TS EAMCET · Physics · Center of Mass Momentum and Collision
At a given instant of time two particles are having the position vectors \(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+57 \hat{\mathbf{k}}\) metres and \(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) respectively. If the velocity of the first particle be \(0.4 \hat{\mathbf{i}} \mathrm{ms}^{-1}\), the velocity of second particle in metre per second if they collide after \(10 \mathrm{sec}\) is
- A \(6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)\)
- B \(0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)\)
- C \(6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)\)
- D \(0.6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\frac{1}{3} \hat{\mathbf{k}}\right)\)
Answer & Solution
Correct Answer
(B) \(0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} s_1+\left(u_1\right) t & =s_2+\left(u_2\right) t \\ & =(4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+(0.4 \hat{\mathbf{i}}) 10 \\ & =(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\left(u_2\right) 10 \\ \Rightarrow \quad u_2 & =\frac{(6…
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