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Previous Years AIEEE/JEE Mains Questions
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]
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]
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]
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.
(1) 56 % (2) 62% (3) 44% (4) 50% [JEE Mains-2015]
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 $2 \mathrm{mv} \hat{\mathrm{j}}+\mathrm{m} 2 \mathrm{v} \hat{\mathrm{i}}=3 \mathrm{m} \overrightarrow{\mathrm{v}}_{\mathrm{f}}$ $\left|\overrightarrow{\mathrm{v}}_{\mathrm{f}}\right|=\sqrt{\frac{8}{9}} \mathrm{v}$ $\mathrm{K}_{\mathrm{f}}=\frac{1}{2} \times 3 \mathrm{m} \times\left(\mathrm{v}_{\mathrm{f}}^{2}\right)=\frac{4 \mathrm{mv}^{2}}{3}$ $\% \Delta \mathrm{K}=\frac{\mathrm{K}_{1}-\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{i}}} \times 100=\frac{5 \mathrm{mv}^{2} / 3}{3 \mathrm{mv}^{2}} \times 100=\frac{5}{9} \times 100=56 \%$
(1) $\frac{5 \mathrm{h}}{8}$ (2) $\frac{3 \mathrm{h}^{2}}{8 \mathrm{R}}$ (3) $\frac{\mathrm{h}^{2}}{4 \mathrm{R}}$ (4) $\frac{3 \mathrm{h}}{4}$ [JEE Mains-2015]
for solid cone c.m. is $\frac{\mathrm{h}}{4}$ from base so $\mathrm{z}_{0}=\mathrm{h}-\frac{\mathrm{h}}{4}=\frac{3 \mathrm{h}}{4}$
(1) (.28, .89) (2) (0, 0) (3) (0, 1) (4) (.89, .28) [JEE Mains-2018]
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 $\mathrm{e}=1 \quad \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{0}$
(1) $\sqrt{2} \mathrm{v}_{0}$ (2)$\frac{\mathrm{v}_{0}}{2}$ (3) $\frac{\mathrm{v}_{0}}{\sqrt{2}}$ (4) $\frac{\mathrm{v}_{0}}{4}$ [JEE Mains-2018]
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