a projectile is fired from a gun has an intial velocity of 9

a projectile is fired from a gun has an intial velocity of 90.0 km/h at an angle of 60.0 deg above the horizontal. when the projectile is at the top of its trajectory, an internal explosion causes it to seperate into two fragments of equal mass. one of the fragments falls straight downward as though it had been realased from rest. how far from the gun does the other fragment land?

B) Rework useing the concept of the center of mass and compute the distance the other fragment lands from the gun?

Solution

a)

At the top of the trajectory,

the speed of the projectile:

u = 90*cos(60 deg) = 45 km/h

Now, after explosion, the speed of the 1st fragment,u1 = 0

speed of the second fragment = u2

So, by conservation of momentum,

m*45 = (m/2)*(u1) + (m/2)*(u2)

So, u2 = 90 km/h

Now, height reached, h = ((90/3.6)*sin(60))^2/(2*9.8)

So, h = 2.96 m

Now, for the second fragment,

using the equation of motion

s = ut + 0.5*at^2

For vertical motion,

s = -2.96 m

u = 0

a = -9.8 m/s2

So, 2.96 = 0.5*9.8*t^2

So, t = 0.78 s

So, distance traveled by the second fragment:

D = (90/3.6)*0.78 = 19.5 m <------answer

b)

In the horizontal direction, there is no force.

So, the center of mass does not change in horizontal direction,

Taking the initial point as the point of projection.

So, Center of mass of the projectile at the initial point = 0 <------- x-component

Now, after time 2t ( t = time at which it reaches the top of trajectory = 0.78 s), the x-component of center of mass

X-COM = ((m/2)*R + (m/2)*(R+D))/(2*m/2) = R + D/2

where D = distance traveled by the second fragment from the point where 1st fragment drops on to the ground

R = (90/3.6)*cos(60 deg)*(0.78) = 9.75 m

Now, distance traveled by the projectile assuming it had not broken:

D\' = (90/3.6)*cos(60 deg)*2*0.78 = 19.5 m = R+D/2

So, 19.5 = 9.75 + D/2

So, D = 19.5 m <-----------answer