Everything Science Grade 12

6.3 CHAPTER 6. MOTIONIN TWO DIMENSIONS

Calculate the angle θ to find the direction of Player 1’s final veloc-

ity:

sin θ =

vfy 1

vftot

θ = 52. 4 ◦

Therefore Player 1 bounces off Player 2 witha final velocity of

11.36 m·s−^1 at an angle of 52.4◦from the horizontal.

Example 7: 2D Conservation of Momentum: II

QUESTION

In a soccer game, Player 1 is running with the ball at 5 m·s−^1 across the pitch

at an angle of 75 ◦from the horizontal. Player 2 runs towards Player 1 at

6 m·s−^1 an angle of 60 ◦to the horizontal and tackles Player 1. In the tackle,

the two players bounceoff each other. Player 2moves off with a velocity in the

opposite x-direction of 0.3 m·s−^1 and a velocity in the y-direction of 6 m·s−^1.

Both the players have amass of 80 kg. What is the final total velocity ofPlayer

1?

SOLUTION

Step 1 : Identify what is required and what is given

The first step is to draw the picture to work out what the situation

is. Mark the initial velocities of both players in the picture.

75 ◦

60 ◦

vix 1

vix 2

viy

2

viy

1

vi

1 =5 ms

−

1

vi 2

=6 ms

− 1

We need to define a reference frame: For y, choose the direction

they are both running inas positive. For x, the direction player 2

is running in is positive.

We also know that m 1 = m 2 = 80 kg. And vfx 2 =-0.3 ms−^1 and

vfy 2 =6 ms−^1.

We need to find the final velocity and angle at which Player 1

bounces off Player 2.