Center of Mass – JEE Main Previous Year Questions with Solutions

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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]

Sol. (1)


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]

Sol. (2)


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]

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]

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) $\frac{3 \mathrm{h}}{4}$

[JEE Mains-2015]

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)

[JEE Mains-2018]

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]

Sol. (1)

Initial


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