A ball with mass m = 0.290 kg and kinetic energy K1 = 1.15 J collides elasticall

Question

A ball with mass m = 0.290 kg and kinetic energy K1 = 1.15 J collides elastically with a second ball of thesame mass that is initially at rest. After the collision, the first ball moves away at an angle of 1= 24.7? with respect to the horizontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

Explanation / Answer

K1 = (1/2) mu^2 => 1.15 = (1/2) * (0.290) u^2 => u = v(7.931034) m/s By t24.7°) = wsin? ... (1) u + 0 = vcos(24.7°) + wcos? => u - vcos(24.7°) = wcos? ... (2) Squarring and adding eqns. (1) and (2), u^2 + v^2 - 2uvcos(24.7°) = 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(24.7°) = 0 => v = ucos(24.7°) => Kinetic energy of the first ball after the collision = (1/2) mv^2 = (1/2) * (0.290) * ucos(24.7°) = (0.100) * v(7.931034) cos(24.7°) ... [plugging the value of u obtained in the first step] = 0.371 J.

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