JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass 2 kg moving with velocity of \(\overrightarrow{\mathrm{v}}_{\mathrm{in}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{ms}^{-1}\) enters into a constant force field of 6 N directed along positive z -axis. If the body remains in the field for a period of \(\frac{5}{3}\) seconds, then velocity of the body when it emerges from force field is
- A \(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
- B \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
- C \(3 \hat{i}+4 \hat{j}-5 \hat{k}\)
- D \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\sqrt{5} \hat{\mathrm{k}}\)
Answer & Solution
Correct Answer
(B) \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \overrightarrow{\mathrm{a}}=\frac{\mathrm{B}}{2} \hat{\mathrm{k}}=3 \hat{\mathrm{k}}, \mathrm{t}=\frac{5}{3} \mathrm{~s} \\ & \overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ &…
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