Chapter 3, Problem 3C

In a head-on, inelastic collision, a 4,000-kg truck going 10 m/s east strikes a 1,000-kg car going 20 m/s west.
(a) What is the speed and direction of the wreckage?
(b) How much kinetic energy was lost in the Collision?

To determine

(a)

The speed and direction of the wreckage.

Answer to Problem 3C

The speed of the wreckage is $4m/s$ along positive x-direction.

Explanation of Solution

Given:

Given that in a head-on, inelastic collision, a $4000kg$ truck going $10m/s$ east strikes a $1000kg$ car going $20m/s$ west

Formula used:

According to the law of conservation of momentum, when the two objects collide in an isolated system, the total momentum before and after the collision is same

According to the law of conservation of energy, the total energy of an isolated system remains same. In other words, it means that energy can neither be created nor be destroyed.

Calculation:

We have

Mass of the truck, $m_{Truck}=4000kg$

Mass of the car, $m_{Car}=1000kg$

Initial Speed of the truck, $u_{Truck}=10m/s$

Initial Speed of the car, $u_{Car}=-20m/s$

Give that the collision is head-on and inelastic

Thus, the final speed of truck and car will be same

$v_{Truck}=v_{Car}=v$

We know that according to the conversation of momentum,

$m_{Truck}{u}_{Truck}+m_{Car}{u}_{Car}=m_{Truck}{v}_{Truck}+m_{Car}{v}_{Car}$

$m_{Truck}{u}_{Truck}+m_{Car}{u}_{Car}=m_{Truck}v+m_{Car}v$

$m_{Truck}{u}_{Truck}+m_{Car}{u}_{Car}=\left(m_{Truck}+m_{Car}\right)v$

$v=\frac{m_{Truck}{u}_{Truck}+m_{Car}{u}_{Car}}{m_{Truck}+m_{Car}}$

Substituting the values, we get

$v=\frac{\left(4000kg\right)\left(10m/s\right)+\left(1000kg\right)\left(-20m/s\right)}{4000kg+1000kg}$

$v=\frac{40000kgm/s-20000kgm/s}{5000kg}$

$v=\frac{20000kgm/s}{5000kg}$

$v=4m/s$.

Conclusion:

Hence, the speed of the wreckage is $4m/s$ along positive x-direction.

To determine

(b)

In a head-on, inelastic collision, a $4000kg$ truck going $10m/s$ east strikes a $1000kg$ car going $20m/s$ west. Also, speed of the wreckage is $4m/s$ along positive x-direction Find the kinetic energy lost in the collision.

Answer to Problem 3C

The kinetic energy lost in the collision is $360kJ$.

Explanation of Solution

Given:

Given that in a head-on, inelastic collision, a $4000kg$ truck going $10m/s$ east strikes a $1000kg$ car going $20m/s$ west. Also, speed of the wreckage is $4m/s$ along positive x-direction

Formula used:

According to the law of conservation of momentum, when the two objects collide in an isolated system, the total momentum before and after the collision is same.

According to the law of conservation of energy, the total energy of an isolated system remains same. In other words, it means that energy can neither be created nor be destroyed.

Kinetic energy is given as

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

Where, $KE=$ Kinetic Energy

$m=$ Mass of the body

$v=$ velocity of the body.

Calculation:

We have

Mass of the truck, $m_{Truck}=4000kg$

Mass of the car, $m_{Car}=1000kg$

Initial Speed of the truck, $u_{Truck}=10m/s$

Initial Speed of the car, $u_{Car}=-20m/s$

Give that the collision is head-on and inelastic

Thus, the final speed of truck and car will be same

$v_{Truck}=v_{Car}=v=4m/s$

The kinetic energy lost in the collision is calculated as

$K{E}_{Lost}=K{E}_{Intial}-K{E}_{Final}$

$K{E}_{Lost}=\left(\frac{1}{2}{m}_{Truck}{\left(u_{Truck}\right)}^2+\frac{1}{2}{m}_{Car}{\left(u_{Car}\right)}^2\right)-\left(\frac{1}{2}\left(m_{Truck}+m_{Car}\right)v^2\right)$

Substituting the values, we get $K{E}_{Lost}=\left(\frac{1}{2}\left(4000kg\right){\left(10m/s\right)}^2+\frac{1}{2}\left(1000kg\right){\left(-20m/s\right)}^2\right)-\left(\frac{1}{2}\left(4000kg+1000kg\right){\left(\left(4m/s\right)\right)}^2\right)$

$K{E}_{Lost}=\frac{1}{2}\left[\left(\left(4000kg\right){\left(10m/s\right)}^2+\left(1000kg\right){\left(-20m/s\right)}^2\right)-\left(\left(4000kg+1000kg\right){\left(\left(4m/s\right)\right)}^2\right)\right]$

$K{E}_{Lost}=\frac{1}{2}\left[800000-80000\right]J$ $K{E}_{Lost}=\frac{1}{2}\times 720000J$

$K{E}_{Lost}=360000J$

$K{E}_{Lost}=360kJ$.

Conclusion:

Hence, the kinetic energy lost in the collision is $360kJ$.