Question

plz ans this que with solution...

Answer 1

let the velocity of ball B just after the collision be V2 ....
and .. the maximum height attained by ball B is 3.2m so... it's velocity at highest point is 0....
so, by the conservation of mechanical energy at lowest and highest point ...
KEi + PEi = KEf + PEf ( i = initial and , f = final)
=> (1/2)m(V2)^2 + mg(0) = (1/2)m(0)^2 + mg(3.2). [ h= 3.2m]
after solving .....
V2 = 8 m/s ...
=> velocity of ball B just after the collision is 8m/s
now , by the conservation of linear momentum before and after the collision...
mU1 + mU2 = mV1 + mV2
=> U1 + U2 = V1 + V2 (mass of both the ball is same)
where , U1 = velocity of A before collision,
U2 = velocity of B before collision,
V1 = velocity of A after collision and ,
V2 = velocity of B after collision...
so, U1 = 10m/s , U2 = 0 , V1 = V1 and, V2 = 8m/s
so, 10 + 0 = V1 + 8
=> V1 = 2 m/s ...
velocity of ball A after collision is 2m/s ...
now , by the Newton's law of collision...
-> (V2 - V1)/(U1 - U2) = e ... where , e= coefficient of restitution...
=> (8-2)/(10-0) = e
=> e = 6/10
=> e = 3/5 ..... (Answer )