# 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 : Option 1) 1.5 J Option 2) 0.6 J Option 3) 0.3 J Option 4) 0.8 J

As we discussed in

Potential Energy stored in the spring -

- wherein

x= elongation or compression of spring from natural position

Initial energy of spring

.......(1)

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.

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-