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Two man of mass m1 and m2 respectively standing on the opposite end of a cart of mass M and length 2L.the cart is on frictionless surface and m1>m2. If each man walks towards each other and exchange their position ,,then what is the distance the cart move???

Two man of mass m1 and m2 respectively standing on the opposite end of a cart of mass M and length 2L.the cart is on frictionless surface and m1>m2. If each man walks towards each other and exchange their position ,,then what is the distance the cart move???

Grade:10

1 Answers

Indrajit dodiya
18 Points
5 years ago
here no external force is there
so linear momentum of the system is conserved
initially cart is at rest so COM of system is at rest.
according to law of conservation of linear momentum COM of system is rest during motion.
assume m1 is moving towards right and m2 is moving towards left
m1>m2 so cart is also moving towards left
assume cart moves distance x towards left
0=m_{1}\left ( 2L-x \right )-m_{2}\left ( 2L+x \right )-M\left ( x \right )
m_{1}\left ( 2L-x \right )=m_{2}\left ( 2L+x \right )+M\left ( x \right )
2m_{1}L-m_{1}x=2m_{2}L+m_{2}x+Mx
simplify and answer is
x=\frac{\left ( m_{1}-m_{2} \right )2L}{m_{1}+m_{2}+M}

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