Yes, it does. the answer comes as 8L/3 and this gives the centr of mass cordintes as (2L,0) The trick is if we take assume velocities (like I did) it is important to define wrt to ground. The relation that i derived for the velocities is correct but it is wrt to each other.
Hence
V(s)xt is not...
V - velocity of unbroken piece.
V(s)- velocity of smaller piece
V(l)- Velocity of larger peice.
t(f)- time of ascent or descent
all velocities in x axis.
before explosion momentum = MV
after explosion momentum = M/4 x[V-V(s)]+ 3M/4[V(l)+V]
equating and removing M and solving we get
3V(l)=...
My apologies. I had made an error in typing. It is the smaller piece that returns back to the launching station.
Thinking in terms of the conservation of momentum of center of mass. My answer seems correct. The centre of mass must be at the distance of L as it falls freely.
But shouldn't it...
An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and...
the time will be in sec and that is the time taken for the ball to reach the top of the truck.
Now use s= ut+1/at^2 where u=0 and t is the calculated time
since the truck is not accelerating we don't need to use the equation of motion.
s= ut+1/at^2
convert 55km/hr into m/s and find out the time for the truck to cross.
s=vxt
Homework Statement
An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of...