Question

MA ball explodes into three pieces. Two pieces of equal mass fly off to the north and to the east with a speed of 28 m/s each. Find the velocity of the third piece given that its mass is equal to the sum of the other two pieces.

Answer 1

mass of original piece = M_a

mass of two equal pieces = m = M_a/4

mass of third piece = m' = M_a/2

consider east-west direction as X-axis and north-south direction as Y-axis

Consider the x-direction

initial momentum = 0

velocity of one of two equal piece along X-direction = 28 m/s

V_x = speed of third particle along X-axis

Using conservation of momentum :

0 = m (28) + m' V_x

0 = ( M_a/4) (28) + (M_a/2) V_x

V_x = - 3.5 m/s      along negative X-direction (west)

Consider the Y-direction

initial momentum = 0

velocity of one of two equal piece along Y-direction = 28 m/s

V_y = speed of third particle along Y-axis

Using conservation of momentum :

0 = m (28) + m' V_x

0 = ( M_a/4) (28) + (M_a/2) V_y

V_y = - 3.5 m/s      along negative Y-direction (south)

Net velocity = V = sqrt (V_x² + V_y²) = sqrt ((-3.5)² + (-3.5)²) = 4.95 m/s

direction : along west-south direction