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The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is :
- a)
- b)0
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The block of massMmoving on the frictionless horizontal surface collid...
According to conservation of energy
MkL2 = p2 (p = mu)
The correct answer is: 
Most Upvoted Answer
The block of massMmoving on the frictionless horizontal surface collid...
The kinetic energy of the body is transfered to the spring and hence it compresses.
the work done by the spring=the kinetic energy of block
=> 1/2M(V) ^2=1/2k(L^2)
multiplying by M both side we get
(MV) ^2=kM(L^2)
taking root on both side we get
MV=L√(Mk).
the work done by the spring=the kinetic energy of block
=> 1/2M(V) ^2=1/2k(L^2)
multiplying by M both side we get
(MV) ^2=kM(L^2)
taking root on both side we get
MV=L√(Mk).
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The block of massMmoving on the frictionless horizontal surface collid...
The all Sur face'the fact of all images and for mula
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root under mk ans
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The block of massMmoving on the frictionless horizontal surface collides with the spring of spring constantkand compresses it by lengthL. The maximum momentum of the block after collision is :a)b)0c)d)Correct answer is option 'C'. Can you explain this answer?
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