# a body of mass 1kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3. the twonpieces of equal masses fly off perpendicular to each other with a speed of 30m/s each.what is the velocity of the heavier fragment? plz tell me the answer quickly with full description

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a body of mass 1kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3. the twonpieces of equal masses fly off perpendicular to each other with a speed of 30m/s each.what is the velocity of the heavier fragment?

plz tell me the answer quickly with full description

## 6 Answers

ACCORDING TO LAW OF CONSERVATION OF LINEAR MOMENTUM

M1V1=mv+mv+3mv

v=30ms-1

30kg*0=30m+30m+3xm

assuming masses m,m and 3m and x=velocity of hevier fragment

o=60m+3xm

-60m/3m=x

-20=x

THEREFORE THE HEVIER FRAGMENT FLIES OFF WITH A SPEED OF 20ms-1 in the perpendicular direction.

plz approve.....!!

Thank you.

Ah! That is an easy one. Look, The figure will make clear that movement of COM of the two equally weighed particles is along the dotted axiz (please consult the figure) which is at an angle 45° from both the particles m1 and m2. Now, we consider it as a single particle of mass 400gms (ATQ) [mass of m1+m2] moving with a velocity of [30 cos45°] along the dotted line.

So, the third particle (m3) which is of mass thrice to the original m1 and m2, and 3/2 of the mass of their centre of mass, must move along the red line with a velocity that balances the net momentum of the rest two particles to zero. Now acc. to Law of Conserv. of Momentum,

Let, m* and v* be the mass and velocity of centre of mass of m1 and m2 respectively.

m3.v3=m*.v*

.6 is the mass of heavier particle.

.4 is the mass of COM of the two lighter particles.

v3=Unknown

v*= 30cos45°

Solving the above eqn using the given statistics, we get

**v3=10√2**

**which is our required answer**

**CHEGCK**

ration of masses = 1:1:3

m_{1} = k

m_{2 = k}

m_{3} = 3k

m_{1} + m_{2} + m_{3} = 1kg

k = 1/5

m_{1}, m_{2},m_{3} are _{1/5 , 1/5 , 3/5 }kg respectively ...

initial momentam = 0

final momentam = m_{1}v_{1 + m}2_{v}2 + m_{3}v_{3}

since there is no force acting on the system so linear momentam remains conserved

m_{1}v_{1} + m_{2}v_{2} + m_{3}v_{3} = 0

after putting value of m_{1},m_{2},m_{3 }we get

v_{3} = -(v1+v_{2})/3 ....................1

now it is given that velocity of particles of same masses are

perpendicular with magnitude 30m/s ...so if v_{1} is 30i then

v_{2} can be 30j .... (i & j are unit vectors along +x & +ve y axis)

v_{3} = -(30i+30j)/3

v_{3} = -10i - 10j ....................3

magnitude of v_{3} = root(10^{2}+10^{2})

=10root2 m/s

approve if u like my ans

ration of masses = 1:1:3

m_{1} = k

m_{2 = k}

m_{3} = 3k

m_{1} + m_{2} + m_{3} = 1kg

k = 1/5

m_{1}, m_{2},m_{3} are _{1/5 , 1/5 , 3/5 }kg respectively ...

initial momentam = 0

final momentam = m_{1}v_{1 + m}2_{v}2 + m_{3}v_{3}

since there is no force acting on the system so linear momentam remains conserved

m_{1}v_{1} + m_{2}v_{2} + m_{3}v_{3} = 0

after putting value of m_{1},m_{2},m_{3 }we get

v_{3} = -(v1+v_{2})/3 ....................1

now it is given that velocity of particles of same masses are

perpendicular with magnitude 30m/s ...so if v_{1} is 30i then

v_{2} can be 30j .... (i & j are unit vectors along +x & +ve y axis)

v_{3} = -(30i+30j)/3

v_{3} = -10i - 10j ....................3

magnitude of v_{3} = **√**(10^{2}+10^{2})

=10**√**2 m/s = 14.14 m/s

Hope it helps.

Thanks and regards,

Kushagra