NEETPhysicsCenter of Mass, Momentum and Collision
A bomb of mass 4m at rest explodes into 3 pieces of masses m , m , and 2m . The two pieces of mass m move with speed v perpendicular to each other. Match the quantities in List-I with their corresponding values in List-II. List-I List-II (A) Magnitude of momentum of the heaviest fragment (I) v 2 (B) Speed of the heaviest fragment (II) 2 m v (C) Kinetic energy of the heaviest fragment (III) 3 2 m v^2 (D) Total kinetic
Options
- A(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- B(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- C(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct answer
D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Step-by-step solution
By the law of conservation of linear momentum, the initial momentum of the bomb is zero. Thus, the vector sum of the momenta of the three fragments must be zero. Let the two fragments of mass m move along the x and y axes. Their momenta are p ₁ = m v i and p ₂ = m v j . The momentum of the heaviest fragment (mass 2m ) is p ₃ = -( p ₁ + p ₂) = -m v i - m v j . Magnitude of momentum of the heaviest fragment: p₃ = (mv)^2 + (mv)^2 = 2 m v . (Matches II) Speed of the heaviest fragment: v₃ = p₃ 2m = 2 m v 2m = v 2 . (Mat