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A bomb of mass 1 kg initially at rest, explodes into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other. Each with a speed of $30 ms − 1$ . what is the velocity of the heavier fragment?

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102 ms − 1 at 450

with each of the other two fragments

20 ms − 1 at 1350

with each of the other two fragments

20 ms − 1 at 45 ∘

with each of the other two fragments

102 ms − 1 at 1350

with each of the other two fragments

answer is B.

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Detailed Solution

Since there is no external force acting on the body. The total linear momentum is conserved.

As the lighter fragments are moving perpendicularly to each other.

The heavier fragment will move in the opposite direction of the resultant momentum of the lighter fragments.

$0 = m 5 (302) + 3 m 5 V ⇒ V = − 102 m s − 1$

If we assume the lighter fragments are moving along east and north directions, the heavier fragment will move along south-west.

So it makes 135^o with each of the other fragments.

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A bomb of mass 1 kg initially at rest, explodes into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other. Each with a speed of 30 ms −1 . what is the velocity of the heavier fragment?

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A bomb of mass 1 kg initially at rest, explodes into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other. Each with a speed of $30 ms − 1$ . what is the velocity of the heavier fragment?