A 6.4 kg block with a speed of 5.4 m/s collides with a 12.8 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 12.8 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 6.4 kg block immediately after the collision

The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

Explanation:

Given;

mass of first block, m₁ = 6.4 kg

initial speed of first block, u₁ =  5.4 m/s

mass of second block, m₂ = 12.8 kg

initial speed of second block, u₂ = 3.6 m/s

final speed of second block, v₂ = 4.5 m/s

To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁  +  m₂u₂ = m₁v₁  + m₂v₂

where;

v₁ is the final speed of 6.4 kg block

v₂ is the final speed of 12.8 kg block

Substitute the given values of m₁, u₁, m₂,  u₂ , v₂ and calculate  v₁

6.4 x 5.4  +  12.8 x 3.6 = 6.4v₁  +  12.8 x 4.5

34.56 + 46.08  = 6.4v₁ + 57.6

80.64 = 6.4v₁  +  57.6

80.64 - 57.6 = 6.4v₁

23.04 = 6.4v₁

v₁ = 23.04 / 6.4

v₁  = 3.6 m/s

Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach