Question

A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms–1. The kinetic energy of the other mass is

Answer 1

 

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A bomb of mass 16 kg at rest explodes into...

A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms-1. The kinetic energy of the other mass is ?

4 years ago

Answers : (9)

Hello student,

From the given data

m1v1 = m2v2
KE1=1/2(m2 v22 )=1/2(4)(144)= 288J

So the kinetic energy of other mass is 288J

Thanks and Rdegards

Shaik Aasif

askIITians faculty

4 years ago

Hi Shaik! I think the answer is 96 J and not 288 J.
The formula is KE = ½ (mv2) .
m = 12 kg and v= 4 m/s 
So, according to the formula, KE = ½ (12)(42) 
                                                     = ½ (12)(16) = 96J
Instead of squaring the velocity, you squared the mass. 
 

3 years ago

Shaik is right but Tishya, u are forgetting that the KE is of the other mass.

3 years ago

According to me the answer is 486 J.By conservation of linear momentumM1v1=M2v"2v"=9m/sUsing this velocity we can calculate the kinetic energy of 12 kg mass which is =1/2*(12)(9)(9).

one year ago

Hello everyone, From the given datam1v1 = m2v2KE1=1/2(m2 v22 )=1/2(4)(144)= 288JSo the kinetic energy of other mass is 288.This is the perfect answer.!

one year ago

momentum before explosion ismv=0(since bomb is at rest)______1where m=16kgmomem. after explosion ism1v1+m2v2 _______2where m1=12kg,v1=4m/s m2=4kg,v2=?now by law of conservation of linear momentum the momentum before and after explosion will be same therefore from 1 and 2 we havem1v1+m2v2=mvm1v1+m2v2=0m1v1=-m2v212*4=4*v2(v2 is the velocity of mass fragment whose KE is to cal.)v2=12m/sthus kinetic energy of other part of mass isKE=1/2m2v2^2KE=1/2*4*144KE=288J

Answer 2

Answer: The kinetic energy of the other mass is 288 J.

Explanation:

Given that,

Mass of first piece = 12 kg

Velocity of first piece $v_{1}=4\ m/s$

Mass of the other piece = 4 kg

Using conservation of momentum

$m_{1}u_{1}= m_{1}v_{1}+m_{2}v_{2}$

$0 = 12\times4+4\times v$

$v = -12\ m/s$

The other piece will move with 12 m/s in opposite direction.

The kinetic energy of the other mass is

$K.E = \dfrac{1}{2}mv^2$

$k.E = \dfrac{1}{2}\times4\times12\times12$

$K.E = 288 J$

Hence, The kinetic energy of the other mass is 288 J.