Question

A projectile of mass 5.5 kg is fired with an initial speed of 178 m/s at an angle of 63◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.2 kg and 2.3 kg . The 2.3 kg fragment lands on the ground directly below the point of explosion 2.2 s after the explosion.

(a) Determine the velocity of the 1-kg fragment immediately after the explosion.
(b) Find the distance between the point of firing and the point at which the 1-kg fragment strikes the ground.
(c) Determine the energy released in the explosion.

Answer 1

a.
`v_(f2)=305.94m/s`

b.
`x=7839.828m`

c.
`E=149.5kJ`

Step-by-step explanation:

Momentum conserved and using Newton's law equation to determine the time of the motion in axis y' and axis x'

`V_x=v*con(63)`

`V_x=178m/s*cos(63)=80.82m/s`

`V_y=v*sin(63)-g*t`

a.

Momentum p'

`m*V=m_1*v_(f1)+m_2*v_(f2)`

`v_(f2)=(m*V)/(m_2)= (5.5kg*178m/s)/(3.2kg)`

`v_(f2)=305.94m/s`

b.

`t=(v*sin(63))/(9.8m/s^2)=(178m/s*sin(63))/(9.8m/s^2)`

`t=16.2s`

`x=(v+v_(f2))*t`

`x=(178m/s+305.94m/s)*16.2s=7839.828m`

c.

`E=K_f-K_i`

`E=(1)/(2)*3.2kg*(305.94m/s)^2-(1)/(2)*5.5kg*(178m/s)^2`

`E=149474.05J`

`E=149.5kJ`