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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???
10 months ago

Indrajit dodiya
18 Points
							here no external force is thereso linear momentum of the system is conservedinitially 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 leftm1>m2 so cart is also moving towards leftassume 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}$

10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions