Question

Block mass 5. 0 sliding to the sight with velocity 7. 5 m/s and collides with block 2 of mass 1. 5 kg moving with 3. 0 m/s. The collision is perfectly elastic what is the velocity of block i after the comision ? positive velocity to the seg indicates motion to the left

Answer 1

AI-generated answer

To find the velocity of block i after the collision, we can use the principles of conservation of momentum and kinetic energy.

First, let's calculate the initial momentum of block 1 (mass = 5.0 kg) and block 2 (mass = 1.5 kg).

The initial momentum of block 1 (p1i) is given by:

p1i = mass1 * velocity1 = 5.0 kg * 7.5 m/s = 37.5 kg·m/s

The initial momentum of block 2 (p2i) is given by:

p2i = mass2 * velocity2 = 1.5 kg * 3.0 m/s = 4.5 kg·m/s

Since the collision is perfectly elastic, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write:

p1i + p2i = p1f + p2f

Substituting the values, we have:

37.5 kg·m/s + 4.5 kg·m/s = p1f + p2f

Now, let's calculate the final momentum of block 1 (p1f) and block 2 (p2f). Since the collision is elastic, the final momentum of each block will be the same as their initial momentum:

p1f = p1i = 37.5 kg·m/s

p2f = p2i = 4.5 kg·m/s

Therefore, the velocity of block i after the collision is the same as its initial velocity:

velocityi = p1f / mass1 = 37.5 kg·m/s / 5.0 kg = 7.5 m/s

Since the velocity is positive, it indicates motion to the right.

Please let me know if you found this information helpful or if you have any further questions.

Step-by-step explanation: