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D University× D Fisica para × D Physics for × eChegg Stuc × LON-CAPA × LON-CAPA × LON-CAPA × Q LON-CAPA × whatsApp × - C D loncapa.hep.uprm.edu/enc/72/f3684898c7a2b024c7200eda7b59de8f9ffce3396907e2f163c79bc9a0740f4731f868a348cb4f58abb29d15aa2e74ef0a76362399ed58fe1 601091 268b876e1fe8 × Q ERICK I PEREZ(Student section: 011) Fisi3171-Agosto-2017 Messages Courses Help Logout Main Menu Contents Grades course contents» » Cap5 Leyes de Newton » Force Due to Spring >, Timer Note., Evaluate 4-Feedback-Print Force Due to Spring Points:1 Two masses, M1-2.0 kg and M2-0.29 kg, are at rest on an air-track with a compressed spring between them. The spring is released; the force on M as a function of time is shown. Find the speed of M2 when t= 60.0 ms Submit Anewer Tries 0/40 Post Discussio Send Feedback Activate Windows Go to Settings to activate Windows 1:08 PM O Type here to search 11/20/2017

Explanation / Answer

Impulse = area under the F vs t curve

I = (40 x 0.06) + (10 x 0.06 / 2) { from 0 to 60 ms}

I = 2.7 kg m/s

from newton's third law, impulse will be same on both,

and imulse = change in momentum

2.7 = (0.29)(v - 0)

v= 9.31 m/s

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