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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2026 · 23 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

In a perfectly inelastic collision, two spheres made of the same...

In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg, moving in opposite directions with speeds of 10 m/s10 \mathrm{~m} / \mathrm{s} and 30 m/s30 \mathrm{~m} / \mathrm{s}, respectively, strike each other and stick together. The rise in temperature (in ∘C{ }^{\circ} \mathrm{C}), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material 31cal/kg.∘C31 \mathrm{cal} / \mathrm{kg}.{ }^{\circ} \mathrm{C} and 1cal=4.2 J1 \mathrm{cal}=4.2 \mathrm{~J})

Options

  1. A1.151.15
  2. B1.751.75
  3. C1.44
  4. D1.95

Correct answer

C. 1.44

Step-by-step solution

Using conservation of momentum:
m1v1+m2v2=(m1+m2)vfm_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f where v2=−30v_2 = -30 m/s (opposite direction).
15(10)+25(−30)=40vf⇒vf=−1515(10) + 25(-30) = 40v_f \Rightarrow v_f = -15 m/s
Initial kinetic energy:
KEi=12(15)(10)2+12(25)(30)2=750+11250=12000KE_i = \frac{1}{2}(15)(10)^2 + \frac{1}{2}(25)(30)^2 = 750 + 11250 = 12000 J
Final kinetic energy:
KEf=12(40)(15)2=4500KE_f = \frac{1}{2}(40)(15)^2 = 4500 J
Heat produced:
Q=12000−4500=7500Q = 12000 - 4500 = 7500 J =75004.2=1785.71= \frac{7500}{4.2} = 1785.71 cal
Temperature rise:
ΔT=Qmtotal⋅c=1785.7140×31=1.44°C\Delta T = \frac{Q}{m_{total} \cdot c} = \frac{1785.71}{40 \times 31} = 1.44°C
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