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 speed v0. the third piece has twice the mass as the other two. what are the speed and direction of the third piece? give the direction as an angle east of north.

Answer 1

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

$P_1 + P_2 + P_3 = 0$

here we know that two parts fly off with same speed v0 along south and west directions

$mv_0 (-\hat j) + mv_0 (-\hat i) + 2m \vec v = 0$

now we will have

$\vec v = \frac{v_0}{2}\hat i + \frac{v_0}{2}\hat j$

so here magnitude of the speed of the third part will be

$|v| = \sqrt{(\frac{v_0}{2})^2+(\frac{v_0}{2})^2}$

$|v| = \frac{v_0}{\sqrt2}$

and the direction of the third part will be along North East direction

Answer 2

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