Q.
The block of mass M moving on the friction less horizontal surface, collides with the spring of the spring constant K and compresses it by length L. The maximum momentum of the block after collision is
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
zero
b
ML2K
c
MK⋅L
d
KL22M
answer is C.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Elastic Energy stored in spring = 12 KL2 here K is force constant, L is length of compression ∴ K.E. of the Block E = 12 KL2 ∴ p2 = 2M⋅E; here p = momentum, m=mass; E = KE ∴ p = 2ME substitute E value p = 2M⋅KL22 p = MK⋅ L