Irodov 1.163: Breaking Off at a Shared Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.163

A body of mass $M$ , whose upper profile is a specially shaped raised surface, rests on a smooth horizontal plane and carries a small disc of mass $m$ on its flat top. The disc is set moving horizontally with velocity $v$ and rides up along $M$ ’s curved surface, a face that curves smoothly upward until it becomes exactly vertical at the lip, which is the point where the disc eventually leaves $M$ . The plane under $M$ is smooth and the disc’s contact with $M$ is frictionless, so $M$ is free to slide as the disc climbs. Find the height, measured relative to the disc’s initial level, to which the disc rises after it breaks off from $M$ .

The disc m starts on M's flat top with velocity v; M's face curves smoothly up into a near vertical wall before a peaked lip, then slopes back down to the floor, on which M slides freely. — The disc $m$ starts on $M$ ’s flat top with velocity $v$ ; $M$ ’s face curves smoothly up into a near vertical wall before a peaked lip, then slopes back down to the floor, on which $M$ slides freely.

Given

Find

Find the height

h

the disc reaches, relative to its starting level

Solution

Step 1The horizontal velocities match at breakoff, fixed by momentum conservation

Both the floor under $M$ and the curved face the disc rides on are frictionless, so while the disc stays in contact, the normal force between the two bodies is the only thing coupling their motions. $M$ ’s face curves up from the flat starting shelf into a wall that is exactly vertical at the lip, the point where the disc leaves the body, so right there the shared tangent between disc and surface is vertical and the disc’s velocity relative to $M$ has no horizontal component left: the disc and $M$ already share one common horizontal velocity, call it $V$ , at the instant of breakoff, even though the disc still carries a nonzero vertical velocity of its own at that same instant.

The floor is smooth, so the disc plus $M$ system feels no external horizontal force at any time, and horizontal momentum is conserved throughout, in particular between the start (disc at $v$ , $M$ at rest) and the breakoff instant, where the disc’s horizontal velocity component equals $M$ ’s velocity $V$ :

mv = (m+M)V.

V = \dfrac{m}{m+M}v

(1)

Step 2Free flight to the apex, where both bodies move at

V

again

Once the disc leaves $M$ at the lip it is a free projectile: with no more contact force, its horizontal velocity stays fixed at the $V$ of (1) for the rest of the flight, and only its vertical velocity changes, under gravity. $M$ likewise feels no more horizontal push once contact ends, so it also coasts on at that same $V$ from then on.

Both surfaces were frictionless throughout, on the curve before breakoff and in free flight after it, so mechanical energy is conserved across the whole process, from the start to the apex, the disc’s kinetic energy splitting into the system’s kinetic energy at $V$ plus the disc’s gained potential energy $mgh$ :

\frac{1}{2}mv^2 = \frac{1}{2}(m+M)V^2+mgh.

Solving for $h$ and substituting the boxed $V$ (1),

\begin{align*}h &= \frac{v^2}{2g}-\frac{(m+M)V^2}{2mg} = \frac{v^2}{2g}-\frac{(m+M)}{2mg}\left(\frac{m}{m+M}v\right)^2 \\ &= \frac{v^2}{2g}-\frac{m v^2}{2g(m+M)} = \frac{v^2}{2g}\left(1-\frac{m}{m+M}\right).\end{align*}

h = \dfrac{Mv^2}{2g(M+m)}

(2)

Check

As $M\to\infty$ the body cannot be budged and the disc might as well be climbing a fixed ramp: the formula correctly reduces to the textbook $h\to v^2/2g$ , the full height a projectile of speed $v$ would reach on a rigid slope.

Remark

The process looks like an inelastic collision, disc and $M$ share one velocity $V$ at breakoff, yet no energy is lost: the ”missing” kinetic energy $\frac{1}{2}mv^2-\frac{1}{2}(m+M)V^2$ reappears exactly as the disc’s gravitational potential energy $mgh$ at the apex, since both surfaces were frictionless throughout. The mass ratio $M/(M+m)$ in $h$ is one minus the ratio $m/(m+M)$ that fixed $V$ , both tracing back to the identical momentum split.