Question

In a volcanic eruption, a 2.40 × 103 kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 340 m directly north of the point of the explosion.

Part A:

Where will the other fragment land? Neglect any air resistance.

Express your answer using three significant figures.

Δr = __m

Answer 1

both the heavier and the lighter particle has velocity in the horizontal direction because at the highest point the momentum is zero and so the vertical velocity is also zero.

applying the conservation of momentum

let m1= mass of lighter partcle

v1=velocity of heavier particle

m2=mass of heavier particle

v2=velocity of heavier particle.

m1v1+m2v2=0

v2= -m1v1/m2

we know that d=v*t

so the distance heavier particle moves is

d=v2t..........................eqn 1

replacing the value of v2 we get in eqn 1

d=-m1v1/m2*t

= -v1t(m1/m2)

=-340/3 ( since d=v1* t = distance lighter particle moved)

= -113.333m ( neglect the negative sign because it shows direction )

=113.333m to the south

so r=113.333m (to the south)