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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)
    v
  • b)
    √2v
  • c)
    2√2v
  • d)
    3√2v
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A shell of mass m is at rest initially. It explodes into three fragmen...
A) v

Since the total momentum of the system is conserved, we can use the principle of conservation of momentum to solve this problem.

Initially, the shell is at rest, so the total momentum is zero. After the explosion, the three fragments fly off in different directions. Let's assume the masses of the three fragments are m1, m2, and m3 respectively, and their speeds are v1, v2, and v3 respectively.

According to the principle of conservation of momentum, we have:

0 = m1v1 + m2v2 + m3v3

Since the mass of the third fragment is 1/2 of the other two fragments, we can write:

m1 = 2m
m2 = 2m
m3 = m/2

Substituting these values into the conservation of momentum equation, we have:

0 = (2m)v1 + (2m)v2 + (m/2)v3

0 = 4mv1 + 4mv2 + (m/2)v3

0 = 8mv1 + 8mv2 + mv3/2

0 = 8m(v1 + v2) + mv3/2

Since v1 = v2 = v, we can simplify the equation as:

0 = 16mv + mv3/2

Multiplying through by 2:

0 = 32mv + mv3

Rearranging the equation:

mv3 = -32mv

Dividing through by mv:

v3 = -32

Since the speed cannot be negative, the speed of the third (lighter) fragment is:

v3 = 32

So the answer is b) 32v.
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Community Answer
A shell of mass m is at rest initially. It explodes into three fragmen...
Momentum of the system would remain conserved.
Initial momentum =0
Final momentum should also be zero.
Let masses be 2m,2m, and m
Momentum along x-direction =
Net momentum =
Now, 2√2mv = mv′
v′ = 2√2v
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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 isa)vb)√2vc)2√2vd)3√2vCorrect answer is option 'C'. Can you explain this answer?
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