A ball of mass m moving at a speed v makes a head on collision with a identical ball at rest .The kinetic energy of the balls after the collision is 3/4 of the original .Find the coefficient of restitution.
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➠ Mass of 1st ball => m
➠ speed => v
➠ Mass of 2nd ball => m
➧ Let final velocities of 1st and 2ndball are v1 and v2 respectively.
➧ Using law of conservation of momentum,
m (v1 + v2) = mv
➠ v1 + v2 = V ..........①
➧ Also,
v1 - v2 = ev ..............②
➧ Given that,
➧ Final K.E = ¾ Initial K.E
➠ ½ mv1 ^2 + ½ mv2 ^2 = ¾ × ½
➠ v1 ^2 + v2 ^2 = ¾ v2
➠ (v1 + v2)^2 + (v1 - v2)^2 / 2 = 3/4 v2
➠ (1 + e^2) v^2 / 2 = 3/4
➠ 1 + e^2 = 3 / 2
➠ e^2 = 1 / 2
➠ e = 1 / √2...✔
✮❀ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ❀✮
╰━─━─━─━─≪✠≫─━─━─━─━╯
➠ Mass of 1st ball => m
➠ speed => v
➠ Mass of 2nd ball => m
➧ Let final velocities of 1st and 2ndball are v1 and v2 respectively.
➧ Using law of conservation of momentum,
m (v1 + v2) = mv
➠ v1 + v2 = V ..........①
➧ Also,
v1 - v2 = ev ..............②
➧ Given that,
➧ Final K.E = ¾ Initial K.E
➠ ½ mv1 ^2 + ½ mv2 ^2 = ¾ × ½
➠ v1 ^2 + v2 ^2 = ¾ v2
➠ (v1 + v2)^2 + (v1 - v2)^2 / 2 = 3/4 v2
➠ (1 + e^2) v^2 / 2 = 3/4
➠ 1 + e^2 = 3 / 2
➠ e^2 = 1 / 2
➠ e = 1 / √2...✔
maheswari34:
thanks bro
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