Question

A 1.2 kg block slides along a frictionless surface at 1.2 m/s . a second block, sliding at a faster 4.4 m/s , collides with the first from behind and sticks to it. the final velocity of the combined blocks is 3.2 m/s .

Answer 1

The question involves calculating the final velocity of two colliding blocks using the conservation of momentum, which states that the total momentum before and after a collision remains constant in the absence of external forces.

Step-by-step explanation:

The question pertains to the conservation of momentum in a collision between two blocks. When the two blocks collide and stick together, the final combined mass is the sum of the individual masses, and the final combined velocity can be found using the principle of conservation of momentum. According to this principle, the total momentum before the collision equals the total momentum after the collision, given that no external forces act on the system.

For the scenario described, we can calculate the final velocity after the collision using the following equation:

M1 × v1 + M2 × v2 = (M1 + M2) × v_final

Where M1 and M2 are the masses of the blocks, v1 and v2 are their initial velocities, and v_final is the final velocity of the combined mass.