A particle of mass m is given speed u at angle theta at topmost point It breaks into two equal parts one comes to rest  find point of landing of second mass

Asked by ashishgarud539 | 15th Dec, 2021, 04:12: PM

Expert Answer:

Let's consider that the horizonatl range of particle from where it is projected is R. 
At highest point the horizontal velocity before splitting is = ucosθ
By law of conservation of linear momentum, 
mucosθ = m/2(v)+m/2(0)
Thus, v = 2ucosθ 
As velocity becomes twice the initial velocity range will also become twice and time will depend only on vertical component of velocity. 
The point of landing of other particle = horizontal distance travelled by particle after breaking into two parts 
d = R/2 + R = 3R/2 

Answered by Shiwani Sawant | 16th Dec, 2021, 07:46: PM