Two bodies A and B of mass m and 2 m respectively are placed on a smooth floor. They are connected by a spring of negligible mass. A third body C of mass m is placed on the floor. The body C moves with a velocity v 0 along the line joining A and B and collides elastically with A . At a certain time after the collision it is found that the instantaneous velocities of A and B are same and the compression of the spring is x 0 . The spring constant k will be :
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Please explain the conservation on energy equatiom in brief.... I didn't get that!!😢
Using conservation of momentum.
m v o = 3 m v where v is instantaneous velocity.
Using conservation of energy
0 . 5 m v o 2 = 0 . 5 × 3 m v 2 + 0 . 5 k x o 2
Did i do correctly @Nishant Rai
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yes, but please do write how you got those 2 equations, i mean by using COLM and COE.
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Can u post the solution for the elctric field one u recently posted
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Note: Since the colliding bodies C & A are of the same mass m , and they collide elastically, so their velocities after collision are exchanged.