Question

A ball of mass $160g$ is thrown up at an angle of ${60}^o$ to the horizontal at a speed of $10m{s}^{-1}$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly: $(g=10m{s}^{-2})$

• A. $1.73kg{m}^2/s$
• B. $3.0kg{m}^2/s$
• C. $3.46kg{m}^2/s$
• D. $6.0kg{m}^2/s$

Answer 1

The correct option is B $3.0kg{m}^2/s$

The initial components of velocity are,
V = $5i+5\times 3^{0.5}j$
At the highest point, only the horizontal component will remain.
Hence, $v=5i$
The vector corresponding to the highest point is given by,
T = $\frac{2usin\theta }{g}$ = $3^{0.5}$
x = $5\times \frac{3^{0.5}}{2}$
$y=ut-$ $\frac{1}{2}g{t}^2$
$y=7.5-3.75=3.75$
Hence, the angular momentum is given by the cross product of the radius vector with the velocity vector.
L = m ($\overrightarrow{r}\times \overrightarrow{v}$)
$=3$