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Physics
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Center of Mass, Momentum and Collision

NEET · Physics · Center of Mass, Momentum and Collision

A moving body of mass M with velocity v_0 i explodes into two...

A moving body of mass MM with velocity v0i^v_0 \hat{i} explodes into two fragments of masses m1m_1 and m2m_2. Immediately after the explosion, the fragment of mass m1m_1 comes to rest. Match the mass ratios given in List-I with the corresponding final velocity of the second fragment given in List-II.

List-IList-II
(A) 1:11:1(I) 1.5v0i^1.5 v_0 \hat{i}
(B) 1:21:2(II) 2v0i^2 v_0 \hat{i}
(C) 2:12:1(III) 3v0i^3 v_0 \hat{i}
(D) 3:13:1(IV) 4v0i^4 v_0 \hat{i}

Choose the correct answer from the options given below:

Options

  1. A(A) - I, (B) - II, (C) - III, (D) - IV
  2. B(A) - II, (B) - III, (C) - I, (D) - IV
  3. C(A) - II, (B) - I, (C) - III, (D) - IV
  4. D(A) - II, (B) - I, (C) - IV, (D) - III

Correct answer

C. (A) - II, (B) - I, (C) - III, (D) - IV

Step-by-step solution

By the principle of conservation of linear momentum, the total initial momentum equals the total final momentum.

(m1+m2)v0i^=m1(0)+m2v2(m_1 + m_2) v_0 \hat{i} = m_1 (0) + m_2 v_2

v2=m1+m2m2v0i^=(m1m2+1)v0i^v_2 = \dfrac{m_1 + m_2}{m_2} v_0 \hat{i} = \left(\dfrac{m_1}{m_2} + 1\right) v_0 \hat{i}

For (A) m1:m2=1:1⇒m1m2=1⇒v2=(1+1)v0i^=2v0i^m_1:m_2 = 1:1 \Rightarrow \dfrac{m_1}{m_2} = 1 \Rightarrow v_2 = (1 + 1) v_0 \hat{i} = 2 v_0 \hat{i} (Matches II)

For (B) m1:m2=1:2⇒m1m2=0.5⇒v2=(0.5+1)v0i^=1.5v0i^m_1:m_2 = 1:2 \Rightarrow \dfrac{m_1}{m_2} = 0.5 \Rightarrow v_2 = (0.5 + 1) v_0 \hat{i} = 1.5 v_0 \hat{i} (Matches I)

For (C) m1:m2=2:1⇒m1m2=2⇒v2=(2+1)v0i^=3v0i^m_1:m_2 = 2:1 \Rightarrow \dfrac{m_1}{m_2} = 2 \Rightarrow v_2 = (2 + 1) v_0 \hat{i} = 3 v_0 \hat{i} (Matches III)

For (D) m1:m2=3:1⇒m1m2=3⇒v2=(3+1)v0i^=4v0i^m_1:m_2 = 3:1 \Rightarrow \dfrac{m_1}{m_2} = 3 \Rightarrow v_2 = (3 + 1) v_0 \hat{i} = 4 v_0 \hat{i} (Matches IV)

Therefore, the correct matching is (A) - II, (B) - I, (C) - III, (D) - IV.

Answer: (A) - II, (B) - I, (C) - III, (D) - IV
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