Irodov 1.166: Merge Two Momenta, Component by Component

Problem Irodov 1.166

A particle of mass $m_1=1.0$ g moves with velocity $\vect v_1=3.0\vect i-2.0\vect j$ (SI units, m/s) and collides perfectly inelastically with a second particle of mass $m_2=2.0$ g moving with velocity $\vect v_2=4.0\vect j-6.0\vect k$ (SI units, m/s), the two particles sticking together on impact. Find the velocity $\vect v$ of the combined particle formed by the collision, both as a vector and as a modulus $v=|\vect v|$ .

Before: the two particles approach with v_1 and v_2. After: they stick together and share the common velocity v. Schematic; the true motion is three dimensional. — Before: the two particles approach with $\vect v_1$ and $\vect v_2$ . After: they stick together and share the common velocity $\vect v$ . Schematic; the true motion is three dimensional.

Given

$m_1=1.0$ g
$\vect v_1=3.0\vect i-2.0\vect j$
$m_2=2.0$ g
$\vect v_2=4.0\vect j-6.0\vect k$

Find

\vect v

and

v=|\vect v|

Solution

Step 1Momentum conservation across the perfectly inelastic collision

Writing $m=m_1+m_2$ for the combined mass and $\vect v$ for the common velocity after impact,

m_1\vect v_1+m_2\vect v_2 = (m_1+m_2)\vect v.

Solving for $\vect v$ ,

\vect v = \frac{m_1\vect v_1+m_2\vect v_2}{m_1+m_2}.

Step 2Evaluate component by component

Since $m_1$ and $m_2$ enter only as the ratio $m_1{:}m_2=1.0{:}2.0$ , the gram units cancel and need not be converted; substituting the given vectors and masses,

\begin{align*}\vect v &= \frac{(1.0)(3.0\vect i-2.0\vect j)+(2.0)(4.0\vect j-6.0\vect k)}{1.0+2.0} \\ &= \frac{3.0\vect i-2.0\vect j+8.0\vect j-12.0\vect k}{3.0} \\ &= \frac{3.0\vect i+6.0\vect j-12.0\vect k}{3.0}\end{align*}

\vect v = 1.0\vect i+2.0\vect j-4.0\vect k

(1)

in m/s.

Step 3The modulus

The modulus follows from the sum of the squared components,

\begin{align*}v &= \sqrt{(1.0)^2+(2.0)^2+(-4.0)^2} \\ &= \sqrt{1.0+4.0+16.0} \\ &= \sqrt{21}\end{align*}

v \approx 4.6\text{ m/s}

(2)

Check

The $x$ component of $\vect v$ comes only from $\vect v_1$ (the only vector with an $\vect i$ term), scaled by $m_1/(m_1+m_2)=1/3$ : $3.0\times(1/3)=1.0$ , matching Step 2. The $z$ component similarly comes only from $\vect v_2$ ’s $-6.0\vect k$ , scaled by $m_2/(m_1+m_2)=2/3$ : $-6.0\times(2/3)=-4.0$ , also matching, so each independent component checks separately.

Remark

Momentum conservation is a vector statement that holds component by component in any fixed Cartesian basis, which is why this three dimensional, perfectly inelastic collision needs no picture of the true collision geometry to solve: the $\vect i$ , $\vect j$ and $\vect k$ components combine completely independently, each exactly as a one dimensional inelastic collision of the same two masses would.