TS EAMCET · Physics · Laws of Motion
A block is between two surfaces as shown in the figure. Find the normal reaction at both surfaces. [Assume, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A \(N_1=37.2 \mathrm{~N}\) and \(N_2=9.6 \mathrm{~N}\)
- B \(N_1=38.2 \mathrm{~N}\) and \(N_2=8.6 \mathrm{~N}\)
- C \(N_1=40 \mathrm{~N}\) and \(N_2=4 \mathrm{~N}\)
- D \(N_1=37.5 \mathrm{~N}\) and \(N_2=9.9 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(N_1=37.2 \mathrm{~N}\) and \(N_2=9.6 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
The given situation is shown in the figure. Resolving applied force, we have following forces on block Given, \(\tan \theta=\frac{3}{4}, \sin \theta=\frac{3}{5}\) and \(\cos \theta=\frac{4}{5}\) So, horizontal component,…
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Momentum before collision,
\(
p_1=4 m v_1-2 m v_2
\)
Momentum after collision,
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Conservation of momentum gives,
\(
\begin{gathered}
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\Rightarrow \quad \frac{v_1}{v_2}=\frac{-32}{4}=-8 \text { or }\left|\frac{v_1}{v_2}\right|=8
\end{gathered}
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