A railway carriage of mass 10000kg moving with a speed 15/s hits a stationary carriage of same mass. after the collision the carriage get coupled and ove together. what is their common velocity after collision?

So the formula for linear momentum, #rho#, is #rho=mv#.

So before the collision, the momentum of the system is #10000*15=150000kgms^-1#.

After the collision, the momentum, according to the law of conservation of momentum, remains the same. That is, #rho_1=rho_2#, or #m_1v_1=m_2v_2#

Here, #v_2=(m_1v_1)/m_2#

#m_1v_1=150000#, we know this from our prior calculations. Now, since the second carriage gets stuck to the first one, the mass of the system, #m_2#, is equal to #20000kg#

#v_2=150000/20000#

#v_2=7.5ms^-1#

Applying the Law of Conservation of Momentum, identify the given data and set-up the needed formula to determine the common velocity of the carriages after the collision.

#"Momentum before" (P_("before"))="Momentum after"(P_(after))#

#where:#

#"Momentum(P)"="mass(m)"xx"velocity(v)"#
#P=mv#

#(P_("before")):#
The masses and the velocities of the carriages before the collision

#m_1=10,000kg; v_1="15m/s"#
#m_2=10,000kg; v_2="0m/s " ( "at rest")#

#(P_(after)):#
The masses and the velocities of the carriages after the collision

#m_1=10,000kg; v_1'=?#
#m_2=10,000kg; v_2'=?#

Take note that after the collision, both carriages moved together of same direction with the moving carriage. So that, a common velocity has been identified as follows:

#v_1'=v_2'=v_c="common velocity"#

Now, set-up the equation and plug in values in both conditions: before and after the collisions as described below.

#P_("before")=P_(after)#

#m_1v_1+m_2v_2=m_1v_1'+m_2v_2'#

but:

#v_1'=v_2'=v_c#

#m_1v_1+m_2v_2=m_1v_c'+m_2v_c'#

#(10,000kg)("15m/s")+cancel((10,000kg)("0m/s"))0=(10,000kg)(v_c)+(10,000kg)(v_c)#

#150,000"kg.m/s"+0=(20,000kg)v_c#; divide both sides by 20,000kg to isolate the variable #v_c#

#v_c=(150,000cancel(kg)."m/s")/(20,000cancel(kg))#

#v_c=7.5"m/s"#