Question

A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 5.12 m/s at an angle of 31.5° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision. magnitude m/s direction ° counter-clockwise from the original direction of motion

Answer 1

Answer:

Magnitude : 3.14 m/s

Direction : 301.31 deg

Explanation:

Assuming that the ball moves in positive x-direction initially.

$m$ = mass of each ball

$\underset{v_{1i}}{\rightarrow}$ = initial velocity of first ball before collision = $6\hat{i} + 0\hat{j}$

$\underset{v_{2i}}{\rightarrow}$ = initial velocity of second ball before collision = $0\hat{i} + 0\hat{j}$

$\underset{v_{1f}}{\rightarrow}$ = final velocity of first ball after collision = $5.12 Cos31.5\hat{i} + 5.12 Sin31.5\hat{j}$

$\underset{v_{2f}}{\rightarrow}$ = final velocity of second ball after collision = ?

Using conservation of momentum

$m$ $\underset{v_{1i}}{\rightarrow}$ + $m$ $\underset{v_{2i}}{\rightarrow}$ =  $m$ $\underset{v_{1f}}{\rightarrow}$ + $m$ $\underset{v_{2f}}{\rightarrow}$

($6\hat{i} + 0\hat{j}$) + ($0\hat{i} + 0\hat{j}$) = ($5.12 Cos31.5\hat{i} + 5.12 Sin31.5\hat{j}$) + $\underset{v_{2f}}{\rightarrow}$

$\underset{v_{2f}}{\rightarrow}$ = ($6\hat{i} + 0\hat{j}$) - ($5.12 Cos31.5\hat{i} + 5.12 Sin31.5\hat{j}$)

$\underset{v_{2f}}{\rightarrow}$ = $1.63\hat{i} - 2.68\hat{j}$

magnitude of final velocity of second ball is given as

magnitude = sqrt((1.63)² + (- 2.68)²) = 3.14 m/s

Direction is given as

θ = 360 + tan⁻¹(- 2.68/1.63)

θ = 301.31 deg