Question

Two bodies of equal mass 1Kg are collided. After collision one of body is moving with a velocity of 4ms^-1.If the total KE of systemafter collision is 12.5J, then velocity of second body is​

Answer 1

mass of each body, m = 1kg

after collision, one body is moving with velocity, v = 4m/s

Let velocity of 2nd body is u m/s

total kinetic energy of system after collision is 12.5 J

so, kinetic energy of first body + kinetic energy of 2nd body = 12.5 J

or, 1/2 mv² + 1/2mu² = 12.5

or, 1/2 × 1 × (4)² + 1/2 × 1 × u² = 12.5

or, 8 + u²/2 = 12.5

or, u²/2 = 4.5

or, u² = 9 => u = 3 m/s

hence, velocity of 2nd body, after collision is 3m/s

Answer 2

Given,

let the mass of two bodies be A and B respectively.

Such that A= B = 1kg.

After collision the velocity of body A be 4m/s(v)

So, let the velocity of body B after collision be x m/s.

Also,total Kinetic energy in the whole system= 12.5J

So, according to the question=>

Kinetic energy of Body A + Kinetic energy of Body B = Total Kinetic energy in the system.

=>1/2(m×v^2) + 1/2{m×(x^2)} = 12.5j

=>1/2{mv^2 + mx^2} = 12.5

=>m{v^2 + x^2} = 12.5×2

=>(4)^2 + x^2 = 25

(since m= 1 kg)

=>16 + x^2 = 25

=>x^2 = 9

=>x= 3m/s

Hence velocity of second body is 3m/s after collision.

Some information=>

Kinetic energy is the energy that is present in an moving object or body.Any object that is in motion is said to be using kinetic energy.

Formula of kinetic energy=> 1/2(m×v^2)

S.I Unit => Joules