Question

A body is projected obliquely from the ground such that its horizontal range is maximum. If the change in its linear momentum, as it move 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. $\sqrt{2}P$
• C. $2P$
• D. $2\sqrt{2}P$

Answer 1

Answer: (D)

$2\sqrt{2}P$

Given,

Initial velocity in vertical direction, $V_y=U\sin \theta$

Maximum height of projectile, $H_{max}=\frac{U^2{\sin }^2\theta }{2g}$

Half of max is $x=\frac{1}{2}{H}_{max}=\frac{U^2{\sin }^2\theta }{4g}$

Apply kinematic equation from half of maximum height to max height.

$v^2-u^2=2ax$

$v^2-U^2{\sin }^2\theta =2(-g)\left(\frac{U^2{\sin }^2\theta }{4g}\right)$

$v=\frac{U\sin \theta }{\sqrt{2}}......\left(1\right)$

Momentum $P=m{v}_2-m{v}_1$

Where,

$v_2=$velocity at max height

$v_1=$velocity at half of max height.

$P=\frac{mU\sin \theta }{\sqrt{2}}......\left(2\right)$

Now,

Change in velocity in the horizontal direction is zero so momentum change horizontally is zero.

Change in velocity in vertical direction is $U\sin \theta \left(initial\right)$$to-U\sin \theta$ (when it comes back to ground).

Change in momentum $\Delta P=m{V}_y-m{V}_3$

Where,

$V_y=$initial velocity of projectile.

$V_3=$ final velocity when it comes to the ground

$\Delta P=m(-U\sin \theta )-mU\sin \theta$

$\Delta P=-2mU\sin \theta =2\sqrt{2}\frac{mU\sin \theta }{\sqrt{2}}$

From equation (2)

$\Delta P=-2\sqrt{2}\frac{mU\sin \theta }{\sqrt{2}}=-2\sqrt{2}P$ (negative sign is just for direction)

Change in momentum is $2\sqrt{2}P$