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An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg1 \text{ kg} moves with a speed of 12 m s−112 \text{ m s}^{-1} and the second part of mass 2 kg2 \text{ kg} moves with 8 m s−18 \text{ m s}^{-1} speed. If the third part flies off with 4 m s−14 \text{ m s}^{-1} speed, then its mass is:

A

3 kg

B

5 kg

C

7 kg

D

17 kg

Step-by-Step Solution

  1. Conservation of Momentum: Since the explosion is caused by internal forces, the external force on the system is zero. Thus, the total linear momentum is conserved. The rock is initially at rest (Pinitial=0P_{initial} = 0). P⃗final=p⃗1+p⃗2+p⃗3=0  ⟹  p⃗3=−(p⃗1+p⃗2)\vec{P}_{final} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) (Refer to NCERT Class 11, Section 5.7 regarding Conservation of Momentum ).
  2. Calculate Momenta of First Two Parts:
  • Part 1: p1=m1v1=1×12=12 kg m s−1p_1 = m_1 v_1 = 1 \times 12 = 12 \text{ kg m s}^{-1}.
  • Part 2: p2=m2v2=2×8=16 kg m s−1p_2 = m_2 v_2 = 2 \times 8 = 16 \text{ kg m s}^{-1}.
  1. Resultant Momentum (p12p_{12}): Since the first two parts move at right angles (90∘90^\circ), the magnitude of their resultant momentum is: p12=p12+p22=122+162=144+256=400=20 kg m s−1p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ kg m s}^{-1}
  2. Determine Mass of Third Part: The momentum of the third part must balance this resultant (p3=p12=20 kg m s−1p_3 = p_{12} = 20 \text{ kg m s}^{-1}). Given v3=4 m s−1v_3 = 4 \text{ m s}^{-1}: m3=p3v3=204=5 kgm_3 = \frac{p_3}{v_3} = \frac{20}{4} = 5 \text{ kg}
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