I've come to a problem today that solves this way,
5. A cannonball explodes into two pieces at a height of h = 100 m when it has a horizontal
velocity Vx=24 m/s.
The masses of the pieces are 3 kg and 2 kg. The 3-kg piece falls vertically to the ground 4 s after
the explosion. How far does the other piece fall?
The book solves:
The momentum is conserved during the explosion.
m1 = 2 kg, m2 = 3 kg, Vx = 100 m/s, Vy = 0, v2x=0.
The 3-kg piece falls down from the height of 100 m in 4 s, so 0-100=V2y*t-0.5gt^2 and get V2y=-5m/s
Now we can calculate the velocity components for the 2 kg body using
v2x=120/2=60 m/s, v2y = 3·5/2=7.5 m/s.
Its motion is a projectile motion. The time it takes for the piece to hit the ground is the positive
solution of the second order equation
t=3.8s, and d=227m.
I'm very confused because I solved it another way, and I get a different answer. Please help me figure out which is right.
For law of conservation of momentum, I solve
(m1+m2)Vi=m1Vf1+m2Vf2, or (3+2)24=3*0+2Vf2 and solve for Vf2=60m/s.
I did this to figure out the velocity in the x-direction only, since the 3kg piece fall straightly without any x-velocity.
Then I used one of the kinematic equations to solve for the time the 3-kg piece fall to the ground, which is about 4.5 second using Ddistance=V0t+0.5gt*2.
Finally, I use the time (4.5) times the x-velocity in the first equation (60m/s) to find out how far the 2-kg would have travelled and reached an answer of 271 meters, very different from the answer got in the textbook.
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