Will rate life saver Two blocks A and B of masses mA = 1.00 kg and mB = 2.00 kg

Question

Will rate life saver

Two blocks A and B of masses mA = 1.00 kg and mB = 2.00 kg on a horizontal, frictionless table head toward each other with initial speeds vA = 3.00 m /s and VB = 2.00 m /s. A light spring is attached to block A, so when the two block collide, they compress the spring and bounce off with velocities v'A and v'B. Find the final velocities v'A and v'B. What are the velocities of the blocks when the spring reaches its maximum compression. How much potential energy was stored in the spring at the instant of maximum compression?

Explanation / Answer

applying conservation of momentum on system of mA , mB and spring

m1u1 + m2u2 + m3u3 = m1v1 + m2v2 + m3v3

=> 1 * 3 + 2 * (-2) = 1 * v'A + 2 * v'B
=>  1 * v'A  + 2 * v'B = -1
=> v'A = -1 - 2 v'B ...............................(1)

applying conservation of energy
1/2 m1u12 + 1/2 m2u22 = 1/2 kx2
1 * 9 + 2 * 4 = kx2
kx2 = 17 .........................................(2)

after compression ,
1/2 kx2 = 1/2 m1v12 + 1/2 m2v22
kx2 = 1 * v'A2 + 2v'B2
=>17   =1 * ( 1 + 4 v'B2 + 4v'B ) + 2v'B2
=> 17 = 6 v'B2 + 4 v'B + 1
=> v'B = -2 m/s or v'B = 1.33 m/s

considering velocity v'B = 1.33 m/s as the blocks are reversed
v'A = -1 - 2.66 = - 3.66 m/s

B) at maximum compression the velocities of the blocks will be zero

C)potential energy = 1/2 kx2 = 17/2 = 8.5 joules

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