newton's laws of motion

Question

a bomb of mass 1 kg initially at rest explodes and breaks into 3 fragments of masses in ratio 1:1:3. the 2 pieces of equal masses fly off perpendicular to each other with speed of 15 m/s each.the speed of heavier fragment is?

Answer 1

as with your previous question we need to set up a conservation of momentum equation

 

the fragments are .2kg, .2kg and .6kg

 

for convenience, use the x-axis for the direction of fragment 1

 

the speed of fragment 1 is 15m/s horizonitally

 

then speed of fragment 2 is 15m/s vertically (because the direction is perpendicular

 

momentum = mv

 

momentum of fragments 1 and 2 combined is sq. root[(2)(.2)² (15)² ] = 3√2

 

momentum fragment 3 = -(.6)v = 3√2

 

v = 7.1 m/s at an angle of 225 degrees in the third quadrant

Answer 2

This is a conservation of (vector) momentum problem.  The idea is that the system momentum before the explosion (zero) must be equal to the system momentum after the explosion.  It can be solved by choosing a coordinate system for the three after explosion momentum vectors and then requiring that the vector sum of these three momenta be zero.

 

A simple coordinate system is to chose the x axis to be along the direction of motion of the first fragment, and the y axis to be along the direction of motion of the second fragment.   Since the total mass is 1 kg, and the mass ratios are  1:1:3,  it can be seen that the mass of the first fragment is 1/5, the mass of the second fragment is also 1/5 and the mass of the third fragment is 3/5 kg.

 

In the simple coordinate system, the first two fragment momenta are:

p1  = ( (1/5) 15, 0)   = (3, 0)

p2  =( 0, (1/5) 15 )  =  (0 , 3)

 

In order for the vector sum of p1, p2 and p3 to be zero, we need   p3 = (-3, -3) .

The magnitude of p3 is thus   3 sqrt(2)

Since the mass of the third fragment is 3/5,   the speed of the third fragment is

 (5/3) 3 sqrt(2)  =   5 sqrt(2) =  7.07 m/s