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 :
(1) 1/9
(2) 8/9
(3) 4/9
(4) 5/9

Answer 1

Answer:

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 = [(m1-m2)u1/(m1+m2)]+[(2m2u2)/(m1+m2)]

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

= 2mu/6m

= u/3

similarly, v2 = [(m2-m1)u2/(m1+m2)]+\[(2m1u1)/(m1+m2)]

= (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

After the collision the fraction of energy lost by the colliding body A is $\mathbf{\frac{8}{9}}$. Option second is correct.

Explanation:

Given data

1. Body A

$\mathbf{\textrm{Mass of block A} \ (M_{A})= 4m}$

$\mathbf{\textrm{Intial velocity of A} \ (U_{A})= u}$

$\mathbf{\textrm{Initial kinetic energy of block A} \ (K_{i})=\frac{1}{2}\times 4m\times u^{2}}$         ...1)

$\mathbf{\textrm{Final velocity of A} = V_{A}}$

2. Body B

$\mathbf{\textrm{Mass of block B} \ (M_{B})= 2m}$

$\mathbf{\textrm{Intial velocity of B} \ (U_{B})= 0}$

$\mathbf{\textrm{Final velocity of B} = V_{B}}$

3.Here is given collision perfectly elastic collision. So coefficient of restitution (e) is one.

4.From conservation of momentum

Initial momentum =Final momentum

$\mathbf{M_{A}U_{A}+M_{B}U_{B}=M_{A}V_{A}+M_{B}V_{B}}$

$\mathbf{{4m\times u}+2m\times 0=4m\times V_{A}+2m\times V_{B}}$

Which can be written as

$\mathbf{2u=2V_{A}+V_{B}}$

Means

$\mathbf{2V_{A}+V_{B}=2u}$        ...2)

5.Now from law Newton's coefficient of restitution

$\mathbf{e=\frac{\textrm{Velocity of separation}}{\textrm{Velocity of approach}}}$

$\mathbf{e=\frac{V_{B}-V_{A}}{U_{A}-U_{B}}}$

Where e=1

$\mathbf{1=\frac{V_{B}-V_{A}}{u}}$

So

$\mathbf{V_{B}-V_{A}=u}$      ...3)

6.On solving equation 2) and equation 3), we get

$\mathbf{V_{A}=\frac{u}{3}}$

7.  $\mathbf{\textrm{Final kinetic energy of block A} \ (K_{f})=\frac{1}{2}\times 4m\times \left ( \frac{u}{3} \right )^{2}}$

$\mathbf{\textrm{Final kinetic energy of block A} \ (K_{f})=\frac{1}{2}\times 4m\times \frac{u^{2}}{9}}$        ...4)

$\mathbf{\textrm{Change in kinetic energy of block A} \ (\Delta K)=K_{i}-K_{f}}$      ...5)

From equation 1) ,equation 4) and equation 5)

$\mathbf{\textrm{Change in kinetic energy of block A} \ (\Delta K)=\frac{4mu^{2}}{2}\times \frac{8}{9}}$      ...6)

8. Now fraction of change in kinetic energy to initial kinetic energy

$\mathbf{\frac{\Delta K}{K_{i}}=\frac{\frac{4mu^{2}}{2}\times \frac{8}{9}}{\frac{4mu^{2}}{2}}}$

So

$\mathbf{\frac{\Delta K}{K_{i}}=\frac{8}{9}}$