Weight of the rail car "(m_1)" =40tons
Velocity "(u_1)" =4Km/hour
Mass of the wagon"(m_2)" =100 tons
Velocity of wagon"(u_2)" =1.2k/hour
Let the final velocity of the rail wagon("v_1") and the final velocity of wagon ("v_2" )
Now, applying the conservation of the linear momentum.
"m_1u_1+m_2u_2=m_1v_1+m_2v_2"
"40v_1+100v_2=40 \\times4-100\\times1.2"
"\\Rightarrow 40v_1+100v_2=160-120=40"
"\\Rightarrow 2v_1+5v_2=2--------(i)"
The collision is elastic,
So, e=1
"e=\\dfrac{v_2-v_1}{u_1-u_2}"
"\\Rightarrow 1=\\dfrac{v_2-v_1}{4-1.2}"
"v_2-v_1=2.8----(ii)"
From equation (i) and (ii)
"v_1=\\dfrac{-12}{7} km\/hour"
"v_2=\\dfrac{7.6}{7}km\/hour"
Comments
Yes, since the collision is elastic, the car and wagon will not merge after impact.
So it would be wrong to assume that the car and the wagon merge after impact?
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