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
Menu
Grade: 12th pass
        A ball of mass 1kg is dropped from height 9.8m strikes with ground and rebound at height of 4.9m if the time of contact b/w ball and ground is 0.1 sec then find impulse and average force acting on the ball?
3 years ago

Answers : (1)

V Kartik
25 Points
							Velocity of the ball when it hit the ground can be calculated as                                                                                                     12mv2=mgh⇒v=2gh−−−√=2x10x20−−−−−−−√⇒v=20m/s in downward direction.on striking the ground let the ball rebound with the velocity u in upward direction. it can be calculated as                                                                                                     12mu2=mgh⇒u=2gh−−−√=2x10x5−−−−−−√⇒u=10m/s in upward directionChange in momentum of the ball = final momentum of the ball −initial momentum of the ball                                                    = mu⇀−mv⇀⇒m( u +v)    as the direction of u and v are opposite∆P=1(10+20) = 30kgm/s
						
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details