Two pendulum bobs of masses m and 2m collide elastically at the lowest point in their motion, when the centres are at the same level as shown in the figure. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision ?

Two pendulum bobs of masses m and 2m collide elastically at the lowest

1.	Two pendulum bobs of masses m and 2m collide elastically at the lowest point in their motion, when the centres are at the same level as shown in the figure. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision ?
Answer» Solution :GIVEN, `m _(1) = m , m _(2) = 2m , u _(1) = sqrt (2gH) and u _(2) = sqrt (2gH)` Since, the collision is elastic, their VELOCITIES after collision are `v _(1) = ((m -2m)/( m + 2m )) (- sqrt (2 gH)) + ((4m)/( m + 2m )) sqrt (2 gH)` `= (sqrt( 2 gH))/(3) + (4 sqrt (2 gH))/( 3 ) = 5/3 sqrt (2 gH) and v _(2) = ((2m)/( m + 2m )) (- sqrt (2 gH))+ ((2m- m )/( m + 2m )) (sqrt (2gH))` `= (sqrt ( 2 gH))/( 3)- (2 sqrt ( 2gH))/( 3) =- ( sqrt (2 gH))/(3)` i.e., the velocities of the balls after the collision are as shown in fig. Therefore the heights to which the balls rise after the collision are: `h _(1) = ((v _(1) ) ^(2))/( 2g) `(using `v ^(2) - u ^(2) - 2 gh)` or `h _(1) = ((( 5)/(3) sqrt (2 gH ) ) ^(2) )/( 2g) = (25)/(9) H and h _(2) ((v _(2)) ^(2))/(2g) = (( sqrt (2gH)) ^(2))/( 2g) = (H)/(9)` Note the collision is elastic, mechanical energy of both the balls will remain conserved, or `E _(i) or E _(f)` `IMPLIES (m + 2m) gH = mgh _(1) + 2m gh _(2) implies 3 mgH =(mg) ((25)/(9) H) + (2MG) ((H)/(9))`