Question

A 10,000-kg railroad car travels down the track at 30 m/s and hits a second stationary railroad car of 20,000 kg. The cars become coupled and continue traveling. Calculate the final velocity of the cars.

Answer 1

Answer:

10 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum: the initial momentum has to be equal to the final momentum, so we can write the following

$p_i = p_f$

$m_1 u_1 + m_2 u_2 = (m_1 +m_2 )v$

where

$m_1 = 10,000 kg$ is the mass of the first car

$u_1=30 m/s$ is the initial velocity of the first car

$m_2 = 20,000 kg$ is the mass of the second car

$u_2 = 0$ is the initial velocity of the second car

$v$ is the final velocity of the two combined cars after the collision

Re-arranging the equation and substituting the numbers, we find

$v=\frac{m_1 u_1 +m_2 u_2}{m_1+m_2}=\frac{(10,000)(30)-0}{10,000+20,000}=10 m/s$

Answer 2

$m_{1}= 10,000 [kg] \\m_{2}=20,000[kg] \\v_{1}=30[ \frac{m}{s} ] \\v_{2}=0[ \frac{m}{s} ] \\\\m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3} \\v_{3}= \frac{m_{1}v_{1}+m_{2}v_{2}}{(m_{1}+m_{2})} \\\\v_{3}=10[ \frac{m}{s} ]$