12. Let's say you have a 1.0 kg mass at rest, struck by a 2.0 kg mass traveling at -2.0 m/s. They collide and stick together, thus combining their masses to 3.0 kg. What is the
velocity of the combined mass and what type of collision is this?

Recall Newton’s third law: When two objects of masses m1 and m2 interact (meaning that they apply forces on each other), the force that object 2 applies to object 1 is equal in magnitude and opposite in direction to the force that object 1 applies on object 2. Let:

F⃗ 21= the force on m1 from m2

F⃗ 12= the force on m2 from m1

Then, in symbols, Newton’s third law says

F⃗ 21m1a⃗ 1==−F⃗ 12−m2a⃗ 2.

9.10

(Recall that these two forces do not cancel because they are applied to different objects. F21 causes m1 to accelerate, and F12 causes m2 to accelerate.)

Although the magnitudes of the forces on the objects are the same, the accelerations are not, simply because the masses (in general) are different. Therefore, the changes in velocity of each object are different:

dv⃗ 1dt≠dv⃗ 2dt.

However, the products of the mass and the change of velocity are equal (in magnitude):

m1dv⃗ 1dt=−m2dv⃗ 2dt.

9.11

It’s a good idea, at this point, to make sure you’re clear on the physical meaning of the derivatives in Equation 9.3. Because of the interaction, each object ends up getting its velocity changed, by an amount dv. Furthermore, the interaction occurs over a time interval dt, which means that the change of velocities also occurs over dt. This time interval is the same for each object.

Let‘s assume, for the moment, that the masses of the objects do not change during the interaction. (We’ll relax this restriction later.) In that case, we can pull the masses inside the derivatives:

ddt(m1v⃗ 1)=−ddt(m2v⃗ 2)

9.12

and thus

dp⃗ 1dt=−dp⃗ 2dt.

9.13

This says that the rate at which momentum changes is the same for both objects. The masses are different, and the changes of velocity are different, but the rate of change of the product of m and v⃗  are the same.

Physically, this means that during the interaction of the two objects (m1andm2), both objects have their momentum changed; but those changes are identical in magnitude, though opposite in sign. For example, the momentum of object 1 might increase, which means that the momentum of object 2 decreases by exactly the same amount.

In light of this, let’s re-write Equation 9.12 in a more suggestive form:

dp⃗ 1dt+dp⃗ 2dt=0.

9.14

This says that during the interaction, although object 1’s momentum changes, and object 2’s momentum also changes, these two changes cancel each other out, so that the total change of momentum of the two objects together is zero.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer