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 ?

SHAIK AASIF AHAMED
8 years ago
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
Tishya Gaba
11 Points
8 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.

Imrul Kayes
20 Points
8 years ago
Shaik is right but Tishya, u are forgetting that the KE is of the other mass.
Harshvardhan singh
11 Points
6 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).
Jay sanghavi
19 Points
5 years 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.!
13 Points
5 years 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
Ekansh nayak
13 Points
5 years ago
By conservation of linear momentum as in Hz dir the net ext force is zero 12×4=4×vV=12 so K.E=1/2×4×(12)^2=288
Aayushma kafle
13 Points
5 years ago
By conservation of momentum,We know,m1v1=m2v2Given.M1=12.V1=4M2=4V2=x(let)So.....m1v1=m2v2 or,12*4=4*x or,x=V2=12Now...KE=0.5*m2*(V2)² =0.5*4*12² =0.5*4*144 =288
Richa
15 Points
4 years ago
First we find v2of the mass of 4kg
By the conservation of momentum
M1v1=m2v2
V2 =8×6/4 =12
Then we find k.E
1/2 mv2 =1/2×4×(12)^2