Question

A ball of mass 0.25 kg moving on a smooth horigental table with a velocity 10m/s strikes on identikal stationary ball B on the table find velocity of ball Bbjust after impact if the impact is perfect plastic

Answer 1

Answer:

m₁ = mass of first ball = 4 kg

v₁ = velocity of first ball before collision = v

v'₁ = velocity of first ball after collision = v/4

m₂ = mass of second ball = m

v₂ = velocity of second ball before collision = 0

v'₂ = velocity of second ball after collision = ?

using conservation of momentum

m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂

inserting the above values

4 v + m (0) = 4 (v/4) + m v'₂

m v'₂ = 3 v                                    

v'₂ = 3v/m                              eq-1

using conservation of kinetic energy for the elastic collision

m₁ v²₁ + m₂ v²₂ = m₁ v²'₁ + m₂ v²'₂

4 v² + m (0)² = 4 (v/4)² + m v²'₂

4 v² = (0.25) v² + m v²'₂

(3.75) v² = m v²'₂

using eq-1

(3.75) v² = m (3v/m)²

(3.75) v² = m (9 v²)/m²

(3.75) = (9)/m

m = 9/3.75

m = 2.4 kg

Answer 2

Explanation:

Given, m

1

=4kg u

1

=u (say) m

2

=m & u

2

=0 (at rest)

also, v

1

=u/4 and collision is elastic

From collision theory,

v

1

=(

m

1

+m

2

m

1

−m

2

)u

1

+(

m

1

+m

2

2m

2

)u

2

Put values

4

u

=(

4+m

4−m

)u+0⇒4+m=4(4−m)⇒m=

5

12

=2.4kg (Ans)