Sir
Question no. 21 attached
Asked by shobhit
15th May 2018,
9:15 AM
Answered by Expert
Answer:
Initially the spring of force constant K=200 N/m is compressed a distance d=10 cm and tied to the block-A which is at rest.
potential energy stored in spring = (1/2)K×d2 = (1/2)×200×10×10×10-4 = 1 Joule.
initial kinetic energy of block-A = (1/2)×5×2×2 = 10 Joule
total energy before collision = (10+1) = 11 Joule
after collision, let block-A is moving with speed v1 and block-B moving with speed v2;
total kinetic energy = (1/2)×m×(v12+v22) = (1/2)×5×(v12+v22) = 11 Joule
we get from the above relation (v12+v22) = 4.4 m2/s2
Answered by Expert
15th May 2018,
3:38 PM
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