A body A of mass m=0.1 kg has an initial velocity of 3 hati m s- 1 . It collides elastically with another body B of the same mass which has an initial velocity of 5 hatj m s- 1 . After the collision, A moves with a velocity overset arrow v=4( hati + hatj) . The energy of B after the collision is written as (x/10) J . The value of x is

Question

A body A of mass m=0.1 kg has an initial velocity of 3 hati m s- 1 . It collides elastically with another body B of the same mass which has an initial velocity of 5 hatj m s- 1 . After the collision, A moves with a velocity overset arrow v=4( hati + hatj) . The energy of B after the collision is written as (x/10) J . The value of x is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Using the principle of conservation of linear momentum 0.1× 3 hati+0.1× 5 hatj=0.1× 4( hati + hatj)+0.1× overset arrow v overset arrow v=- hati+ hatj KB=(1/2)× 0.1× v2=(1/10)J x=1

Using the principle of conservation of linear momentum 0.1×3i^+0.1×5j^​=0.1×4(i^+j^​)+0.1×v→ v→=−i^+j^​ KB​=21​×0.1×v2=101​J x=1

Using the principle of conservation of linear momentum
$0.1 \times 3 \hat{i} + 0.1 \times 5 \hat{j} = 0.1 \times 4 (\hat{i} + \hat{j}) + 0.1 \times v \to$
$v \to = - \hat{i} + \hat{j}$
$K_{B} = \frac{1}{2} \times 0.1 \times v^{2} = \frac{1}{10} J$
$x = 1$