Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        two blocks of masses 4kg and 2kg are attached to ends of a light spring of spring constant 1200 N/m.The whole system is placed on smooth horizontal surace.When the spring is relaxedstate , 4 Kg block is given a velocity 3m/s , the max extention in spring is	a)10cm b)12cm c) 20cm`
2 years ago

53 Points
```							we will use the concept of reduced mass for this questionreduced mass=(m1*m2)/m1+m2    by this method we can consider the two blocks as one and the spring is connected to a fixed support.therefort by energy conservation,½ mv2 + ½ kx2      where m=reduced mass, v=initial velocity, k= spring constant, x= max extension of springon solving this equation we getx2=1/100 metretherefore x=1/10 =0.1m=10cm   if you agree with my solution please click “Approve” belowthankyou
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question