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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediumsingle choice

A particle of mass ' $m$ ' is projected with velocity $v_{0}$ at an angle ' $α$ ' with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is e.

Options:

Answer:

Solution:

Successive flight times form a GP: $t_{0} = \frac{2 v _{0} s i n α}{g}$ , then $e t_{0}, e^{2} t_{0}, \dots$ $T = \frac{t _{0}}{1 - e} = \frac{2 v _{0} s i n α}{g ( 1 - e )}$ , so (A) is correct. Horizontal distance $= v_{0} cos α \cdot T = \frac{v _{0}^{2} s i n 2 α}{g ( 1 - e )}$ , which is twice option (B), so (B) is wrong. Total ground impulse over n impacts $= m u (1 + e) \frac{1 - e ^{n}}{1 - e} (u = v_{0} sin α)$ ; dividing by $T_{n} = \frac{2 u}{g} \frac{1 - e ^{n}}{1 - e}$ gives average force $\frac{m g ( 1 + e )}{2}$ , so (C) and (D) are wrong. Answer: (A)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 3mℹ️ Source: QB

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