Question 339: Centre of Mass and Rotational Motion

Question

A body falls on a surface of coefficient of restitution 0.6 from a height of $$1 \mathrm{~m}$$. Then the body rebounds to a height of

Accepted Answer

As the body falls from a height
$$\begin{array}{ll} 
& \mathrm{u}=0 \
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gh} \
	herefore & \mathrm{v}^2=2 	imes 9.8 	imes 1=19.6 \
	herefore \quad & \mathrm{v}=\sqrt{19.6} \mathrm{~m} / \mathrm{s}
\end{array}$$
Coefficient of restitution $$\mathrm{e}=\frac{\mathrm{v}}{\mathrm{u}}$$
$$\begin{aligned}
\mathrm{e} & =\frac{	ext { Velocity after collision }\left(\mathrm{v}_{\mathrm{f}}ight)}{	ext { Velocity before collision }\left(\mathrm{v}_{\mathrm{b}}ight)} \
	herefore \quad \mathrm{v}_{\mathrm{f}} & =\mathrm{e} 	imes \mathrm{v}_{\mathrm{b}} \
\mathrm{v}_{\mathrm{f}} & =0.6 	imes \sqrt{19.6} \mathrm{~m} / \mathrm{s}
\end{aligned}$$
After the body rebounds,
$$\begin{aligned}
& v^2=u^2-2 g h \
& \Rightarrow u^2=2 g h \
	herefore \quad & h=u^2 / 2 g
\end{aligned}$$
Here, $$\mathrm{u}=\mathrm{v}_{\mathrm{f}}$$
$$\begin{aligned}
	herefore \quad \mathrm{h} & =\frac{(0.6 	imes \sqrt{19.6})^2}{2 	imes 9.8} \
& =0.36 \mathrm{~m}
\end{aligned}$$

Answer

1 m

Answer

0.36 m

Answer

0.4 m

Answer

0.6 m

Centre of Mass and Rotational Motion

A body falls on a surface of coefficient of restitution 0.6 from a height of $1 \mathrm{~m}$. Then the body rebounds to a height of

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