2 00000 m Two blocks move towards each other on a frictionless surface, the one

Question

2 00000 m Two blocks move towards each other on a frictionless surface, the one on the left having a velocity of v111 and mass m1-7kg, the one on the right having a velocity of V2- 10-and mass m2-5kg. However the one on the left has a spring attached to it of spring constant 1634. The figure shows the initial configuration of the system in purple. The blocks in the middle (green and red) are the same two blocks but at a later time. This configuration shows the spring at maximum compression when they're in contact What is the initial total momentum? 8 m. 8 kg m/s 27 At the instant when the spring is at maximum compression, the two blocks are moving at the same velocity. What is their velocity at this instant? Since the surface is frictionless, you can assume conservation of momentum. 27 m/s What is the maximum compression from cquilibrium of the spring? Assume conscrvation of cnergy. 2.14

Explanation / Answer

1st part :- Initial Total Momentum = 27 kgm/s

2nd part :- This is also the momentum when two blocks are moving with same velocity,

let the velocity be V

momentum of two blocks at same velocity V = (m1+m2)V

So, 27 = (m1+m2)V = 12 V

V = 27/12 = 2.25 m/s

3rd part:-

Initial Energy of the system = Sum of kinetic energy of both masses = 1/2 m1 v1^2 + 1/2 m2 v2^2 = 1/2(7*121 + 5*100) = 1/2(1347) = 673.5 Joules

Final Energy of system = Kinetic Energy of both masses + Spring Potential energy[ energy of compressed spring ]

Kinetic Energy of both masses = 1/2 (m1+m2) 2.25^2 = 1/2 (12) 5.0625 = 30.375

Spring potential Energy = 1/2 k (compression)^2 = 1/2 k X^2 [ Let max. compression be X ]

= 1/2 * 1634 * X^2 = 817 X^2

Now,

By conservation of energy

Initial Energy of the system = Final Energy of system

673.5 = 30.375 + 817 X^2

643.125 = 817 X^2

X^2 = 0.7871

X= 0.887 m

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