Irodov 1.161: The Shot That Stops a Sliding Cannon – Amit & Deepali Sinha Foundation

Problem Irodov 1.161

A cannon of mass $M$ is released from rest on a frictionless incline that makes angle $\alpha$ with the horizontal, and slides freely down the slope. After it has slid a distance $l$ along the incline, it fires a shell horizontally (in the ground frame), the shell carrying away momentum of magnitude $p$ ; the shell’s own mass is negligible next to $M$ . The recoil brings the cannon exactly to rest. Find the duration $\tau$ of the shot (the time the shell takes to leave the barrel).

The cannon, of mass M, has slid a distance l down the incline and reached speed v_0 along the slope when it fires; the shell leaves horizontally with momentum p. — The cannon, of mass $M$ , has slid a distance $l$ down the incline and reached speed $\vect v_0$ along the slope when it fires; the shell leaves horizontally with momentum $\vect p$ .

Given

$M$
$\alpha$
$l$ (distance slid before firing)
$p$ (shell’s momentum, horizontal)
shell mass negligible

Find

duration

\tau

of the shot

Solution

Step 1Speed of the cannon just before firing

Starting from rest and sliding a distance $l$ under constant acceleration $g\sin\alpha$ along the slope,

v_0^2 = 2(g\sin\alpha)l.

v_0 = \sqrt{2gl\sin\alpha}

(1)

Step 2Momentum balance along the incline through the shot

Resolve the shell’s momentum, which is horizontal, along the down-slope direction: since the slope makes angle $\alpha$ with the horizontal, the down-slope component of a horizontal vector of magnitude $p$ is $p\cos\alpha$ . Momentum conservation for the cannon-plus-shell system along the incline, from just before firing to just after, together with gravity’s impulse on the cannon during $\tau$ (the shell’s own weight is negligible), gives

Mv_{\text{final}} = Mv_0 + Mg\sin\alpha\,\tau - p\cos\alpha, \tag{2}

where the term $-p\cos\alpha$ is the along-slope recoil the cannon receives as the shell carries $p\cos\alpha$ away down-slope.

Step 3Impose the stopping condition

The cannon stops as a consequence of the shot, so $v_{\text{final}}=0$ . Setting the momentum balance (2) to zero and substituting the speed (1),

\begin{align*}0 &= Mv_0+Mg\sin\alpha\,\tau-p\cos\alpha\\ p\cos\alpha-M\sqrt{2gl\sin\alpha} &= Mg\sin\alpha\,\tau.\end{align*}

\tau = \dfrac{p\cos\alpha-M\sqrt{2gl\sin\alpha}}{Mg\sin\alpha}

(3)

Check

Dimensionally, $p\cos\alpha$ and $M\sqrt{2gl\sin\alpha}=Mv_0$ are both momenta, so the numerator is a momentum and dividing by $Mg\sin\alpha$ (mass times acceleration) leaves a time, as required. The stopping condition itself is the check on the sign: the shell must carry away more down-slope momentum than the cannon already has, $p\cos\alpha>Mv_0$ , or $\tau$ comes out negative, which correctly flags that a shell too weak, or aimed too far from the down-slope direction, cannot possibly bring a sliding cannon to rest.

Remark

Only the shell’s down-slope projection $p\cos\alpha$ ever enters, because the shell is constrained to leave horizontally, not along the slope: on a shallow incline ( $\alpha\to0$ ) a horizontal shot points almost straight down the slope and nearly all of $p$ helps stop the cannon, while on a steep incline ( $\alpha\to90^\circ$ ) the same horizontal shot points almost across the slope, $\cos\alpha\to0$ , and an ever larger $p$ is needed to stop the same cannon. The gravity term $Mg\sin\alpha\,\tau$ is not a rounding correction either. Measured against the shell’s own down-slope impulse it is $1-Mv_0/(p\cos\alpha)$ , which is small only when $p\cos\alpha$ barely exceeds $Mv_0$ , that is only in the marginal shot that stops the cannon over a vanishing firing time; a more powerful shell drives that ratio toward unity instead, so gravity’s share of the impulse grows rather than shrinks with $p$ . And in every case it is the only place the firing time $\tau$ itself enters the balance, so dropping it would leave no equation to solve for $\tau$ at all.