Question

2) In a completely inelastic explosion, a 1kg pineapple at rest is blown up into three parts. Two of the pieces, mi me have the same mass of 0.25 kg each. m, has a velocity of 2m/s and a direction 30 degrees below the x-axis. m2 has a velocity of 3m/s with a direction 45 degrees above the x-axis. a) What is the initial momentum? b) What is the final momentum of the third piece (both magnitude and direction)?

Answer 1

Part A.

Given that initially pineapple was at rest, So

Initial momentum = Mass*initial velocity

Pi = m*Vi = 1 kg*0 m/sec

Pi = 0 kg-m/sec

Part B.

Since there is no external force applied, So Momentum before and after explosion will remain conserved

Pi = Pf

Since Pi = 0, So

Pf = 0

m1 = mass of 1st part = 0.25 kg

m2 = mass of 1st part = 0.25 kg

m3 = mass of 1st part = 1 - 0.25*2 = 0.5 kg

v1 = speed of m1 = 2 m/sec at 30 deg below x-axis

v1 = 2*cos 30 deg i - 2*sin 30 deg j = 1.732 i - 1 j

v2 = speed of m2 = 3 m/sec at 45 deg above x-axis

v2 = 3*cos 45 deg i + 3*sin 45 deg j = 2.12 i + 2.12 j

So

Pf = 0 = m1*v1 + m2*v2 + m3*v3

v3 = -(m1*v1 + m2*v2)/m3

v3 = -(0.25*(1.732 i - 1 j) + 0.25*(2.12 i + 2.12 j))/0.50

v3 = (-1.732 i + 1 j - 2.12 i - 2.12 j)/2

v3 = (-1.926 i - 0.56 j)

So magnitude of velocity of third piece will be:

|v3| = sqrt ((-1.926)^2 + (-0.56)^2)

|v3| = 2.01 m/sec

Direction of third piece = arctan (-0.56/(-1.926)) = 16.2 deg below negative x-axis = 163.8 deg below +ve x-axis

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