Question

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

Answer 1

Answer:

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.