Irodov 1.168: Perpendicular Recoil, Full Stop Head-On

Problem Irodov 1.168

A particle of mass $m_1$ , moving with some initial speed $v_0$ , undergoes a perfectly elastic collision with a second particle of mass $m_2$ that is initially at rest. Find, as a fraction of the striking particle’s own initial kinetic energy, how much kinetic energy it loses if

it ends up moving at a right angle to its original direction of motion;
the collision is head-on, so both particles move along the same line before and after.

Momentum diagram for part (a): the two final momenta compose, by conservation, to the initial momentum m_1 v_0 (drawn for an illustrative m_2=3m_1; the construction holds for any m_1, m_2). — Momentum diagram for part (a): the two final momenta compose, by conservation, to the initial momentum $m_1\vect v_0$ (drawn for an illustrative $m_2=3m_1$ ; the construction holds for any $m_1$ , $m_2$ ).

Given

$m_1$
$m_2$ (initially at rest)
perfectly elastic collision

Find

Find the fraction

\eta

of the striking particle’s own kinetic energy lost in (a) a

90^\circ

recoil and (b) a head-on collision.

Solution

Step 1Part (a): momentum conservation with a right-angle recoil

Let $v_0$ be the initial speed of $m_1$ , and $v_1'$ , $v_2'$ the final speeds of $m_1$ and $m_2$ , with the initial direction as the $x$ axis and $m_1$ recoiling along $y$ ( $90^\circ$ to $x$ ). Momentum conservation along each axis then reads

\begin{align*} m_2v_2'\cos\theta &= m_1v_0, \\ m_2v_2'\sin\theta &= m_1v_1', \tag{1}\end{align*}

where $\theta$ is the angle at which $m_2$ recoils, an unknown we do not need.

Step 2Eliminating the unknown angle

Squaring each line of (1) and adding them, the two $\theta$ terms collapse through $\cos^2\theta+\sin^2\theta=1$ .

\begin{align*}(m_2v_2')^2\cos^2\theta+(m_2v_2')^2\sin^2\theta &= (m_1v_0)^2+(m_1v_1')^2 \\ (m_2v_2')^2 &= m_1^2\left(v_0^2+v_1'^2\right)\end{align*}

The collision is perfectly elastic, so kinetic energy is also conserved. Isolating $m_2v_2'^2$ and multiplying through by $m_2$ turns that into a second expression for the same square.

\begin{align*}m_1v_0^2 &= m_1v_1'^2+m_2v_2'^2 \\ m_2v_2'^2 &= m_1\left(v_0^2-v_1'^2\right) \\ (m_2v_2')^2 &= m_1m_2\left(v_0^2-v_1'^2\right)\end{align*}

Setting the two expressions for $(m_2v_2')^2$ equal, cancelling the common factor $m_1$ and collecting $v_0^2$ on one side and $v_1'^2$ on the other,

\begin{align*}m_1m_2\left(v_0^2-v_1'^2\right) &= m_1^2\left(v_0^2+v_1'^2\right) \\ m_2\left(v_0^2-v_1'^2\right) &= m_1\left(v_0^2+v_1'^2\right) \\ v_0^2(m_2-m_1) &= v_1'^2(m_1+m_2)\end{align*}

v_1'^2 = v_0^2\,\dfrac{m_2-m_1}{m_1+m_2}

(2)

Step 3Fraction of kinetic energy lost, part (a)

The striking particle’s own kinetic energy falls from $m_1v_0^2/2$ to $m_1v_1'^2/2$ , so the fraction lost is one minus the speed-squared ratio, which (2) supplies.

\eta_a = 1-\frac{v_1'^2}{v_0^2} = 1-\frac{m_2-m_1}{m_1+m_2} = \frac{(m_1+m_2)-(m_2-m_1)}{m_1+m_2}.

\eta_a = \dfrac{2m_1}{m_1+m_2}

(3)

Step 4Part (b): the head-on collision

Now $m_1$ and $m_2$ move along one line after the collision, with final speeds $v_1'$ , $v_2'$ signed along the original direction of motion. Momentum and energy conservation give

m_1v_0 = m_1v_1'+m_2v_2', \qquad m_1v_0^2 = m_1v_1'^2+m_2v_2'^2.

Rearranging each equation isolates the same factor $m_1(v_0-v_1')$ on the right:

\begin{align*}m_2v_2' &= m_1(v_0-v_1'), \\ m_2v_2'^2 &= m_1(v_0-v_1')(v_0+v_1').\end{align*}

Dividing the second line by the first (valid whenever $m_2v_2'\neq0$ , i.e. a collision actually occurs) cancels the shared factor, leaving

v_2' = v_0+v_1'.

Substituting back into the momentum equation and collecting the $v_1'$ terms on one side,

\begin{align*}m_1v_0 &= m_1v_1'+m_2(v_0+v_1') \\ m_1v_0-m_2v_0 &= m_1v_1'+m_2v_1' \\ v_0(m_1-m_2) &= v_1'(m_1+m_2)\end{align*}

v_1' = v_0\,\dfrac{m_1-m_2}{m_1+m_2}

(4)

Step 5Fraction of kinetic energy lost, part (b)

The same fraction-lost bookkeeping as in Step 3, now with (4),

\eta_b = 1-\frac{v_1'^2}{v_0^2} = 1-\left(\frac{m_1-m_2}{m_1+m_2}\right)^2 = \frac{(m_1+m_2)^2-(m_1-m_2)^2}{(m_1+m_2)^2}.

The numerator is a difference of two squares, $(a+b)^2-(a-b)^2=4ab$ with $a=m_1$ , $b=m_2$ :

\eta_b = \dfrac{4m_1m_2}{(m_1+m_2)^2}

(5)

Check

Setting $m_1=m_2$ in part (b) gives $\eta_b=1$ : an elastic head-on collision between equal masses stops the striking particle dead and hands all of its kinetic energy to the target, the familiar Newton’s-cradle result. Part (a) carries its own restriction: the right-hand side of (2) stays nonnegative, and $\eta_a$ stays under $1$ , only when $m_2>m_1$ , so a right-angle recoil of the striking particle is possible only off a heavier target.

Remark

As $m_2\to\infty$ , part (a) gives $\eta_a\to0$ : recoiling at right angles off an effectively immovable target is a pure change of direction, exactly the elastic reflection off a rigid wall, and costs the striking particle no speed at all.