A1.2 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 .

2.0 kg

Explanation:

For a closed system, the momentum is always conserved. Momentum of an object is given by mv where m is the mass and v is the velocity.

Momentum of the system before the collision is given by:

Momentum of the system after the collision is given by:

Mv where M is the combined mass of the 2 blocks as they stick together and v is the velocity.

As the momentum is conserved i.e momentum before collision and after collision is the same:

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2