IIT JEE , AIEEE Forum - Physics , Mechanics

loyalgooner

A block of mass 2.0 kg is moving on a horizontal frictionless surface with a velocity of 1.0 m/s towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end to the stationary block is 100 N/m. Find the maximum compression of the spring.

I do have the figure but I am not able to post it. Can someone tell me how to upload the figure?

Betu

equate the kinetic energy of the moving block to 1/2.kx^2 and find x. this is the maximum displacement

loyalgooner

It wont be the max displacement thats for sure coz the second block will move too so part of the KE of the first block will be converted to the KE of the second block.....This isnt an easy problem.

aditya_arora04

Make the diagram in microsoft paint, and attach the pic. this should serve yr purpose..

For the question,

For max. compression, the stationary block shd not move at all.

So, KE of moving block = (1/2)kx²

1 = 50x²

x = 14 cm approx.

I hope i am correct.

loyalgooner

I did the same as you said but it said that the image wasnt of correct format although it was less than 500 kb in size and also less in resolution. And btw the answer given is not 14.14 cm.

waterdemon

Simple,

First let "V" be the speed of the system after collision.
Applying the law of Conservation of energy, we have:

m₁u₁ + m₂u₂ = (m₁+m₂)V

Here, u₁ = 1m/s and u₂ = 0m/s.

2 = (4)V
V = 0.5m/s.

Now applying the law of Conservation of K.E We get:

(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)(m₁+m₂)V² + (1/2)Kx²

Now we get:
m₁u₁² = (m₁+m₂)V² + Kx²

2 = 4*(1/4) + 100(x)²
1 = (100)x²
Therefore,

x =

(1/100)
x = 0.1m.

Hope you find it useful.
Rate if useful.

Cheers!!!!!!!!!@@@!!!!!!!!!!

loyalgooner

Gee thanx! that was quick! I was kind of thinking of doing it without using momentum principle coz i still havent done that topic.....anyways great work thanx!

waterdemon

Anytime!!!!!!! dude no probs.
Cheers!!!!!@@@!!!!!!!

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