Question

An explosion blows a rock into three parts. two parts go off right angle to each other these two are, 1kg first moveing with velocity 12 m/s and 2 kg, second part moving with velocity of 8 m/s. if the third part flies off with a velocity of 4 m/s, its mass would be

Answer 1

The mass of the 3rd particle will be 5 kg

Explanation :

According to the law of conservation of momentum, the initial and final momentum should be equal. But since the momentum before explosion is zero, hence final momentum after explosion should be also zero.

Since the first two particle are at right angle to each other, hence their resultant will be,

P = √(P₁² + P₂²)

=> P = √(m₁²v₁² + m₂²v₂²)

= √(1² x 12² + 2²x8²)

= √ (144 + 256)

= √ 400

= 20

Hence this should be equal to the momentum of 3rd particle,

=> m₃v₃ = P

=> m₃ x 4 = 20

=> m₃ = 5 kg

Hence the mass of the 3rd particle will be 5 kg

Answer 2

Answer:

The mass of the 3rd particle is 5 kg

Explanation:

$\bf \implies \: P_ 1 + P_2 + P_3 = 0 \\ \\ \bf \implies \: |P_3| = |P_1 +P_2 | \\ \\ \bf \implies \:m \times 4 = \sqrt{ {P_1}^{2} + {P_2}^{2} } \\ \\ \bf \implies \: m \times 4 = \sqrt{ {12}^{2} + {16}^{2} } \\ \\ \bf \implies \:m \times 4 = 20 \: kg \\ \\ \bf \implies \:m = 5kg$