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QUESTION

# sosi

This is the case of oblique elastic collision.

Let the first ball have velocity initial velocity before the collision = u and final velocity after the collision = v at 30.9° above the horizontal

and the second ball have velocity w after the collision at an angle θ below the horizontal.

By the law of conservation of linear momentum applied in the vertical direction,

vsin(30.9°) = wsinθ ... (1) [mass of both the balls being equal cancel out]

Applying the law of conservation of momntum in the horizontal direction,

u + 0 = vcos(30.9°) + wcosθ

=> u - vcos(30.9°) = wcosθ ... (2)

Squarring and adding eqns. (1) and (2),

u^2 + v^2 - 2uvcos(30.9°) = w^2 ... (3)

For elastic collision, kinetic energy is also conserved

=> (1/2) mu^2 = (1/2)mv^2 + (1/2)mw^2

=> u^2 = v^2 + w^2

=> u^2 - v^2 = w^2 ... (4)

Subtracting eqn. (4) from (3),

2v^2 - 2uvcos(30.9°) = 0

=> v = ucos(30.9°)

=> Kinetic energy of the first ball after the collision

= (1/2) mv^2

= (1/2) * (0.200) * ucos(30.9°)

= (0.100) * √(294) cos(30.9°) ... [plugging the value of u obtained in the first step]

= 1.47 J.

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