Physics · System Of Particles

JEE Main 2026 — 23 January, Morning Shift — Question 41

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 $10 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$ , respectively, strike each other and stick together. The rise in temperature (in ${ }^{\circ} \mathrm{C}$ ), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material $31 \mathrm{cal} / \mathrm{kg} .{ }^{\circ} \mathrm{C}$ and $1 \mathrm{cal}=4.2 \mathrm{~J}$ )

Option A:
1.75
Option B:
1.44
Correct
Option C:
1.15
Option D:
1.95

Answer: B

Step-by-step solution

$(\mathrm{K} . \mathrm{E})_{\text {lost }}=\frac{1}{2} \mu \mathrm{~V}_{\text {rel }}^{2}\left(1-\mathrm{e}^{2}\right)$

\begin{aligned} & =\frac{1}{2}\left(\frac{\mathrm{~m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)(10+30)^{2}(1-0) & =\frac{1}{2}\left[\frac{(15)(25)}{40}\right][40]^{2} & =7500 \mathrm{~J} \end{aligned}

$(\mathrm{K.E})_{\text {loss }}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)(\mathrm{S})(\Delta \mathrm{T})$ $\left[\mathrm{S}=31 \times 4.2 \mathrm{~J} / \mathrm{kg}-{ }^{\circ} \mathrm{C}\right]$ $7500=(40)(31)(\Delta \mathrm{T})$ $\Delta \mathrm{T}=\frac{7500}{40 \times 31 \times 4.2}=1.44^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

Practise System Of Particles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: System Of Particles
Topic: Collisions in One Dimension