Irodov 1.96: Momentum Keeps No Memory of Launch – Amit & Deepali Sinha Foundation

Problem Irodov 1.96

A body of mass $m$ is launched with initial velocity $\vect v_0$ at some angle to the horizontal and moves freely under constant gravitational acceleration $\vect g$ , air resistance neglected. Find:

the momentum increment $\Delta\vect p$ acquired over the first $t$ seconds of the flight
the magnitude of the momentum increment $\Delta\vect p$ over the whole flight, from launch until the body returns to its launch height

Given

$m$
$\vect v_0$
$\vect g$

Find

(a)

\Delta\vect p

over the first

t

seconds, (b)

|\Delta\vect p|

over the whole flight

Solution

Step 1Part (a): the impulse of gravity over

[0,t]

With air drag neglected, gravity is the only force on the body, so the momentum increment is the impulse of $m\vect g$ , constant, integrated over the elapsed time.

\Delta\vect p(t) = \int_0^t m\vect g\,d\tau

\Delta\vect p(t) = m\vect g\,t

(1)

The velocity triangle at P: a ghost copy of v_0 (dashed) carried across to P sits tail to tail with v(t), and the third side, joining their tips, is g t pointing straight down. — The velocity triangle at $P$ : a ghost copy of $\vect v_0$ (dashed) carried across to $P$ sits tail to tail with $\vect v(t)$ , and the third side, joining their tips, is $\vect g\,t$ pointing straight down.

Step 2The displacement, for the landing condition

Part (b) needs the total flight time $T$ , which requires the displacement. Dividing (1) by $m$ gives $\vect v(t) = \vect v_0+\vect g t$ , and integrating that once more gives the displacement.

\vect r(t) = \int_0^t\bigl(\vect v_0+\vect g\tau\bigr)\,d\tau = \vect v_0 t+\frac{\vect g t^2}{2}. \tag{2}

Step 3The total flight time

The flight ends at the instant $T$ when the body returns to launch height, so the displacement then has zero component along $\vect g$ . Dotting the displacement (2) with $\vect g$ and keeping the nonzero root isolates $T$ .

\begin{align*}0 &= \vect r(T)\cdot\vect g \\ &= T(\vect v_0\cdot\vect g)+\frac{T^2g^2}{2} \\ &= T\left[(\vect v_0\cdot\vect g)+\frac{Tg^2}{2}\right]\end{align*}

$T=0$ is the launch instant itself; the flight time is the other root, where the bracket vanishes.

T = -\dfrac{2(\vect v_0\cdot\vect g)}{g^2}

(3)

Step 4Part (b): the momentum increment over the whole flight

The increment (1) holds for any elapsed time, so setting $t=T$ and substituting (3) gives the momentum increment over the whole flight; its magnitude is $mg$ times $T$ , since $\vect g\,T$ and $\vect g$ share a direction.

\begin{align*}\left|\Delta\vect p(T)\right| &= mg\,T \\ &= mg\left(-\frac{2(\vect v_0\cdot\vect g)}{g^2}\right)\end{align*}

\left|\Delta\vect p\right| = -\dfrac{2m(\vect v_0\cdot\vect g)}{g}

(4)

Check

Dimensionally $[m(\vect v_0\cdot\vect g)/g] = \text{kg}\cdot(\text{m/s})(\text{m/s}^2)/(\text{m/s}^2) = \text{kg}\cdot\text{m/s}$ , a momentum. The minus sign is necessary rather than decorative: $\vect v_0$ has a component against $\vect g$ at launch, so $\vect v_0\cdot\vect g<0$ and the whole expression comes out positive, as a magnitude must. For a body launched exactly horizontally ( $\vect v_0\cdot\vect g=0$ ), $T=0$ : it never rises above its launch height at all, so there is no later instant at which to measure a nonzero return.

Remark

Part (a)’s result carries no memory of the launch: $\Delta\vect p(t)=m\vect g\,t$ depends only on the elapsed time, never on $\vect v_0$ , because gravity’s impulse rate $d\vect p/dt=m\vect g$ is the same at every instant of every trajectory under this constant field. The initial velocity only ever enters through $T$ in part (b), which sets how long the flight lasts, not the rate at which momentum is changing during it.