Question

a 3.0 kg object at rest explodes into 3 distinct fragments each moving in different directions. The first is 0.50kg and flies west at 2.8m/s, the second is 1.3 kg moving south at 1.5m/s. What is the mass, speed, and direction of the third fragment? How much kinetic energy was released in this explosion?

Answer 1

Initital momentum of the system = 0 <---- as it is at rest

Now, final momentum of the 1st fragment, p1 = m1*v1 = -0.5*2.8 i

final momentum of the 2nd fragment , p2 = m2*v2 = -1.3*1.5 j

Let the velocity of the 3rd particle be v

So, p1 + p2 + m3*v = 0 <---------- Using conservation of momentum

So, -0.5*2.8 i + (-1.3*1.5 j ) + (3 - 0.5 - 1.3)*v = 0

m3 =  (3 - 0.5 - 1.3) = 1.2 kg

So, v = 1.62 j + 1.17 i

So, mass of the 3rd fragment = m3 = 1.2 kg

speed of the 3rd fragment = v = sqrt(1.62^2 + 1.17^2) = 2 m/s <------- answer

direction of the 3rd fragment = atan(1.62/1.17) = 54.2 deg North of east <----- answer

initial KE = 0

final KE = 0.5*0.5*2.8^2 + 0.5*1.3*1.5^2 + 0.5*1.2*2^2 = 5.82 J <------- answer