Question

w
A body of mass m is projected from ground for
horizontal range R. At the highest point, it is
broken into two identical pieces. If one part returns
back to the point of projection, then other part will
have total horizontal range counting from the point of
projection of
3R
(1) 3R
R 2R
5R
(3) R​

Answer 1

>> let ' u ' be the horizontal velocity given to the particle of mass ' m ' . R be the horizontal range .

Till reaching the highest point , horizontal distance travelled is given R / 2

at the highest point ,

it gets break into two parts , each of mass m / 2 .

acxording to the law of conservation of momentum :

mu = mu1 / 2 - mu2 / 2

u1 = 2 u + u 2

both have difderent directiin

particle moving with velocity u2 reached the initial position .

horizontal distance travelled by this particle in time ' t ' us given by :

R / 2 = u 2 t

u2 = R / 2t

distance travelled by particle moving with velocity u1 is given by :

x = ( 2 u + u 2 ) t

x = 2 u t + u2 t. as ( t = T / 2)

x = 2 u T / 2 + u2 t

x = R + R / 2

x = 2 R + R / 2

x = 3R / 2

>> range is given by = initial horizontal distance travelled before breaking + x

>> range = 3 R / 2 + R / 2

= 4 R / 2

= 2 R

so , correct option is 2R .

reagrds ❤❤ mark me brainliest dear