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. 5.14) is a possible result after collision ?
Case (ii)
It can be observed that the total momentum before and after collision in each case is constant. F
or an elastic collision, the total kinetic energy of a system remains conserved before and after collision.
For mass of each ball bearing m, we can write:
Total kinetic energy of the system before collision:
= \(\frac{1 }{ 2 }\) mV2 + \(\frac{1 }{ 2 }\) (2m)0
= \(\frac{1 }{ 2 }\) mV2
Case (i)
Total kinetic energy of the system after collision:
= \(\frac{1 }{ 2 }\) m × 0 + \(\frac{1 }{ 2 }\) (2m) (\(\frac{v }{ 2 }\))2
= \(\frac{1 }{ 4 }\) mV2
Hence, the kinetic energy of the system is not conserved in case (ii).
Case (iii)
Total kinetic energy of the system after collision:
= \(\frac{1 }{ 2 }\) (3m) \((\frac{v }{ 3 })\) 2
\(=\frac{ 1 }{ 2} mV^2\)
Hence, the kinetic energy of the system is conserved in case (iii).
The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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