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 253mdirectly 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.

Answer 1

Given that boulder explodes at the highest point of motion, So there veritcal velocity of boulder is zero, Now given that one of the fragment starts out only with horizontal direction, So due to momentum conservation other fragment should also starts with only horizontal velocity

So Using momentum conservation:

Pix = Pfx

At max point, Pix = 0, So

Pfx = 0

m1*v1x + m2*v2x = 0

v2x = -(m1/m2)*v1x

given that m2 = 3*m1, So m1/m2 = 1/3

v2x = -(1/3)*v1x

v2x/v1x = -(1/3)

Now given that m2 lands at 253 m noth of the point of explosion, So

Range in projectile motion = R = V0*t

Since both fragments are launched from same height so both will reach at ground at same time, So

t = R1/v1x = R2/v2x

R2 = R1*(v2x/v1x)

R2 = 253*(-1/3) = -84.333 m

-ve sign means m2 will land in the opposite direction of explosion point w.r.t. m1

So In three significant figures

other fragment will land at 84.3 m distance

Let me know if you've any query.