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A 1 kg stationary bomb exploded in three parts having mass ratio 1:1:3 . parts having same mass moving perpendicular direction with velocity 30mper second then the velocity of bigger part will be
- a)10 √2 m/s
- b)10/√2 m/s
- c)15 √s m/s
- d)15/s m/s
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A 1 kg stationary bomb exploded in three parts having mass ratio 1:1:3...
Conservation of momentum
1X0=1 X30i +1X30j +3v
3v= -(30i +30j)
v = -(10i +10j)
v ( magnitude)= √(10)^2+(10)^2
=10√2 m/s
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
A 1 kg stationary bomb exploded in three parts having mass ratio 1:1:3...
Let the mass of the three parts be m, m, and 3m respectively.
The total momentum before the explosion is zero since the bomb is stationary. After the explosion, the momentum can be calculated as follows:
Let the velocity of the first two parts be v1 and v2 respectively. Then, the velocity of the third part (the bigger one) can be found using conservation of momentum:
0 = m*v1 + m*v2 + 3m*v3
Simplifying:
v3 = -(v1 + v2)/3
The kinetic energy of the system before the explosion is also zero. After the explosion, the kinetic energy can be calculated as follows:
KE = (1/2)*m*v1^2 + (1/2)*m*v2^2 + (1/2)*3m*v3^2
Simplifying:
KE = (1/2)*m*(v1^2 + v2^2 + 3*(v1+v2)^2/9)
Given that v1 = v2 = 30 m/s, we can simplify further:
KE = (1/2)*m*(2*30^2 + 3*(2*30)^2/9)
KE = 1800/3 * m
KE = 600m
Now, using conservation of energy, we can equate the kinetic energy before and after the explosion:
0 = KE_before - KE_after
0 = 0 - 600m
m = 0
This is a contradiction, which means that the problem is not well-defined. There must be some other information missing or incorrect.
The total momentum before the explosion is zero since the bomb is stationary. After the explosion, the momentum can be calculated as follows:
Let the velocity of the first two parts be v1 and v2 respectively. Then, the velocity of the third part (the bigger one) can be found using conservation of momentum:
0 = m*v1 + m*v2 + 3m*v3
Simplifying:
v3 = -(v1 + v2)/3
The kinetic energy of the system before the explosion is also zero. After the explosion, the kinetic energy can be calculated as follows:
KE = (1/2)*m*v1^2 + (1/2)*m*v2^2 + (1/2)*3m*v3^2
Simplifying:
KE = (1/2)*m*(v1^2 + v2^2 + 3*(v1+v2)^2/9)
Given that v1 = v2 = 30 m/s, we can simplify further:
KE = (1/2)*m*(2*30^2 + 3*(2*30)^2/9)
KE = 1800/3 * m
KE = 600m
Now, using conservation of energy, we can equate the kinetic energy before and after the explosion:
0 = KE_before - KE_after
0 = 0 - 600m
m = 0
This is a contradiction, which means that the problem is not well-defined. There must be some other information missing or incorrect.
Community Answer
A 1 kg stationary bomb exploded in three parts having mass ratio 1:1:3...
Mass = 1 kg
initial momentum = 0
now particle explodes in 3 parts of mass 1/5,1/5 and 3/5 kg.
both mass of 1/5 kg has velocity 30m/s and are moving perpendicular to each other.
take angle bisector of this perpendicular as the x-axis.
momentum of both 1/5kg mass along x axis is= 2mvcos(π/2) = 2× 1/5×30×1/√2 = 30√2/5 kgm/s
initial momentum was zero
so total momentum3/5v + 30√2/5 = 0
v= -10√2
initial momentum = 0
now particle explodes in 3 parts of mass 1/5,1/5 and 3/5 kg.
both mass of 1/5 kg has velocity 30m/s and are moving perpendicular to each other.
take angle bisector of this perpendicular as the x-axis.
momentum of both 1/5kg mass along x axis is= 2mvcos(π/2) = 2× 1/5×30×1/√2 = 30√2/5 kgm/s
initial momentum was zero
so total momentum3/5v + 30√2/5 = 0
v= -10√2
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