Question

A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at 1.00 m/s.
a. What is the speed of the connected cars?
b. How much does the kinetic energy of the system decrease during the collision?

Answer 1

Answer:

a) 1.75m/s

b) 14,062.5J

Explanation:

• a) We solve this problem using linear momentum

$p=mv$

where $p$ is the momentum, $m$ is the mass, and $v$ is the velocity.

the momentum of the first train before the collision:

$p_{1}=m_{1}v_{1}$

where $m_{1}=25,000kg$ and $v_{1}=2.5m/s$

$p_{1}=(25,000kg)(2.5m/s)$

$p_{1}=62,500kgm/s$

the momentum of the second train before the collision:

$p_{2}=m_{2}v_{2}$

where $m_{2}=25,000kg$ and $v_{1}=1m/s$

$p_{2}=(25,000kg)(1m/s)$

$p_{2}=25,000kgm/s$

the total momentum before the collision is:

$p_{1}+p_{2}$

and due to the conservation of linear momentum: the amount of linear momentum before the collision must be the same after the collision.

so due to conservation:

$p_{1}+p_{2}=p_{f}$

the linear momentum of the two train system after the collision

$p_{f}=m_{f}v_{f}$

where $p_{f}$ is the final linear momentum, $m_{f}$ is the final mass of the system, if the two cars end up connected the mass is:

$m_{f}=25,000kg+25,000kg$ (the sum of the mass of the two cars)

$m_{f}=50,000kg$

and $v_{f}$ is the final speed: the speed of the connected cars.

so, going back to the conservation of momentum

$p_{1}+p_{2}=m_{f}v_{f}$

replacing all known values:

$87,500kgm/s=(50,000kg)v_{f}$

clearing for the final speed:

$v_{f}=\frac{87,500kgm/s}{50,000kg} =1.75m/s$

• b) the initial kinetic energy:

$k1+k2$

which is:

$\frac{1}{2} m_{1}v_{1}^2+\frac{1}{2} m_{2}v_{2}$

replacing all known values:

$\frac{1}{2}(25,000kg)(2.5m/s)^2+\frac{1}{2}(25,000kg)(1m/s)^2\\=\frac{1}{2}(25,000kg)(6.25m^2/s^2)+\frac{1}{2}(25,000kg)(1m^2/s^2)\\=78,125J+12,500J\\=90,625J$

anf the final kinetic energy is:

$k_{f}=\frac{1}{2} m_{f}v_{f}^2\\k_{f}=\frac{1}{2} (50,000kg)(1.75m/s)^2\\k_{f}=\frac{1}{2} (50,000kg)(3.0625m^2/s^2)\\k_{f}=76,562.5J$

the difference is:

$90,625J-76,562.5J=14,062.5J$

the kinetic energy of the system decreased 14,062.5J