JEE Advance - Physics (2011 - Paper 1 Offline - No. 17)
The phase space diagram for a ball thrown vertically up from ground is




Explanation
Let the ball of mass m be thrown up with an initial velocity u. Its velocity v and displacement x are related by v2 $$-$$ u2 = $$-$$2gx, where g is the acceleration due to gravity. The momentum (p = mv) is given by
p2 = m2u2 $$-$$ 2m2gx,
which gives
$$p = \pm \sqrt {{m^2}{u^2} - 2{m^2}gx} $$.
At x = 0, the momentum is mu when the ball starts going up and it becomes $$-$$mu when the ball comes back. At the maximum height, x = u2/(2g), the momentum becomes zero.
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