Question

Body A of mass 4m moving with speed u collides with another body B of mass 2m,at rest.the collision is head on and elastic in nature.after the collision the fraction of energy lost by the colliding body A is

Answer 1

body A of mass 4m moves with speed u and body B of mass 2m at rest.

so, initial linear momentum of system = 4mu + 2m × 0 = 4mu

Let final velocity of body A is v1 and body B is v2

then, final linear momentum of system = 4mv1 + 2mv2

as collision is elastic ,

so, v1 = $\frac{(m_1-m_2)u_1}{(m_1+m_2)}+\frac{2m_2u_2}{(m_1+m_2)}$

= (4m - 2m) × u/(4m + 2m) + 2(2m)×0/(4m+2m)

= 2mu/6m

= u/3

similarly, v2 = $\frac{(m_2-m_1)u_2}{(m_1+m_2)}+\frac{2m_1u_1}{(m_1+m_2)}$

= (2m - 4m) × 0/(4m + 2m) + 2(4m)u/(4m+ 2m)

= 8mu/6m

= 4u/3

so, initial energy of body A = 1/2 (4m) u² =2mu²

final kinetic energy of body A = 1/2 × (4m) × u²/9

= 2mu²/9

so, change in kinetic energy = final kinetic energy - initial kinetic energy

= 2mu²/9 - 2mu²

= 2mu² [ 1/9 - 1]

= -16mu²/9 [ here negative sign indicates kinetic energy lost after collision of body A]

so, fractional lost in kinetic energy = change in kinetic energy/initial kinetic energy

= (16mu²/9)/(2mu²)

= 8/9

Answer 2

Answer:

Explanation:

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Body A of mass 4m moving with speed u collides with another body B of mass 2m,at rest.the collision is head on and elastic in nature.after the collision the fraction of energy lost by the colliding body A is

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abhi178

abhi178

body A of mass 4m moves with speed u and body B of mass 2m at rest.

so, initial linear momentum of system = 4mu + 2m × 0 = 4mu

Let final velocity of body A is v1 and body B is v2

then, final linear momentum of system = 4mv1 + 2mv2

as collision is elastic ,

so, v1 = \frac{(m_1-m_2)u_1}{(m_1+m_2)}+\frac{2m_2u_2}{(m_1+m_2)}

= (4m - 2m) × u/(4m + 2m) + 2(2m)×0/(4m+2m)

= 2mu/6m

= u/3

similarly, v2 = \frac{(m_2-m_1)u_2}{(m_1+m_2)}+\frac{2m_1u_1}{(m_1+m_2)}

= (2m - 4m) × 0/(4m + 2m) + 2(4m)u/(4m+ 2m)

= 8mu/6m

= 4u/3

You can do it now by using k

e concept easily