A 5.0 kg block with a speed of 2.8 m/s collides with a 10 kg block that has a speed of 1.8 m/s in the same direction. After the collision, the 10 kg block is observed to be traveling in the original direction with a speed of 2.3 m/s. What is the velocity of the 5.0 kg block immediately after the collision

Answer: v1 = 1.8m/s

The velocity of the 5.0 kg block immediately after the collision is 1.8m/s

Explanation:

Applying the law of conservation of momentum;

Change in momentum of block 1 = change in momentum of block 2

∆p1 = ∆p2

m1 (∆v1) = m2 (∆v2)

m1 (u1-v1) = m2 (v2-u2) ... 1

where,

m1 and m2 are the masses of the 5 kg and 10kg block respectively.

v1 and v2 are the final speeds of the 5 kg and 10kg block respectively.

u1 and u2 are the initial speeds of the 5 kg and 10kg block respectively.

Given;

m1 = 5 kg

m2 = 10kg

u1 = 2.8m/s

u2 = 1.8m/s

v1 = ?

v2 = 2.3 m/s

Substituting the given values into the equation 1

5 (2.8-v1) = 10 (2.3-1.8)

(2.8-v1) = 10 (0.5) / 5

2.8-v1 = 1

v1 = 2.8-1 = 1.8

v1 = 1.8m/s

Therefore, the velocity of the 5.0 kg block immediately after the collision is 1.8m/s