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A ball of mass m kg hits an inclined smooth surface with a velocity VA=3m/s. Find out velocity of rebound

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Given:-

VA=3m/se=0.8

To find:-

Velocity after collision (V1A)

Solution:

The surface is smooth and rigid (µ=0), the component of velocity parallel to the surface will not change.

V1Ax=VAx=3cos60

Now,

e=\dfrac { V^1_{Ay} - V^1_{\text {Surface}}}{V_{\text {surface}}-V_{Ay}} =-\dfrac {V^1_{Ay}}{V_{Ay}} [ \because \text {Surface is rigid}] \\ \therefore V^1_{Ay} =-eV_{Ay} \\ =0.8 × 3 \sin60=-0.2078 \\ \therefore V^1_{Ay}=2.078 m/s (\downarrow )

The resultant velocity,

V^1_A=2.563 m/s (54.182⁰ )

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