PHYSICS HW problem-speed (please help, I am lost)

Question

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 16 m/s. The third piece has twice the mass as the other two.

1. What is the speed of the third piece? (in m/s)
2. What is the direction of the third piece? (answer in degrees north of east)

Answer 1

I hope this helps!

Answer 2

Lauren,

The key here is conservation of total linear momentum: it must be same after explosion as it was before it (zero in this case)

Before the coconut was blown all of its parts were not moving that's why the total momentum of it was zero. But the conservation of total linear momentum the total of linear momentum of all parts of the coconut after it's explosion should also equal to zero.

Now, the two parts that move, one to the West and another to the South with the speed v=16m/s each with a momentum of p=M*v=M*16m/s by size. Their momenta therefore would add to a momentum [a diagonal made from vectors of momenta p (South) and p(West)] and it's abs.value P_1&2=sqrt(2)*p=sqrt(2)*M*16m/s and direction of 45^o SouthWest. The momentum of the third part with the mass of 2M must emerge from the coconut with momentum equal to P_1&2 in size but in opposite (I e. 45^o North of East direction). If it's speed is v₃ then

2M*v₃=P_1&2=sqrt(2)*M*16m/s

Therefore, after dividing by 2M,

v₃=8*sqrt(2) ≈11m/s [rounded to 2sig.figs]

Hope it is helpful

Dr.Ariel B.