Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.
Explanation:
1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.
2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:
m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃
Solving for m₃, we have:
m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.
3) and 4)
As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.
So, we can write two equations, one for the x-component, and other for the y-component, as follows:
pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0
py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0
Replacing by the values, and solving for v₃ₓ and v₃y, we get:
v₃ₓ = 15.4 m/s
v₃y = 12.9 m/s
v = √(15.4)²+(12.9)² = 20.1 m/s
5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.
6) As initial kinetic energy is 0, as the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:
K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)
Replacing by the values, we get:
K= 3.62 KJ
. Re-arranging the formula, we can calculate the time t needed to do this amount of work: