Question

An explosion breaks a 32.0-kg object into three parts. The object is initially moving at a velocity of 25.0 m/s due north. Part (1) has a mass m11 = 3.30-kg and a velocity of 51.0 m/s due east. Part (2) has has a mass m22 = 6.80-kg and a velocity of 71.0 m/s due north. What is the magnitude of the velocity of part (3)?

Answer 1

Using momentum conservation:

In x-direction

Pix = Pfx

Pix = 0, since initially object was moving in north, So

Pfx = m1*v1x + m2*v2x + m3*v3x = 0

given values are:

m1 = 3.30 kg, m2 = 6.80 kg, m3 = 32.0 - 6.80 - 3.30 = 21.9 kg

v1x = speed of m1 in x-direction = 51.0 m/sec (since traveling towards east)

v2x = 0 m/sec (since traveling due north)

v3x = ?

3.30*51.0 + 6.80*0 + 21.9*v3x = 0

v3x = -3.30*51.0/21.9

v3x = -7.685 m/sec

In y-direction

Piy = Pfy

Piy = m*U = 32.0*25.0 = 800 kg-m/sec, since initially object was traveling in north, So

Pfy = m1*v1y + m2*v2y + m3*v3y = 800

v1y = speed of m1 in y-direction = 0 m/sec (since traveling towards east)

v2y = 71.0 m/sec (since traveling due north)

v3y = ?

3.30*0 + 6.80*71.0 + 21.9*v3y = 800

v3y = (800 -6.80*71.0)/21.9

v3y = 14.484 m/sec

Now velocity of third particle will be:

V3 = V3x i + V3y j

V3 = -7.685 i + 14.484 j

Magnitude of velocity of third particle will be:

|V3| = sqrt ((-7.685)^2 + (14.484)^2)

|V3| = 16.4 m/sec

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