# A projectile with a mass of 9.10 g is traveling at a speed of 1.10 km/s, determine the following....

## Question:

A projectile with a mass of 9.10 g is traveling at a speed of 1.10 km/s, determine the following.

A) kinetic energy of the projectile in kilojoules: kJ

B) kinetic energy of the projectile in kilojoules, if its speed is reduced by a factor of two: kJ

C) kinetic energy of the projectile in kilojoules, if its original speed is increased by a factor of three: kJ

## Kinetic Energy:

Kinetic energy is the energy that an object holds due to its motion. The total kinetic energy of an object is composed of its linear and angular kinetic energies resulting from its translation and rotation. The kinetic energy of an object moving in linear motion can be determined using its mass and velocity.

## Answer and Explanation:

Given:

{eq}m_p = 9.10 \ g = 0.0091 \ kg {/eq} Mass of the projectile

{eq}v = 1.10 \frac {km}{s} = 1100 \frac {m}{s} {/eq} Speed of the projectile

Part A) The kinetic energy of the projectile is given by,

{eq}KE = \frac {1}{2} m_p v^2 = \frac {1}{2} (0.0091 \ kg)(1100 \frac {m}{s})^2 = 5505.5 \ J = \boxed {5.51 \ kJ} {/eq}

Part B) The resulting speed of the projectile is,

{eq}v = \frac {1100 \frac {m}{s}}{2} = 550 \frac {m}{s} {/eq}

Solving its kinetic energy,

{eq}KE = \frac {1}{2} (0.0091 \ kg)(550 \frac {m}{s})^2 = 1376.375 \ J = \boxed {1.38 \ kJ} {/eq}

Part C) When the projectile's speed is increased by a factor of three,

{eq}v = 3 * 1100 \frac {m}{s} = 3300 \frac {m}{s} {/eq}

Hence, its resulting kinetic energy is,

{eq}KE = \frac {1}{2} (0.0091 \ kg)(3300 \frac {m}{s})^2 = 49549.5 \ J = \boxed {49.5 \ kJ} {/eq}

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Chapter 4 / Lesson 14