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 => 2.94 = (1/2) * (0.200) u^2 => u = v(294) m/s 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 ? below the horizontal. By the law of conservation of linear momentum applied in the vertical direction, vsin(30.9°) = wsin? ... (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°) + wcos? => u - vcos(30.9°) = wcos? ... (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) * v(294) cos(30.9°) ... [plugging the value of u obtained in the first step] = 1.47 J.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.