Answer: v1 = 1.8m/s
The velocity of the 5.0 kg block immediately after the collision is 1.8m/s
Step-by-step 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