A shell is fired from a gun with a muzzle velocity of 27 m/s, at an angle of 60°

Question

A shell is fired from a gun with a muzzle velocity of 27 m/s, at an angle of 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

Ricardo, mass 82 kg and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the center of the canoe. Ricardo notices that the canoe moved 21.5 cm relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

Explanation / Answer

1)
v0x=cos60*27=13.5 m/s
v0y=sin60*27=23.38 m/s

how long does it take the bullet to reach max height?
v=v0-gt
v0/g=t
t=23.38/g

when the bullet splits, from conservation of momentum:
m*v=0.5m*u
2v=u
so the new vertical speed is u=27m/s.

the time it takes the bullet to land is the same time it took it to rise.
so..
t*v0x+t*2v0x=3tv0x=3*23.38/g*13.5=96.62m

2)
This is a center of mass question. Taking CM about the log assuming the log is directly below the center of the canoe before the seat exchange

(82*3-C*3)/M

where C is Carmelita's mass and M is the total mass of the system.

after the seat exchange
(C*(3+.215)+30*.215-82*(3-.215))/M

set equal and solve for C

82*3-C*3=C*(3+.215)+30*.215-82*(3-.215...

simplify

C=(82*(6-.215)-30*.215)/6.215

75.28 kg

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