A small bomb, of mass 10 kg, is moving toward the North with a velocity of 4.0 m/s. It explodes into three fragments: a 5.0-kg fragment moving west with a speed of 8.0 m/s; a 4.0-kg fragment moving east with a speed of 10 m/s; and a third fragment with a mass of 1.0 kg. What is the velocity of the third fragment?

40 m/s North

Explanation:

m = 10 kg

u = 4 m/s North = 4 j

m1 = 5 kg

v1 = 8 m/s west = - 8 i

m2 = 4 kg

v2 = 10 m/s east = 10 i

m3 = m - m1 - m2 = 10 - 5 - 4 = 1 kg

Let the velocity of third fragment is v3 = v

Using the conservation of momentum

m x u = m1 x v1 + m2 x v2 + m3 x v3

10 x 4 j = 5 x ( - 8i) + 4 x 10 i + 1 x v3

40 j = - 40 i + 40 i + v3

v3 = 40 j

Thus, the third fragment moves in the direction of north with velocity 40 m/s