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 mass 5 kg and 2 kg are connected by a spring of negligible mass and placed on a frictionless horizontol surface. an impulse gives a velociy of 7 m/s to the heavier block in the direction of the lighter block.the velociy of center of mass
3 years ago

Pushkar Prateek
23 Points

Heavier block receives impulse(very large force acted in very short time duration) and the smaller block doesn’t gain velocity in this impulsive period.
Vcm = (M1 V1 + M2 V2) / M1 + M2
= (5x7 + 2x0) / 7
=  5 m/s.
3 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