Question

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away
from another trolley, Y, of mass (2.561 ± 0.001) kg with a speed of (3.2 ± 0.01) ms-1. The second trolley is moving away with a distance of (2.5 ± 0.01) ms-1
What is the absolute uncertainty of the ratio of momentum of the two trolleys
X/Y?

Answer 1

Trolley X:

$$p_x=m_xv_x=2.34\cdot3.2=7.49\text{ Ns}.\\ \Delta p_x=m_x\Delta v_x+\Delta m_x v_x=\\=2.34\cdot0.01+0.01\cdot3.2=0.06\text{ Ns}.\\ p_x=(7.49\pm0.06)\text{ Ns}.$$

Trolley Y:

$$p_y=m_yv_y=2.561\cdot2.5=6.40\text{ Ns}.\\ \Delta p_y=m_y\Delta v_y+\Delta m_y v_y=\\=2.561\cdot0.01+0.001\cdot2.5=0.03\text{ Ns}.\\ p_y=(6.40\pm0.03)\text{ Ns}.$$

The absolute uncertainty of the ratio of momentum of the two trolleys:

$$\Delta p=\frac{\Delta p_x}{p_y}+\frac{p_x\Delta p_y}{p_y^2}=0.01\text{ Ns}.$$