Conservation of Linear Momentum — Question 6

Question 6

A recoil speed measured from a moving platform

A $10\,\mathrm{kg}$ assembly moves right at $3\,\mathrm{m/s}$ before dividing into $4$ and $6\,\mathrm{kg}$ pieces. The lighter piece then moves left at $2\,\mathrm{m/s}$ in the laboratory. External impulse is negligible.

1. Find the heavier piece's laboratory velocity.

2. Check momentum conservation in a frame moving right at $3\,\mathrm{m/s}$.

Question 6 - Solution

A recoil speed measured from a moving platform

(a) Derivation and calculation

The initial total momentum is $P_i=10(3)=30$ kg m/s: $$30=4(-2)+6v\Rightarrow\boxed{v=\frac{38}{6}=\frac{19}{3}\ \mathrm{m/s}}.$$ $$V_C=P_i/(4+6)=3\ \mathrm{m/s}.$$

(b) Result and interpretation

In the inertial frame moving at 3 m/s, $$v_1^{\prime}=-2-3=-5,\qquad v_2^{\prime}=19/3-3=10/3.$$ $$\boxed{P_f^{\prime}=4(-5)+6(10/3)=0=P_i^{\prime}}.$$ Subtracting a constant frame velocity preserves the conservation law for fixed total mass.

Physical check

The relative momenta in the center frame are $-20$ and $+20$ kg m/s.