m1v1 + m2v2 = (m1+m2)vf
solve for v1
solve for v1
Before the collision, the momentum of Block B is given by the product of its mass and velocity:
Momentum of Block B before = mass of Block B × velocity of Block B
= 3.40 kg × (-4.20 m/s) (since Block B is moving to the left)
= -14.28 kg·m/s
The momentum of Block A before the collision would be:
Momentum of Block A before = mass of Block A × velocity of Block A
Let's assume the velocity of Block A before the collision is v. Since it is moving to the right, the velocity of Block A would be positive:
Momentum of Block A before = 6.40 kg × v
Using the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. After the collision, both blocks move to the right with a speed of 0.87 m/s. Therefore, the momentum of the system after the collision is:
Momentum of the system after = (mass of Block A + mass of Block B) × final velocity
The final velocity of both blocks moving to the right with a speed of 0.87 m/s should be positive. Now we can equate the total momentum before and after the collision:
Momentum of Block A before + Momentum of Block B before = Momentum of the system after
6.40 kg × v + (-14.28 kg·m/s) = (3.40 kg + 6.40 kg) × 0.87 m/s
Simplifying the equation:
6.4v - 14.28 = 9.8 × 0.87
6.4v - 14.28 = 8.526
6.4v = 22.806
v = 22.806 / 6.4
v ≈ 3.57 m/s
Therefore, to have the two blocks move to the right with a speed of 0.87 m/s after the collision, the velocity of Block A before the collision should be approximately 3.57 m/s.