Question

A 3.12-kg projectile is fired with an initial speed of 122 m/s at an angle of 31° with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1.04 kg and 2.08 kg. At 3.51 s after the explosion, the 2.08-kg fragment lands on the ground directly below the point of explosion.

Answer 1

    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 = 3.12 kg

    m1 = 2.08 kg

    m2 = 1.04 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*3.66 + (1/2)*9.8*3.51^2

    v1y = (201.4 - (1/2)*9.8*3.51^2)/3.51

    = -40.18 m/s

    Apply conservation of momentum in y-direction

    0 = m1*v1y + m2*v2y

    v2y = -m1*v1y/m2

    = -2.08*(-40.18)/1.04

    = 80.36 m/s

    Apply conservation of momentum in x-direction.

    M*vox = m1*0 + m2*v2x

    v2x = M*vox/m2

    = 3.12*104.6/1.04

    = 313.8 m/s

    so, v2 = v2xi + v2yj

    = 313.8 (m/s) i + 80.36 (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 = 80.36*t - 4.9*t^2

    4.9*t^2 - 80.36*t - 201.4 = 0

    on sloving the above equation

    we get

    t = 18.6 m

    distance travelled before landing = v2x*t

    = 313.8*18.6

    = 5837 m

    distance travelled by first object = R/2

    = vo^2*sin(2*theta)/(2*g)

    = 122^2*sin(2*31)/(2*9.8)

    = 671 m

    so, the distance travelled by second fragmnet, = 671 + 5837

    = 6508 m

    = 6.508 km <<<<<<<<<<------------------Answer

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

    = 0.5*3.12*(104.6)^2

    = 17068 J

    Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

    = 0.5*2.08*40.16^2 + 0.5*1.04*(313.8^2 + 80.36^2)

    = 56240 J

    Energy released in the explosion,

    Kf - Ki = 56240 - 17068

    = 39172 J

    = 39.172 kJ <<<<<<<<<<------------------Answer