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4-1.Newton's Laws of Motion
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A block slides down a smooth inclined plane when released from the top, while another falls freely from the same point
Asliding block will reach the ground first with greater speed
Bfreely falling block will reach the ground first with greater speed
Cboth the blocks will reach the ground at the same time but with different speeds
Dboth the blocks will reach the ground with same speed but the freely falling block first
Solution
In case of sliding motion on an inclined plane,
${\mathrm{h}=\mathrm{s}=\frac{1}{2} \mathrm{g} \sin \theta \mathrm{t}^{2}}$
${\text { or } \mathrm{t}=\mathrm{t_s}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}}$
${\text { and } \mathrm{u}_{\mathrm{s}}=\sqrt{2 \mathrm{gh}}=\sqrt{2(\mathrm{g} \sin \theta) \mathrm{s}}}$
while in case of free fall,
$\mathrm{t}_{\mathrm{F}}=\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$ and $\mathrm{u}_{\mathrm{F}}=\sqrt{2 \mathrm{gh}}$
$\frac{\mathrm{t}_{\mathrm{F}}}{\mathrm{t}_{\mathrm{s}}}=\sin \theta<1 \mathrm{i.e.}, \mathrm{t}_{\mathrm{F}}<\mathrm{t}_{\mathrm{s}}$
i.e., falling body reaches the ground first.
Also $\quad \frac{\mathrm{u}_{\mathrm{F}}}{\mathrm{u}_{\mathrm{s}}}=1,$ i.e., $\mathrm{u}_{\mathrm{F}}=\mathrm{u}_{\mathrm{s}}$
i.e., both reach the ground with same speed (not velocity as for falling body direction is vertical while for sliding body along the plane downwards).
${\mathrm{h}=\mathrm{s}=\frac{1}{2} \mathrm{g} \sin \theta \mathrm{t}^{2}}$
${\text { or } \mathrm{t}=\mathrm{t_s}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}}$
${\text { and } \mathrm{u}_{\mathrm{s}}=\sqrt{2 \mathrm{gh}}=\sqrt{2(\mathrm{g} \sin \theta) \mathrm{s}}}$
while in case of free fall,
$\mathrm{t}_{\mathrm{F}}=\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$ and $\mathrm{u}_{\mathrm{F}}=\sqrt{2 \mathrm{gh}}$
$\frac{\mathrm{t}_{\mathrm{F}}}{\mathrm{t}_{\mathrm{s}}}=\sin \theta<1 \mathrm{i.e.}, \mathrm{t}_{\mathrm{F}}<\mathrm{t}_{\mathrm{s}}$
i.e., falling body reaches the ground first.
Also $\quad \frac{\mathrm{u}_{\mathrm{F}}}{\mathrm{u}_{\mathrm{s}}}=1,$ i.e., $\mathrm{u}_{\mathrm{F}}=\mathrm{u}_{\mathrm{s}}$
i.e., both reach the ground with same speed (not velocity as for falling body direction is vertical while for sliding body along the plane downwards).
Standard 11
Physics
