Question

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Answer 1

here,

m1 = 2 kg

m2 = 4.15 kg

height , h = 6 m

the initial velocity of 1 before the collison , u1 = sqrt(2 * g * h)

u1 = sqrt(2*9.8*6) = 10.85 m/s

the initial velocity of 2 before the collison , u2 = - sqrt(2 * g * h)

u2 = - sqrt(2*9.8*6) = - 10.85 m/s

let the final velocities after the collison be v1 and v2

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

2 * 10.85 - 4.15 * 10.85 = 2 * v1 + 4.15 * v2 ......(1)

and

using conservation of kinetic energy

0.5 * m1 * u1^2 + 0.5 * m2 * u2^2 = 0.5*m1 * v1^2 + 0.5 * m2 * v2^2

2 * 10.85^2 + 4.15 * 10.85^2 = 2 * v1^2 + 4.15 * v2^2 ......(2)

from (1) and (2)

v1 = - 18.4 m/s

v2 = 3.26 m/s

the height gained by v1 , h1 = v1^2 /(2g)

h1 = 18.4^2 /(2*9.81) = 17.3 m

the height gained by v2 , h2 = v2^2 /(2g)

h2 = 3.26^2 /(2*9.81) = 0.54 m