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

Question

Tejas Paresh Lodaya · Answer

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.

Chegck Punk · Answer

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

vikas askiitian expert · Answer

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

Binciya Izara · Answer

mass of the body =1Kg i.e, 1:1:3 (1m,1m,3m) 1m+1m+3m=1Kg 5m=1Kg m=(1/5)Kg -------(1) Total momentum before explosion, P=M*V (V=0,AS IT IS IN REST) Total momentum after explosion, P1=m*v =1m*v (from eq (1 we get) =1*(1/5)*30 =6m/s --------(2) P2=m*v =1*(1/5)*30 =6m/s --------(3) P3=m*v =3m*v --------(4) By law of conservation of mass, total momentum before explotion=total momentum after explosion i.e, 0=P1+P2+P3 -P1-P2=P3 -(P1+P2)=P3 P3=-(P1+P2) (as the questionthe angle between two peices with equal mass is 90 degree so by using addition of vectors R= root of A^2 +B^2) thus, P3=root of( P1^2+P2^2) from (2) and (3) we get, P3=root of (6^2+6^2) =root of (2*6^2) =6 root of(2) from eq. (4) we get 3m*v=P3 3m*v=6 root of (2) 3*(1/5)*v=6 root of (2) v=6 root of (2) *(1/3)*5 =10 root of (2)=10*1.414 =14.14m/s therefore, velocity of the heavier body is 14.14m/s

Shubhangkar Choudhury · Answer

Let V is the velocity of heavier fragmentm=m*=0.2kgM=0.6kgv=v*=30m/sFrom the principal of conservation of momentum MV=mvcos45*+m*v*cos45*Or 0.6V=0.2*30*1/v2+0.2*30*1/v2V=10v2=14.24m/s

Kushagra Madhukar · Answer

Dear student,

Please find the answer to your question.

ration of masses = 1:1:3

m₁ = k

m_{2 = k}

m₃ = 3k

m₁ + m₂ + m₃ = 1kg

k = 1/5

m₁, m₂,m₃ are _{1/5 , 1/5 , 3/5}kg respectively ...

initial momentam = 0

final momentam = m₁v_{1 + m}2_v2 + m₃v₃

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

m₁v₁ + m₂v₂ + m₃v₃ = 0

after putting value of m₁,m₂,m₃we get

v₃ = -(v1+v₂)/3 ....................1

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

perpendicular with magnitude 30m/s ...so if v₁ is 30i then

v₂ can be 30j .... (i & j are unit vectors along +x & +ve y axis)

v₃ = -(30i+30j)/3

v₃ = -10i - 10j ....................3

magnitude of v₃ = √(10²+10²)

=10√2 m/s = 14.14 m/s

Hope it helps.

Thanks and regards,

Kushagra

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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

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