Question

Two objects A and B of identical masses, 5 kg each , collided elastically with velocities 5 m/s and 10 m/s respectively. What will be the velocity of mass B after collision?​

Answer 1

Explanation:

e = (v1 - u2) / ( v1 - u2)

for elastic collision e=1

from law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

m1 = m2 = 5kg

u1 = 5 m/s

u2 = 10 m/s

substitute value in above equation:

5×5 + 5×10 = 5×1 + 5×1

25 + 50 = 5 + 5

75 = 10

75 / 10

7.5 m/s

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Answer 2

Given:

Two objects A and B of identical masses, 5 kg each , collided elastically with velocities 5 m/s and 10 m/s respectively.

To find:

Velocity of B after collision.

Calculation:

Let u1 and v1 be related to object A and u2 and v2 be related to object B;

Collision is elastic , hence coefficient of restitution will be equal to 1.

$\rm{e = \dfrac{v2 - v1}{u1 - u2} }$

$= > \rm{1 = \dfrac{v2 - v1}{u1 - u2} }$

$= > \rm{v2 - v1 = u1 - u2}$

$= > \rm{v2 - v1 = 5 - ( - 10)}$

$= > \rm{v2 - v1 = 15 \: \: \: ......(1)}$

Now , applying Conservation of Momentum:

$\rm{ \therefore \: m(u1) + m(u2) = m(v1) + m(v2)}$

$\rm{ = > \:u1 + u2 = v1 + v2}$

$\rm{ = > \:5 + ( - 10) = v1 + v2}$

$\rm{ = > \: v1 + v2 = - 5 \: \: \: \: ......(2)}$

Adding the equations:

$\rm{ \therefore \: 2(v2) = 15 - 5}$

$\rm{ = > \: 2(v2) = 10}$

$\rm{ = > \:v2= 5 \: m {s}^{ - 1} }$

So, final Velocity of B is 5 m/s opposite to the direction of its initial velocity.

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