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A ball with a velocity of 5 ms−1 impinges at angle of 60˚ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

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प्रश्न

A ball with a velocity of 5 ms−1 impinges at an angle of 60˚ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

योग
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उत्तर

U1 = 5 ms-1

θ = 60°

e = 0.5

v = ?

Initial momentum = final momentum along the original line of m of the con.

∵ The coefficient of restitution is 0.5 (less than 1) the collision is inelastic

Applying component of velocities. The x component of velocity is

u sin θ = v sin Φ → (1)

But the magnitude of the y component is not the same using the coefficient of restitution

`ℓ = (v cos phi)/(u_1 cos theta) = 1` → (2)

squaring and adding (1) & (2)

`u^2 sin^2theta = v^2 sin^2phi`

`e^2 u^2 cos^2theta = v^2 cos^2 phi`

`v^2 = u^2[sin^2theta + e^2cos^2theta]`

v = u`sqrt(sin^-2theta + e^2cos^2theta)`

= 5`sqrt(sin^-2 60 + (0.5)^2 cos^2 60)`

= `5sqrt(3/4 + 0.25 xx 1/4)`

= `5sqrt(0.75 + 0.0625)`

= `5sqrt(0.8125)`

= `5 xx 0.90`

= 4.5 m/s

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अध्याय 4: Work, Energy and Power - Evaluation [पृष्ठ २०५]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 4 Work, Energy and Power
Evaluation | Q IV. 2. | पृष्ठ २०५

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