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Two masses m1 = 2kg and m2 = 5kg are moving on a frictionless surface with velocities 10m/s and 3 m/s respectively. m2 is ahead of m1. An ideal spring of spring constant k = 1120 N/m is attached on the back side of m2. The maximum compression of the spring will be.?

Two masses m1 = 2kg and m2 = 5kg are moving on a frictionless surface with velocities 10m/s and 3 m/s respectively. m2 is ahead of m1. An ideal spring of spring constant k = 1120 N/m is attached on the back side of m2. The maximum compression of the spring will be.?

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Grade:12th pass

2 Answers

Vikas TU
14149 Points
7 years ago
From momentum conservation we get,
2*10 + 5*3 = 2v1 + 5v2
2v1 + 5v2 = 35
v1 = v2 = v = 5 m/s for maximum compression,
Therefore, 
v = 5 m/s.......(1)
From  Energy conservation we get,
0.5*2*100 + 0.5*5*9 = 0.5*2*v^2 + 0.5*5*v^2 + 0.5*1120*x^2..............(2)
put eqn. (1) in eqn. (2) we get,
122.5 – 3.5v^2 = 560x^2
x^2 = (122.5 – 87.5)/560
x = 0.25 meter
 
 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
From momentum conservation we get,
2*10 + 5*3 = 2v1 + 5v2
2v1 + 5v2 = 35
v1 = v2 = v = 5 m/s for maximum compression,
Therefore, 
v = 5 m/s.......(1)
From  Energy conservation we get,
0.5*2*100 + 0.5*5*9 = 0.5*2*v^2 + 0.5*5*v^2 + 0.5*1120*x^2..............(2)
put eqn. (1) in eqn. (2) we get,
122.5 – 3.5v^2 = 560x^2
x^2 = (122.5 – 87.5)/560
x = 0.25 meter
 
Thanks and regards,
Kushagra

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