Respuesta :
As per momentum conservation theory
If net force on a system is zero then initial momentum of the system will be equal to final momentum of that system.
Since initially coconut is at rest so initial momentum is zero
Here it break into three pieces such that two parts of same mass while third part is of double mass
now we can say
[tex]P_1 + P_2 + P_3 = 0[/tex]
here we know that two parts fly off with same speed v0 along south and west directions
[tex]mv_0 (-\hat j) + mv_0 (-\hat i) + 2m \vec v = 0[/tex]
now we will have
[tex]\vec v = \frac{v_0}{2}\hat i + \frac{v_0}{2}\hat j[/tex]
so here magnitude of the speed of the third part will be
[tex]|v| = \sqrt{(\frac{v_0}{2})^2+(\frac{v_0}{2})^2}[/tex]
[tex]|v| = \frac{v_0}{\sqrt2}[/tex]
and the direction of the third part will be along North East direction
Consider east-west direction along the x-axis with east pointing towards positive x-axis.
Consider north-south direction along the y-axis with north pointing towards positive y-axis.
V = initial velocity of three pieces of coconut together before blast = 0 m/s
m₁ = mass of the piece moving towards south = m
v₁ = velocity of the piece moving towards south = 0 i - v₀ j
m₂ = mass of the piece moving towards west = m
v₂ = velocity of the piece moving towards west = - v₀ i + 0 j
m₃ = mass of the third piece = 2 m
v₃ = velocity of the third piece
using conservation of momentum
(m₁ + m₂ + m₃ ) V = m₁ v₁ + m₂ v₂ + m₃ v₃
inserting the values
(m + m + 2 m ) (0) = m (0 i - v₀ j) + m (- v₀ i + 0 j) + (2m) v₃
0 = m (0 i - v₀ j) + m (- v₀ i + 0 j) + (2m) v₃
0 = - (mv₀) j - (m v₀) i + (2m) v₃
0 = - (v₀) j - (v₀) i + (2) v₃
v₃ = (0.5) v₀ i + (0.5) v₀ j
speed : Sqrt(((0.5) v₀)² + ((0.5) v₀)²) = (0.71) v₀
θ =angle east of north = tan⁻¹((0.5) v₀)/(0.5) v₀)) = tan⁻¹(1) = 45 degree
