*Simulator***Previous Years AIEEE/JEE Mains Questions**

Q. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be :-

**Directions :**Question number 9 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best discribes the two statements.**[AIEEE – 2009]**
Q.

**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. (1) Statement–1 is true, Statement–2 is false (2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–1 (3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1 (4) Statement–1 is false, Statement–2 is true**[AIEEE – 2010]**
Q. This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements.

**Statement – I :**A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as $\mathrm{f}\left(\frac{1}{2}\mathrm{mv}^{2}\right)$then$\mathrm{f}=\left(\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}\right)$**Statement – II :**Maximum energy loss occurs when the particles get stuck together as a result of the collision. (1) Statement–I is true, Statement–II is true, Statement–II is a correct explanation of Statement–I. (2) Statement–I is true, Statement–II is true, Statement–II is a not correct explanation of Statement–I. (3) Statement–I is true, Statement–II is false. (4) Statement–I is false, Statement–II is true**[JEE Mains-2013]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**(4) Energy loss will be maximum when collision will be perfectly inelastic (By momentum) Maximum energy loss $=\mathrm{K}_{\mathrm{i}}-\mathrm{K}_{\mathrm{f}}$ $=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \mathrm{v}_{\mathrm{f}}^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \frac{\mathrm{m}^{2} \mathrm{v}^{2}}{(\mathrm{m}+\mathrm{M})^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}\left[1-\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}}\right]$ $=\left(\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}\right) \frac{1}{2} \mathrm{mv}^{2}$ statement 1 is false.

Q. A particle of mass m moving in the x direction with speed 2u is hit by another particle of mass 2m moving in the y direction with speed u. If the collisions perfectly inelastic , the percentage loss in the energy during the collision is close to :
(1) 56 % (2) 62% (3) 44% (4) 50%

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**Sol.**(1) Before collison Kinetic energy $=\frac{1}{2} \mathrm{m}(2 \mathrm{v})^{2} \times \frac{1}{2} 2 \mathrm{m}(\mathrm{v})^{2}$ $=3 \mathrm{mv}^{2}$ After collison Applying momentum conservation for inelastic collision $2 \mathrm{mv} \hat{\mathrm{j}}+\mathrm{m} 2 \mathrm{v} \hat{\mathrm{i}}=3 \mathrm{m} \overrightarrow{\mathrm{v}}_{\mathrm{f}}$ $\left|\overrightarrow{\mathrm{v}}_{\mathrm{f}}\right|=\sqrt{\frac{8}{9}} \mathrm{v}$ $\mathrm{K}_{\mathrm{f}}=\frac{1}{2} \times 3 \mathrm{m} \times\left(\mathrm{v}_{\mathrm{f}}^{2}\right)=\frac{4 \mathrm{mv}^{2}}{3}$ $\% \Delta \mathrm{K}=\frac{\mathrm{K}_{1}-\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{i}}} \times 100=\frac{5 \mathrm{mv}^{2} / 3}{3 \mathrm{mv}^{2}} \times 100=\frac{5}{9} \times 100=56 \%$

Q. Distance of the centre of mass of a solid uniform cone from its vertex is $\mathrm{Z}_{0}$. If the radius of its base is R and its height is h then $\mathrm{Z}_{0}$ is equal to :-
(1) $\frac{5 \mathrm{h}}{8}$
(2) $\frac{3 \mathrm{h}^{2}}{8 \mathrm{R}}$
(3) $\frac{\mathrm{h}^{2}}{4 \mathrm{R}}$
(4) $\frac{3 \mathrm{h}}{4}$

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**Sol.**(4) for solid cone c.m. is $\frac{\mathrm{h}}{4}$ from base so $\mathrm{z}_{0}=\mathrm{h}-\frac{\mathrm{h}}{4}=\frac{3 \mathrm{h}}{4}$

Q. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $\mathrm{P}_{\mathrm{d}}$ ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $\mathrm{P}_{\mathrm{c}}$. The values of $\mathrm{P}_{\mathrm{d}}$ and $\mathrm{P}_{\mathrm{c}}$ are respectively :
(1) (.28, .89)
(2) (0, 0)
(3) (0, 1)
(4) (.89, .28)

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**Sol.**(4) Let initial speed of neutron is $\mathrm{v}_{0}$ and kinetic energy is K. 1st collision : by momentum conservation $\mathrm{mv}_{0}=\mathrm{mv}_{1}+2 \mathrm{mv}_{2} \Rightarrow \mathrm{v}_{1}+2 \mathrm{v}_{2}=\mathrm{v}_{0}$ by $\mathrm{e}=1 \quad \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{0}$

Q. In a collinear collision, a particle with an initial speed $\mathrm{V}_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :
(1) $\sqrt{2} \mathrm{v}_{0}$
(2)$\frac{\mathrm{v}_{0}}{2}$
(3) $\frac{\mathrm{v}_{0}}{\sqrt{2}}$
(4) $\frac{\mathrm{v}_{0}}{4}$

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**Sol.**(1) Initial

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