Chapter 9: Q53P (page 251)

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviation MVC. Suppose a 1000 kg car slides into a stationary 500 kgmoose on a very slippery road, with the moose being thrown through the windshield (a common MVC result). (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? A similar danger occurs in Saudi Arabia because of camel–vehicle collisions (CVC). (b) What percent of the original kinetic energy is lost if the car hits acamel? (c) Generally, does the percent loss increase or decrease if the animal mass decreases?

Short Answer

Expert verified

Percent loss on K.E. due to collision with moose is 33%
Percent loss on K.E. due to collision with camel is 23%
As the mass of the animal decreases, the percent loss in K.E. will also decrease.

Step by step solution

Step 1: Given Data

Mass of the car is, $m_{c} = 1000 k g$

Mass of the moose is, $m_{m} = 500 k g$

Mass of the camel is, $m_{k} = 300 k g$

Determining the concept

Usetheprinciple of conservation of momentum to relate the velocities. Usingtheformula for kinetic energy, find the percent loss in kinetic energy due to collision.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$P_{i} = P_{f}$

$P = m v$

$K = \frac{1}{2} m v^{2}$

where, m is mass, v is velocity, P is linear momentum and K is kinetic energy.

(a) Determining the percent loss on K.E. due to collision with moose

Let, the initial velocity of the moose be zero as compared to the car. Also, as afterthecollision moose will stick to the car’s windshield,thefinal velocity of boththecar andthemoose will bethesame.Applyingtheprinciple of conservation of momentum,

Total momentum $\vec{P_{i}}$ before collision = Total momentum after collision $\vec{P_{f}}$

For the given situation,

Total initial momentum = Initial momentum of car + Initial momentum of moose.

role="math" localid="1661489273723" $\vec{P_{i}} = \vec{P_{i c}} + \vec{P_{i m}}$

As, initially moose is at rest, $\vec{P_{i m}} = 0$

role="math" localid="1661489287893" ${\vec{P}}_{i} = {\vec{P}}_{i c} = m_{c} v_{i c} . . . . . \dots (1)$

total final momentum = final momentum of car + final momentum of moose

$\vec{P_{f}} = m_{c} v_{f c} + m_{m} v_{f m} v_{f m} = v_{f c} = v_{f} \vec{P_{f}} = (m_{c} + m_{m}) v_{f} . . . . . . . . (2)$

Equating equation 1 and 2

$m_{c} v_{i c} = (m_{c} + m_{m}) v_{f} . . . . . . . . . (3) m_{c} = 2 m_{m} . . . . . . . . . . . . (4) 2 m_{m} v_{i c} = (2 m_{m} + m_{m}) v_{f} 2 m_{m} v_{i c} = 3 m_{m} v_{f}$

Cancellingand rearranging the equation,

$v_{i c} = \frac{3}{2} v_{f} . . . . . . . (5)$

Now, to find out initial and final K.E. energy of the system,

$k_{i} = \frac{1}{2} m_{c} v_{i c}^{2} k_{i} = \frac{1}{2} (m_{c} + m_{m}) v_{f}^{2}$

Taking ratio of $K_{i} a n d K_{f}$

role="math" localid="1661489775229" $\frac{K_{f}}{K_{i}} = \frac{\frac{1}{2} (m_{c} + m_{m}) v_{f}^{2}}{\frac{1}{2} m_{c} v_{i c}^{2}}$

Using equation (4), (5),

role="math" localid="1661489894632" $\frac{K_{f}}{K_{i}} = \frac{\frac{1}{2} (2 m_{m} + m_{m}) v_{f}^{2}}{\frac{1}{2} (2 m_{m}) (\frac{3}{2} v_{f})}$

Simplifying the equation,

$\frac{K_{f}}{K_{i}} = \frac{2}{3}$

Fraction of energy lost $= 1 - \frac{2}{3}$

Fraction of energy lost $\frac{1}{3}$

Percentage of loss in K.E. $= \frac{1}{3} \times \frac{100}{100}$

Percentage of loss in K.E. $= \frac{100}{3} %$

Percentage of loss in K.E. $= 33 %$

Hence, percent loss on K.E. due to collision with moose $= 33 %$

(b) Determining the percent loss on K.E. due to collision with camel

Modify equation 3 forthe camel,

$m_{c} v_{i c} = (m_{c} + m_{k}) v_{f} m_{c} = 1000 k g a n d m_{k} 300 k g m_{c} = \frac{10}{3} m_{k} . . . . . . . . . . (6) \frac{10}{3} m_{k} v_{i c} = (\frac{10}{3} m_{k} + m_{k}) v_{f} \frac{10}{3} m_{k} v_{i c} = \frac{13}{3} m_{k} v_{f}$

Cancelling and rearranging the equation,

$v_{i c} = \frac{13}{10} v_{f} . . . . . . . . . (7)$

Now, to find out initial and final K.E. energy of the system,

role="math" localid="1661490606474" $K_{i} = \frac{1}{2} m_{c} v_{i c}^{2} K_{f} = \frac{1}{2} (m_{c} + m_{k}) v_{f}^{2}$

Taking ratio of $K_{i} a n d K_{f}$

$\frac{K_{f}}{K_{i}} = \frac{\frac{1}{2} (m_{c} + m_{k}) v_{f}^{2}}{\frac{1}{2} m_{c} v_{i c}^{2}}$

Using equation (1) and (2),

$\frac{K_{f}}{K_{i}} = \frac{\frac{1}{2} (\frac{10}{3} m_{k} + m_{k}) v_{f}^{2}}{\frac{1}{2} (\frac{10}{3} m_{k}) {(\frac{13}{10} v_{f})}^{2}}$

Simplifying the equation,

$\frac{K_{f}}{K_{i}} = \frac{10}{13}$

Fraction of energy lost $= 1 - \frac{10}{13}$

Fraction of energy lost $= \frac{3}{13}$

Percentage of loss in K.E $. = \frac{3}{13} \times \frac{100}{100}$

Percentage of loss in K.E. $= \frac{300}{13} %$

Percentage of loss in K.E. $= 23.07 % \approx 23 %$

Hence, percent loss on K.E. due to collision with camel is 23%

(c) Determining if the percent loss increase or decrease if the animal mass decreases

As the mass of the animal decreases,thepercent loss in K.E. will also decrease.

Therefore, by applying the principle of conservation of momentum and taking ratio of final and initial kinetic energy, the percent loss in kinetic energy during the collision can be calculated.