6.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision ?

6.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision ?
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1 Answer
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6.16 The mass of the ball bearing = m
Before the collision, the total K.E. of the system = K.E. of the stationary ball bearing + K.E. of the striking ball bearing = 0 +
After the collision, the K.E. of the total system is
(a) Case (i) = 0 + =
(b) Case (ii) = 0 +
(c) Case (iii) = =
Case (ii) is possible since K.E. is conserved in this case.
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(b) conserving energy between “O” ans ”A”

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
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(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.
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