question_answer

An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of \[12\,m{{s}^{-1}}\] and 2 kg second part moving with a velocity of \[8\,m{{s}^{-1}}\]. If the third part flies off with a velocity of  \[4\,m{{s}^{-1}}\], its mass would be [AIPMT (S) 2009]

A) 5 kg

B) 7 kg

C) 17 kg

D) 3 kg

Correct Answer: A

Solution :

Key Idea Apply law of conservation of linear momentum.

Momentum of first part \[=1\times 12=12kg\,m{{s}^{-1}}\]

Momentum of the second part \[=2\times 8=16\,kg\,m{{s}^{-1}}\]

\[\xrightarrow[\text{(ii)}{{\text{H}}_{\text{2}}}\text{O,heat}]{\text{(i)}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{room}\,\text{temperature}}Z;\]            Resultant momentum

\[=\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=20\,kg\,m{{s}^{-1}}\]

The third part should also have the same momentum.

Let the mass of the third part be M, then

\[4\times M=20\]

\[M=5\,kg\]

Alternative:

\[Mv\,\cos \,\theta =12\]

\[Mv\,\sin \,\theta =16\]

\[\tan \theta =\frac{16}{12}=\frac{4}{3}\]

\[M=\frac{12\times 5}{4\times 3}=\frac{60}{12}=5\,kg\]