Question 256: Laws of Motion

Question

A 1 kg stationary bomb is exploted in three parts having mass 1 : 1 : 3 respectively. Parts having same mass move in perpendicular direction  with velocity 30 m/s,  then the velocity of bigger part will be

Accepted Answer

Apply conservation of linear momentum.
Total momentum before explosion
= total momentum after explosion
$$0 = {m \over 5}{v_1}\widehat i + {m \over 5}{v_2}\widehat j + {{3m} \over 5}{\overrightarrow v _3}$$
$${{3m} \over 5}{\overrightarrow v _3} =  - {m \over 5}\left[ {{v_1}\widehat i + {v_2}\widehat j} ight]$$
$${\overrightarrow v _3} = {{ - {v_1}} \over 3}\widehat i - {{{v_2}} \over 3}\widehat j$$
$$ 	herefore $$ $${v_1} = {v_2} = 30$$ m/sec
$${\overrightarrow v _3} =  - 10\widehat i - 10\widehat j;{v_3} = 10\sqrt 2 $$ m/sec

Answer

$$10\sqrt 2 $$ m/sec

Answer

$${{10} \over {\sqrt 2 }}$$  m/sec

Answer

$$15\sqrt 2 $$ m/sec

Answer

$${{15} \over {\sqrt 2 }}$$  m/sec

Laws of Motion

A 1 kg stationary bomb is exploted in three parts having mass 1 : 1 : 3 respectively. Parts having same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be

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