6. A 3.27-kg projectile is fired with an initial speed of 119 m/s at an angle of

Question

6. A 3.27-kg projectile is fired with an initial speed of 119 m/s at an angle of 28degree with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1.05 kg and 2.22 kg. At 3.66 s after the explosion, the 2.22-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 1.05-kg fragment Immediately after the explosion (b) Find the distance between the point of firing and the point at which the 1.05-kg fragment strikes the ground km (c) Determine the energy released in the explosion. kJ Choose the system to include the projectile and Earth so that no external forces act to change the momentum of the system during the explosion. With this choice of system you can also use conservation of energy to determine the elevation of the projectile when It explodes. You will also find it useful to use constant-acceleration equations In your description of the motion of the projectile and Its fragments. Neglect air resistance. 7. The figure below shows the result of a collision between two objects of unequal mass.

Explanation / Answer

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

vox = vo*cos(31)

= 119*cos(28)

= 105.1 m/s

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

= 119^2*sin^2(28)/(2*9.8)

= 159.2 m

let M = 3.27 kg

m1 = 2.22 kg

m2 = 1.05 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

159.2 = v1y*3.66 + (1/2)*9.8*3.66^2

v1y = (159.2 - (1/2)*9.8*3.66^2)/3.66

= -25.6 m/s

Apply conservation of momentum in y-direction

0 = m1*v1y + m2*v2y

v2y = -m1*v1y/m2

= -2.22*(-25.6)/1.05

= 54.1 m/s

Apply conservation of momentum in x-direction.

M*vox = m1*0 + m2*v2x

v2x = M*vox/m2

= 3.27*105.1/1.05

= 327.3 m/s

so, v2 = v2xi + v2yj

= 327.3 (m/s) i + 54.1 (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

-159.2 = 54.1*t - 4.9*t^2

4.9*t^2 - 54.1*t - 159.2 = 0

on sloving the above equation

we get

t = 13.4 m

distance travelled before landing = v2x*t

= 327.3*13.4

= 4386

distance travelled by first object = R/2

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

= 119^2*sin(2*28)/(2*9.8)

= 599 m

so, the distance travelled by second fragmnet, = 599 + 4386

= 4985 m

= 4.98 km <<<<<<<<<<------------------Answer

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

= 0.5*3.27*(105.1)^2

= 18060 J

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

= 0.5*2.22*25.6^2 + 0.5*1.05*(327.3^2 + 54.1^2)

= 58505 J

Energy released in the explosion,

Kf - Ki = 58505 - 18060

= 40445 J

= 40.4 kJ <<<<<<<<<<------------------Answer

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