A 3.03-kg projectile is fired with an initial speed of 122 m/s at an angle of 31

Question

A 3.03-kg projectile is fired with an initial speed of 122 m/s at an angle of 31 degree with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.90 kg and 2.13 kg. At 3.84 s after the explosion, the 2.13-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 0.90-kg fragment immediately after the explosion. v1=m/s i +m/s j (b) Find the distance between the point of firing and the point at which the 0.90-kg fragment strikes the ground. km (c) Determine the energy released in the explosion. kJ

Explanation / Answer

at the heighest point, the projectile has only x component of velcoity.

vox = vo*cos(31)

= 122*cos(31)

= 104.6 m/s

height above the ground when explosion takes place, h = vo^2*sin^2(theta)/(2*g)

= 122^2*sin^2(31)/(2*9.8)

= 201.4 m

let M = 3.03 kg

m1 = 2.12 kg

m2 = 0.9 kg

let v1 and v2 are speed of m1 and m2 after the explosion.

v1x = 0 (beacuse it is mving down stright)

Apply, h = v1y*t + (1/2)*g*t^2

201.4 = v1y*3.84 + (1/2)*9.8*3.84^2

v1y = (201.4 - (1/2)*9.8*3.84^2)/3.84

= -33.63 m/s

Apply conservation of momentum in y-direction

0 = m1*v1y + m2*v2y

v2y = -m1*v1y/m2

= -2.13*(-33.63)/0.9

= 79.6 m/s

Apply conservation of momentum in x-direction.

M*vox = m1*0 + m2*v2x

v2x = M*vox/m2

= 3.03*104.6/0.9

= 352.1 m/s

so, v2 = v2xi + v2yj

= 352.1 (m/s) i + 79.6 (m/s) j <<<<<<<<<-------------Answer

let t is time taken for the second part to fall down.

Apply, -h = v2y*t - 0.5*g*t^2

-201.4 = 79.6*t - 4.9*t^2

4.9*t^2 - 79.6*t - 201.4 = 0

on sloving the above equation

we get

t = 18.47 m

distance travelled before landing = v2x*t

= 352.1*18.47

= 6503

distance travelled by first object = R/2

= vo^2*sin(2*theta)/(2*g)

= 122^2*sin(2*31)/(2*9.8)

= 670 m

so, the distance travelled by second fragmnet, = 670 + 6503

= 7173 m

= 7.17 km <<<<<<<<<<------------------Answer

c) at the top point, Ki = 0.5*M*vox^2

= 0.5*3.03*(104.6)^2

= 16576 J

Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*2.13*33.63^2 + 0.5*0.9*(79.6^2 + 352.1^2)

= 59844 J

Energy released in the explosion,

Kf - Ki = 59844 - 16576

= 43268 J

= 43.3 kJ <<<<<<<<<<------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.