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A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is
- a)√2 v
- b)2√2 v
- c)3√2 v
- d)v
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A shell of mass m is at rest initially. It explodes into three fragmen...
To solve this problem, we can use momentum conservation. Initially, the shell is at rest, so the total momentum is zero. After the explosion, the three fragments fly off with some velocities. Let's call the velocities of the fragments v₁, v₂, and v₃.
According to momentum conservation, the total momentum after the explosion should also be zero. Therefore, we have:
m₁v₁ + m₂v₂ + m₃v₃ = 0
Since the fragments have masses in the ratio 2:2:1, we can express the masses as m₁ = 2m, m₂ = 2m, and m₃ = m, where m is the mass of the shell.
Substituting these values into the momentum conservation equation, we get:
2m(v₁ + v₂) + m(v₃) = 0
Dividing both sides of the equation by m:
2(v₁ + v₂) + v₃ = 0
Now, we know that the fragments with equal mass (v₁ and v₂) fly off with the same speed v. Therefore, we can rewrite the equation as:
2(2v) + v₃ = 0
Simplifying the equation, we get:
4v + v₃ = 0
Rearranging the equation to solve for v₃, we have:
v₃ = -4v
Therefore, the speed of the third (lighter) fragment is -4v.
According to momentum conservation, the total momentum after the explosion should also be zero. Therefore, we have:
m₁v₁ + m₂v₂ + m₃v₃ = 0
Since the fragments have masses in the ratio 2:2:1, we can express the masses as m₁ = 2m, m₂ = 2m, and m₃ = m, where m is the mass of the shell.
Substituting these values into the momentum conservation equation, we get:
2m(v₁ + v₂) + m(v₃) = 0
Dividing both sides of the equation by m:
2(v₁ + v₂) + v₃ = 0
Now, we know that the fragments with equal mass (v₁ and v₂) fly off with the same speed v. Therefore, we can rewrite the equation as:
2(2v) + v₃ = 0
Simplifying the equation, we get:
4v + v₃ = 0
Rearranging the equation to solve for v₃, we have:
v₃ = -4v
Therefore, the speed of the third (lighter) fragment is -4v.
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A shell of mass m is at rest initially. It explodes into three fragmen...
By conservation of momentum:
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A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is a)√2 vb)2√2 vc)3√2 vd)vCorrect answer is option 'B'. Can you explain this answer?
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