A ball of mass m collides elastically with an | vedclass

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(D) In an elastic collision between two identical masses where one is initially at rest,the conservation of linear momentum and kinetic energy leads t

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

(D) In an elastic collision between two identical masses where one is initially at rest,the conservation of linear momentum and kinetic energy leads to the following vector equation: $\vec{v}_1 = \vec{v}_1' + \vec{v}_2'$.Since the collision is elastic,$\frac{1}{2}mv_1^2 = \frac{1}{2}m(v_1')^2 + \frac{1}{2}m(v_2')^2$,which simplifies to $v_1^2 = (v_1')^2 + (v_2')^2$.This equation represents the Pythagorean theorem,implying that the velocity vectors $\vec{v}_1'$,$\vec{v}_2'$,and $\vec{v}_1$ form a right-angled triangle. Thus,the angle between the two balls after the collision must be $90^o$.Regarding energy transfer,$100 \%$ transfer occurs only in a head-on collision (impact parameter $= 0$). For a non-zero impact parameter,the balls move at an angle,and the kinetic energy is shared between them,meaning $100 \%$ transfer cannot take place in such cases.Therefore,both statements $(A)$ and $(C)$ are correct.

$A$ ball of mass $m$ collides elastically with an identical ball at rest with some impact parameter.

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