Question

A 20.0-kg projectile is fired at an angle of 60.0∘ above the horizontal with a speed of 80.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. Ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

Answer 1

Part a)

`R = 847.5 m`

Part b)

`\Delta E = 16000 J`

Step-by-step explanation:

As we know that projectile is launched at an angle of 60 degree with the horizontal

so at highest point of its path the velocity of projectile is only along horizontal direction

so we will have

`v_x = v cos60`

`v_x = 80 cos60 = 40 m/s`

now it explodes into two equal parts so by momentum conservation we will have

`mv_x = (m)/(2) v`

`v = 2 v_x`

`v = 2(40) = 80 m/s`

Part a)

time of flight of the projectile is given as

`T = (2vsin\theta)/(g)`

now total distance moved by it is given as

`R = (vsin60)/(g)(vcos60) + (vsin60)/(g)(2vcos60)`

`R = (80^2(0.866)(0.5))/(9.81) + (2(80^2)(0.866)(0.5))/(9.81)`

`R = 847.5 m`

Part b)

Energy released = change in kinetic energy

`\Delta E = (1)/(2)(m)/(2)(2vcos60)^2 - (1)/(2)m(vcos60)^2`

`\Delta E = (1)/(4)mv^2 - (1)/(8)mv^2`

`\Delta E = (1)/(8)(20)(80^2)`

`\Delta E = 16000 J`