Knockout JEE Main April 2021 (One Month)
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A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest.
After approaching half the distance from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms -1. The total initial energy of the spring is :
1.5 J
0.6 J
0.3 J
0.8 J
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As we discussed in
Potential Energy stored in the spring -
- wherein
x= elongation or compression of spring from natural position
Initial energy of spring
After approaching half the distance its energy will be
This momentarily comes to result. Its kinetic energy at that moment
After coming to rest, total kinetic energy is transfered to
mass.
K.E. of second mass
0.45 J =
Option 1)
1.5 J
This is an incorrect option.
Option 2)
0.6 J
This is the correct option.
Option 3)
0.3 J
This is an incorrect option.
Option 4)
0.8 J
This is an incorrect option.