/(2) * 0.011 * 380^2\n\nK.E_i = 0.0055* 144400\n\nK.E_i = 794.2 \; J$

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:

$K.E = (1)/(2) (M_b + M_p)V^2\n\nK.E = (1)/(2) *(0.011 + 10) * 0.42^2\n\nK.E = (1)/(2) * 10.011 * 0.1764\n\nK.E = 5.0055 * 0.1764$

Kinetic energy = 0.883 Joules.

Answer 2

Answer:

Answer:

a) h = 0.0088 m

b) Kb = 794.2J

c) Kt = 0.88J

Explanation:

By conservation of the linear momentum:

$m_b*V_b = (m_b+m_p)*Vt$

$Vt = (m_b*V_b)/(m_b+m_p)$

$Vt=0.42m/s$

By conservation of energy from the instant after the bullet is embedded until their maximum height:

$1/2*(m_b+m_p)*Vt^2-(m_b+m_p)*g*h=0$

$h =(Vt^2)/(2*g)$

h=0.0088m

The kinetic energy of the bullet is:

$K_b=1/2*m_b*V_b^2$

$K_b=794.2J$

The kinetic energy of the pendulum+bullet:

$K_t=1/2*(m_b+m_p)*Vt^2$

$K_t=0.88J$

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