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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]
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 explanationof 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 ofStatement–I.(3) Statement–I is true, Statement–II is false.(4) Statement–I is false, Statement–II is true [JEE Mains-2013]
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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% [JEE Mains-2015]
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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}$$=3 \mathrm{mv}^{2}$After collisonApplying momentum conservation for inelastic collision$2 \mathrm{mv} \hat{\mathrm{j}}+\mathrm{m} 2 \mathrm{v} \hat{\mathrm{i}}=3 \mathrm{m} \overrightarrow{\mathrm{v}}_{\mathrm{f}}$$\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) $\frac{3 \mathrm{h}}{4}$ [JEE Mains-2015]
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Sol. (4)
for solid cone c.m. is $\frac{\mathrm{h}}{4}$ from baseso $\mathrm{z}_{0}=\mathrm{h}-\frac{\mathrm{h}}{4}=\frac{3 \mathrm{h}}{4}$
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) [JEE Mains-2018]
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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) $\frac{\mathrm{v}_{0}}{4}$ [JEE Mains-2018]
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Sol. (1)Initial
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Worst solutions
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