
If it is a completely inelastic collision then m1v1+m2v2=m1v+m2vv=m1+m2m1v1+m2v2 K⋅E=2 m1p12+2 m2p22 as p1 and p2 both simultaneously cannot be zero therefore total KE cannot be lost.
JEE Main 2010 — Physics Mechanics
Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions. Of the four choices given after the statements, choose the one that best describes the two statements. Of the four choices given after the statements, choose the one that best describes the two statements.
Held on 30 Apr 2010 · Verified 6 Jul 2026.
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement1
Statement-1 is false, Statement-2 is true.
Statement-1 is true, Statement-2 is false.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.
The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.
Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. 
A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)
Work through every JEE Main Mechanics PYQ, year by year.