Q.

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass. Spring Constant of spring is K = 140 N/M and blocks are placed on a frictionless horizontal surface. An impulse gives a speed of 14 m/s to 10 kg block in the direction of the lighter block. Then, the maximum compression in the spring will be

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

1 m

b

2 m

c

3 m

d

4 m

answer is B.

(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

12m1m3m1+m2v1-v22=12kx212×4014(14)2=12×140x24=x2X = 2m
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass. Spring Constant of spring is K = 140 N/M and blocks are placed on a frictionless horizontal surface. An impulse gives a speed of 14 m/s to 10 kg block in the direction of the lighter block. Then, the maximum compression in the spring will be