Question

A firework of mass 0.25 kg is launched at an angle of 75 degrees above the horizontal, pointing due west. The initial speed of the firework is 69 m/s. At the top of its trajectory, it explodes into two pieces of equal mass. One of the pieces takes 2.1 s to fall to the ground, and lands on the ground a distance of 53 m due west from the launch position. Neglect air resistance. (a) Find the velocity of both pieces just after the explosion. (b) Where does the other piece land? (c) Sketch a picture of the resulting motion and convince yourself that it is plausible.

Answer 1

V_o = initial velocity of launch = 69 m/s

= angle of launch = 75 deg

R = normal horizontal range of the firework when no explosion take place

R = V²_o Sin(2)/g

R = (69)² Sin(150)/9.8 = 243 m

consider the motion of one piece after explosion in vertical direction

Y = vertical height travelled = V²_o Sin²)/(2g) = (69)² Sin²75/9.8 =226.7

t = time of travel = 2.1 sec

a = acceleration = 9.8 m/s²

V_1y = velocity of piece in vertical direction

using the equation

Y = V_1y t + (0.5) a t²

226.7 = V_1y (2.1) + (0.5) (9.8) (2.1)²

V_1y = 97.7 m/s

consider the motion along the horizontal direction

X = horizontal distance travelled = (R/2) - 53 = (243/2) - 53 = 68.5 m

V_1x = X/t = 68.5/2.1 = 32.6 m/s

Using conservation of momentum along the horizontal direction

m₁ V_1x + m₂ V_2x = m V_x

(0.25/2) (-32.6) + (0.25/2) V_2x = (0.25) (69 Cos75)

V_2x = 68.3 m/s

Using conservation of momentum along the vertical direction

m₁ V_1y + m₂ V_2y = m V_y

(0.25/2) (- 97.7) + (0.25/2) V_2y = (0.25) (0)

V_2y = 97.7 m/s

for first mass :

V₁ = magnitude of velocity = sqrt(V_1x² + V_1y²) = sqrt((- 32.6)² + (- 97.7)²) = 103 m/s

₁ = direction = tan^-1(97.7/32.6) = 71.5 deg below horizontal towards east

for second mass :

V₂ = magnitude of velocity = sqrt(V_2x² + V_2y²) = sqrt((68.3)² + (97.7)²) = 119.2 m/s

₂ = direction = tan^-1(97.7/68.3) = 55.03 deg above horizontal towards west

b)

r_cm = location of center of mass of the system = 243 m

r₁ = location of center of mass of first piece = 53 m

r₂ = location of center of mass of second piece = ?

m₁ r₁ + m₂ r₂ = m r

(0.25/2) (53) + (0.25/2) r₂ = (0.25) (243)

r₂ = 433 m