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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.?`
2 years ago

Vikas TU
9499 Points
```							From momentum conservation we get,2*10 + 5*3 = 2v1 + 5v22v1 + 5v2 = 35v1 = 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^2x^2 = (122.5 – 87.5)/560x = 0.25 meter
```
2 years ago
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