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Question: A particle of mass 1 g moving with a velocity \({\overrightarrow{v}}_{1} = 3\widehat{i} - 2\widehat{...

A particle of mass 1 g moving with a velocity v→1=3i^−2j^ms−1{\overrightarrow{v}}_{1} = 3\widehat{i} - 2\widehat{j}ms^{- 1}experiences a perfectly elastic collision with another particle of mass 2 g and velocity v→2=4j^−6k^ms−1{\overrightarrow{v}}_{2} = 4\widehat{j} - 6\widehat{k}ms^{- 1}. The velocity of the particle is:

A

2.3ms−12.3ms^{- 1}

B

4.6ms−14.6ms^{- 1}

C

9.2ms−19.2ms^{- 1}

D

6ms−16ms^{- 1}

Answer

4.6ms−14.6ms^{- 1}

Explanation

Solution

From conservation of momentum

m1v→1+m2v→2=(m1+m2)v→m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2} = (m_{1} + m_{2})\overrightarrow{v}

1×(3i^−2j^)+2×(4j^−6k^)=(1+2)v→1 \times (3\widehat{i} - 2\widehat{j}) + 2 \times (4\widehat{j} - 6\widehat{k}) = (1 + 2)\overrightarrow{v}

⇒3i^+6j^−12k^=3v→.⇒v→=i^+2j^−4k^\Rightarrow 3\widehat{i} + 6\widehat{j} - 12\widehat{k} = 3\overrightarrow{v}. \Rightarrow \overrightarrow{v} = \widehat{i} + 2\widehat{j} - 4\widehat{k}

v=∣v→∣=1+4+16=4.6ms−1v = |\overrightarrow{v}| = \sqrt{1 + 4 + 16} = 4.6ms^{- 1}.