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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassCollisionMedium2 minQB

PhysicsMediuminteger

Passage / Comprehension

A particle of mass m collides elastically with the pan of mass M = 2m of a spring balance. The pan is in equilibrium before collision. The spring constant is k and the particle speed before collision is $v_{0}$ . The figure shows the collision geometry. Answer the question using this common setup.

Source diagram for Phase II Advanced shared passage group 2

Maximum height attained by the particle (from the point of collision) after collision is found to be $\frac{v _{0}^{2}}{β g}$ . Find the value of $β$ where $β$ is the least integer.

Answer:

Solution:

The incident vertical speed is

$u_{y} = v_{0} cos 6 0^{\circ} = \frac{v _{0}}{2}$

For a one-dimensional elastic collision with the pan of mass 2m, the particle rebounds with vertical speed

$∣ v_{y} ∣ = \frac{m - 2 m}{m + 2 m} u_{y} = \frac{v _{0}}{6}$

Its maximum rise above the collision point is

$h = \frac{v _{y}^{2}}{2 g} = \frac{v _{0}^{2}}{72 g}$

$β = 72$

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Collision

⏱ 2mℹ️ Source: QB

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