Question

An object of mass 5 kg is projected with a velocity of 20 m/s at an angle of $60^{o}$ to the horizontal. At the highest point of its path the projectile explodes and breaks up into two fragments of masses 1 kg and 4 kg. The fragments separate horizontally after the explosion. The explosion releases internal energy such that the kinetic energy of the system at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground.

Answer 1

Answer:

Explanation:

initial velocity = 20m/s

projection = 60o

velocity at highest point = u/2 = 10m/s

v= 10

since the system explodes due to internal forces do momentam will remain conserved ...

Pi = Pf             ..........1

Pi = Mv = 5(10)  = 50     ............2

finally after explosion both particle are moving in horizontally , let v1 , v2 are the velocity

of particles of mass 1kg & 4kg respectively ...

final momentam (Pf) = v1 + 4v2      ...............3

momentam is conserved so from eq 1

v1 + 4v2 = 50         ...........4

it is given that after explosion KE of particle is doubled

(KE)f = 2(KE)i ( ke before explosion)

(KE)i = Mv2/2 = 5(10)2/2 = 250J   so  

(KE)f = 500 j =(v12 + 4v22)/2        ...............5

solving 4 & 5 we get v

v1 ,v2 = (-10,15)  

now time taken by particle to reach the highest point = T/2 = usin@/g   =  time taken by particle to reach to ground                                                                                                              

distance covered by both particles = d = (speed)(time)

taking g = 10m/s2

d1 = v1t = 10/g                       (distance  covered by particle 1)

d2  = v2t  = 15/g  = 1.5              (distance covered vy particle 2)

total distance bw particle when they reach the ground  = d1+d2 = 2.5usin@

 d = 25root3 meters

   =43.25m

Answer 2

Answer:

Explanation:

Hopes this helps you....