Question

If the magnitude of the initial velocity of the ball v0 = 7.94 ± 0.03, and the "gun" is tilted 31 ± 0.4º upwards, what is the uncertainty in the horizontal component of velocity σvx?

Answer 1

Answer:

0.05

Explanation:

Given a variable C which is the product of two variables A, B:

$C=A\cdot B$

Then the absolute error on C is given by:

$\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}$

where $\sigma_A, \sigma_B, \sigma_C$ are the uncertanties on the measure of A, B and C, respectively.

In this problem, the horizontal component of the velocity $v_x$ is given by

$v_x = v_0 cos \theta$

Therefore, the uncertainty on vx is given by:

$\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta}$ (1)

where we have:

$v_0 = 7.94$

$\sigma_{v_0}=0.03$

$\theta=31^{\circ}$, so

$cos \theta=cos 31^{\circ}=0.857$

The uncertainty on $cos \theta$ is given by:

$\sigma_{cos \theta}=|sin \theta|\sigma_\theta$

where:

$|sin \theta| = |sin 31^{\circ}|=0.515$

and

$\sigma_\theta=0.4^{\circ}=0.007 rad$

So

$\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036$

Also,

$v_x = v_0 cos \theta = (7.94)(cos 31^{\circ})=6.81 m/s$

So, combininb everything into (1), we find:

$\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05$