Answer
(a) The proton's speed is $~1.99\times 10^6~m/s~~$ at an angle of $~~45^{\circ}~~$ clockwise from the y-axis.
(b) The proton's speed is $~1.47\times 10^6~m/s~~$ at an angle of $~~16.2^{\circ}~~$ clockwise from the y-axis.
Work Step by Step
(a) In the laboratory frame:
$v_x = 1.41 \times 10^6~m/s$
$v_y = 1.41 \times 10^6~m/s$
We can find the proton's speed:
$v = \sqrt{(1.41 \times 10^6~m/s)^2+(1.41 \times 10^6~m/s)^2}$
$v = 1.99\times 10^6~m/s$
We can find the angle clockwise from the y-axis:
$tan~\theta = \frac{1.41 \times 10^6~m/s}{1.41 \times 10^6~m/s}$
$tan~\theta = 1$
$\theta = tan^{-1}~(1)$
$\theta = 45^{\circ}$
The proton's speed is $~1.99\times 10^6~m/s~~$ at an angle of $~~45^{\circ}~~$ clockwise from the y-axis.
(b) In the rocket frame:
$v_x = 1.41 \times 10^6~m/s - 1.00\times 10^6~m/s = 0.41\times 10^6~m/s$
$v_y = 1.41 \times 10^6~m/s$
We can find the proton's speed:
$v = \sqrt{(0.41 \times 10^6~m/s)^2+(1.41 \times 10^6~m/s)^2}$
$v = 1.47\times 10^6~m/s$
We can find the angle clockwise from the y-axis:
$tan~\theta = \frac{0.41 \times 10^6~m/s}{1.41 \times 10^6~m/s}$
$tan~\theta = 0.2908$
$\theta = tan^{-1}~(0.2908)$
$\theta = 16.2^{\circ}$
The proton's speed is $~1.47\times 10^6~m/s~~$ at an angle of $~~16.2^{\circ}~~$ clockwise from the y-axis.
