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

Answer:

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:

p1 + p2 = 1.2*1.2 + x*4.4 (where x is the mass of the 2nd block);

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.

Momentum = (1.2+x)*3.2

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

(1.2+x)*3.2 = 1.2*1.2 + 4.4*x

Evaluate x to get the mass of the 2nd block.