Question

A ball of mass 300 g is projected horizontally with a speed of 15 m/s from the top of a tower 20 m high. The magnitude of the rate of average change of angular momentum about the foot of the tower till the instant the ball strikes the ground is, $(g={10m/s}^2)$

• A. zero
• B. $300kg{m}^2{/s}^2$
• C. $60kg{m}^2{/s}^2$
• D. $45kg{m}^2{/s}^2$

Answer 1

The correct option is B $300kg{m}^2{/s}^2$

Denoting the horizontal by $\hat{i}$

and the vertical downward as $\hat{j}$

we get initial momentum as $P_i=m\times 15\hat{i}$

Just before hitting the ground the $vertical$ velocity will be $V_v=\sqrt{2gh}\hat{j}=20\hat{j}m/s$

and the $horizontal$ velocity will be $still$ $15m/s$ so the new $momentum$ will be $P_f=15m\hat{i}+20m\hat{j}$

so the change in momentum will be $\Delta P=P_f-P_i=20m\hat{j}$

Now the change in $angular$ momentum will be $R\times \Delta P=20R$ $\left(magnitude\right)$ , $R$ is range.

time of flight will be $T=\sqrt{\frac{2h}{g}}=2sec$ , where $h=20meter$

so range will be $R=U_{horizontal}t=15\times 2=30meter$

so the rate will be $\frac{R\times \Delta P}{t}=300kg{m}^2/s^2$