A body is projected obliquely from the ground such that its horizontal range is maximum. If the change in its linear momentum…

Q: A body is projected obliquely from the ground such that its horizontal range is maximum. If the change in its linear momentum, as it moves from half the maximum height to maximum height, is P, the change in its linear momentum as it travels from the point of projection to the landing point on the ground will be

(a) P

(b) √2 P

(c) 2P

(d) 2√2 P

Ans: (d)

Sol: Maximum height attained is

$ \displaystyle H = \frac{u^2 sin^2 \theta}{2 g} $

Hence , Half of the maximum height is

$ \displaystyle \frac{1}{2}H = \frac{u^2 sin^2 \theta}{4 g} $

$ \displaystyle v_y^2 = u_y^2 + 2a_y S_y $

$ \displaystyle v_y^2 = (usin\theta)^2 – 2 g (\frac{u^2 sin^2 \theta}{4 g}) $

$ \displaystyle v_y = \frac{u sin\theta}{\sqrt 2} $

Change in momentum from half of maximum height to maximum height :

$ \displaystyle P = m\frac{u sin\theta}{\sqrt 2} $

Required change in momentum

P’ = -2 m u sinθ

P’ = -2(√2 P)

P’ = – 2√2 P