Peoples Physics Book Version-2

http://www.ck12.org Chapter 8. Momentum Conservation Version 2

8.5 Examples

Example 1

Question: Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed
of 2.5m/s. Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial
speed of block B?

Solution: To find mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg
and the mass of block A is 8.0kg.

10kg− 8 .0kg= 2 .0kg

Now that we know the mass of both blocks we can find the speed of block B. We will use conservation of momentum.
This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This
problem is one dimensional, because all motion happens along the same line. Thus we will use the equation

(mA+mB)vf=mA×vA+mB×vB

and solve for the velocity of block B.

(mA+mB)vf=mA×vA+mBvB⇒

(mA+mB)(vf)−(mA)(vA)

mB

=vB

Now we simply plug in what we know to solve for the velocity.

( 2 .0kg+ 8 .0kg)( 2 .5m/s)−( 8 .0kg)(0m/s)

2 .0kg

= 12 .5m/s

Example 2

Question: Chris and Ashley are playing pool on a frictionless table. Ashley hits the cue ball into the 8 ball with a
velocity of 1.2m/s. The cue ball(c)and the 8 ball (e) react as shown in the diagram. The 8 ball and the cue ball
both have a mass of.17kg. What is the velocity of the cue ball? What is the direction (the angle) of the cue ball?

Answer: We know the equation for conservation of momentum, along with the masses of the objects in question as
well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation