Question
Figure shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
✓
Answer
H = 1m, h = 0.5mApplying law of conservation of Energy for point A & B
$\text{mgH}=\frac{1}{2}\text{mv}^2+\text{mgh}$
$\Rightarrow\text{g}=\frac{1}{2}\text{v}^2+0.5\text{g}$
$\Rightarrow\text{v}^22(\text{g}-0.59)=\text{g}$
$\Rightarrow\text{v}=\sqrt{\text{g}}=3.1\text{m}/\text{s}$
After point B the body exhibits projectile motion for which
$\theta=0^\circ,\text{v}=-0.5$
So, $-0.5=(\text{u}\sin\theta)\text{t}-\Big(\frac{1}{2}\Big)\text{gt}^2$
$\Rightarrow0.5=4.9\ \text{t}^2$
$\Rightarrow\text{t}=0.31\text{sec}$
So, $\text{x}=(\text{v}\cos\theta)\text{t}$
$=3.1\times3.1=1\text{m}$
So, the particle will hit the ground at a horizontal distance in from B.
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