MA ball explodes into three pieces. Two pieces of equal mass fly off to the north...

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.

Solutions

Expert Solution

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

genius_generous answered 1 year ago

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