Perfectly Inelastic Collision

Question

Two blocks are sliding along a frictionless track. Block A (mass 3.30 kg) is moving to the right at 1.70 m/s. Block B (mass 3.96 kg) is moving to the left at 2.04 m/s. Assume the system to be both Block A and Block B.

What is the total momentum of the system before the collision?

Left

Right

Tries 2/10

What is the total momentum of the system after the collision?

Left

Right

Tries 3/10

The two blocks stick together and move off together. What is the magnitude and direction of the velocity of the blocks after the collision?

Left

Right

Answer 1

Momentum before = 3.30•1.7 + (3.96•-2.04) = -2.47N•s (- sign indicates direction to the left)

Momentum after is equal to momentum before (Law of conservation of momentum).

-2.47 = V(m₁ + m₂)

-2.47/7.26 = V = -0.34m/s (- sign indicates direction to the left).

I hope this helps.