# (a) What is the kinetic energy of a 1400 kg car travelling at a speed of 30 ms^{-1} (b) From what...

## Question:

(a) What is the kinetic energy of a {eq}1400 \ kg {/eq} car travelling at a speed of {eq}30 m s^{-1} {/eq}
(b) From what height would the car have to be dropped to have this same amount of kinetic energy before impact ?
(c) Does your answer to part b depend on the car's mass?
(i) yes
(ii) no

## Conservation of Energy:

The conservation of energy is given by:

{eq}KE + PE = 0 {/eq}

where {eq}KE {/eq} is the kinetic energy and {eq}PE {/eq} is the potential energy.

{eq}KE = \frac{1}{2}mv^{2} {/eq}

where {eq}m {/eq} is the mass of the object, {eq}v {/eq} is the velocity of the object before impact.

The potential energy {eq}PE {/eq} is given by the equation:

{eq}PE = -mgh {/eq}

where {eq}g {/eq} is the acceleration due to gravity, {eq}h {/eq} is the height at which the object is placed before falling.

Given:

{eq}m = 1400 \space kg {/eq}

{eq}v = 30 \space ms^{-1} {/eq}

(a) The kinetic energy {eq}KE_{car} {/eq} of the car is given by the equation:

{eq}KE_{car} = \frac{1}{2}mv^{2} {/eq}

Plugging in the given values:

{eq}KE = \frac{1}{2}\cdot 1400 \cdot 30^{2} {/eq}

{eq}KE = 6.3 \times 10^{5} \space J = 630 \space kJ {/eq}

(b) To solve the height, the conservation of energy and potential energy equations must be used hand-in-hand:

{eq}KE + PE = 0 {/eq}

{eq}PE = -mgh {/eq}

The two equations yield:

{eq}KE -mgh = 0 {/eq}

{eq}KE= mgh {/eq}

Plugging in the values yield:

{eq}6.3 \times 10^{5}= 1400 \cdot 9.8 \cdot h {/eq}

Solving for h:

{eq}h = 46 \space m {/eq}

The car would need to drop 46 meters to reach an equivalent kinetic energy equal to travelling at 30 ms^{-1}.

(c) (ii) No, the answer on part b does not depend on the car's mass. A ball with a mass of 1 gram would still need a height of 46 meters to reach a velocity of 30 m/s. Using the law of conservation of energy:

{eq}KE + PE = 0 {/eq}

{eq}\frac{1}{2}mv^{2} -mgh = 0 {/eq}

{eq}\frac{1}{2} mv^{2} = mgh {/eq}

The mass values cancel out and we are left with a familiar formula used in free-falling bodies:

{eq}v^{2} = 2gh {/eq}