Question

Consider a 0.52 g puck that moves as shown in the graph above. Consider each point to be a picture of the puck and consider that 60 pictures were taken each second. Plot the x position versus time and the y position versus time and use a linear trend line to find the x component of the velocity and the y component of the velocity. Attach you graphs. What are the x component of the velocity and the y component of the velocity? Write the velocity of the puck in i and j notation and in magnitude and angle notation. What is the momentum of the puck? (Write it both in i and j notation and as a magnitude and an angle.)

Answer 1

From graph, its clear that relation between x and y coordinate is

$y=5x$

Since 60 pictures were taken per second hence time interval between two consecutive picture will be 1/60 sec. i.e.

$\Delta t=\frac{1}{60}$

In this same time interval, x coordinate change by 0.1 cm while y coordinate change by 0.5 cm i.e.

$\Delta x=0.001 m\\ \Delta y=0.005 m$

(a) Velocities

$v_x=\frac{\Delta x}{\Delta t}=\frac{.001}{1/60}=0.06 m/sec\\ v_y=\frac{\Delta y}{\Delta t}=\frac{.005}{1/60}=0.30 m/sec$

(b) Velocities in terms of i and j

$V=v_x\hat{i}+v_y\hat{j}=(0.06\hat{i}+0.3\hat{j})m/sec$

Magnitude of velocity

$V=|V|=|v_x\hat{i}+v_y\hat{j}|=(\sqrt{0.06^2+0.3^2})m/sec=0.3059 m/sec$

In terms of magnitude and angle

$\tan{\theta}=\frac{v_y}{v_x}\Rightarrow \theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{0.3}{0.06})=78.7^o$

therefore

$\vec{V}=V\cos{\theta}\hat{i}+V\sin{\theta}\hat{j}=0.3059\cos{(78.7)}\hat{i}+0.3059\sin{(78.7)}\hat{j}$

(c) Momentum in i and j

$\vec{P}=m(v_x\hat{i}+v_y\hat{j})=0.52\times10^{-3}\times(0.06\hat{i}+0.3\hat{j})kgm/sec\\ =(3.12\hat{i}+15.6\hat{j})kgm/sec$

In terms of magnitude and angle

$\vec{P}=m(V\cos{\theta}\hat{i}+V\sin{\theta}\hat{j})=0.52\times 10^{-3}\times(0.3059\cos{(78.7)}\hat{i}+0.3059\sin{(78.7)}\hat{j})\\ =0.0063\cos{(78.7)}\hat{i}+0.0063sin{(78.7)}\hat{j}$