Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?
Option 1 - <p>100N</p>
Option 2 - <p>200N</p>
Option 3 - <p>300N</p>
Option 4 - <p>400N</p>
6 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
P
Answered by
8 months ago
Correct Option - 2
Detailed Solution:
= Change in momentum of each ball
Similar Questions for you
T1 = m (g + a)
T2 = m (g - a)
Apparent weight = mg – ma
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...
Didn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering





