Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:12

6 Answers

Tejas Paresh Lodaya
33 Points
10 years ago

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
26 Points
10 years ago

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
509 Points
10 years ago

ration of masses = 1:1:3

m1 = k

m2 = 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 = m1v1 + m2v2 + 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,m3 we 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 = root(102+102)

                       =10root2 m/s

approve if u like my ans

Binciya Izara
24 Points
5 years ago
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
13 Points
3 years ago
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
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the answer to your question.
 

ration of masses = 1:1:3

m1 = k

m2 = 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 = m1v1 + m2v2 + 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,m3 we 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)

                       =102 m/s = 14.14 m/s

Hope it helps.

Thanks and regards,

Kushagra

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free