JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2020
- Medium
- Numerical
A body \(A,\) of mass \(m=0.1\; kg\) has an initial velocity of \(3 \hat{\mathrm{i}}\; \mathrm{ms}^{-1} .\) It collides elastically with another body, \(\mathrm{B}\) of the same mass which has an initial velocity of \(5 \hat{\mathrm{j}} \;\mathrm{ms}^{-1}\). After collision. A moves with a velocity \(\overline{\mathrm{v}}=4(\hat{\mathrm{i}}+\hat{\mathrm{j}})\). The energy of \(\mathrm{B}\) after collision is written as \(\frac{\mathrm{x}}{10} \;\mathrm{J}\) The value of \(x\) is
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Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
By conservation of linear momentum :
\((0.1)(3 \hat{\mathrm{i}})+(0.1)(5 \hat{\mathrm{j}})=(0.1)(4)(\hat{\mathrm{i}}+\hat{\mathrm{j}})+(0.1) \overline{\mathrm{v}}\)
\(\Rightarrow \overrightarrow{\mathrm{v}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}\)
Speed of \(\mathrm{B}\) after collision \(|\overrightarrow{\mathrm{v}}|=\sqrt{2}\)
Now, kinetic energy \(=\frac{1}{2} \mathrm{mV}^{2}=\frac{1}{2}(0.1)(2)=\frac{1}{10}\)
\(\therefore \mathrm{x}=1\)
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