Question

Question :
A ball of mass 0.3 kg coming towards a player with a velocity of 10 ms–1 is kicked back with a velocity of 20 ms–1 in the opposite direction. If the impact lasts for 1/30 th of a second, what is the average force involved?

Answer 1

I got 90 N somehow :P

but the answer is wrong

Answer 2

I took u = 10 and v = 20

Answer 3

Did you use the impulse equation?
\[\vec I=\sum \vec F \Delta t\]

Answer 4

and t = 1/30

so what we have :
a = (v-u)/t = (10)/(1/30) = 300 m/s^2

Answer 5

F = 0.3 kg * 300 m/s^2 = 90 N

Answer 6

@Xishem that is absolutely NOT in our course.. so this will not be used here...

any easier method like I did?

Answer 7

Hmmm. One sec.

Answer 8

oh k

Answer 9

Why not use the momentum-impulse theorem?\[mv_f-mv_f=Ft\]

Answer 10

Your problem is that you aren't accounting for the fact that the second velocity is in the negative x direction.

Answer 11

Typo fixed:\[mv_f-mv_i=Ft\]

Answer 12

You can use F=ma where a is the average acceleration. You'll get:\[F=ma \rightarrow F=m(\frac{v_f-v_i}{\Delta t})\]

Answer 13

right

Answer 14

That's what you did, just make sure you use the correct signs, and you end up getting:\[F=(0.3kg)(\frac{(-20\frac{m}{s}-10\frac{m}{s})}{\frac{1}{30}s})=-270N\]

Answer 15

Yep ^^

Answer 16

oh why -ve there?

Answer 17

I don't think he wants to use impulse theory because I don't think they've covered momentum yet?

Answer 18

@mathslover: Because the ball changed directions...

Answer 19

@Xishem: You and I both used the same formula :)

Answer 20

The initial velocity is 10m/s in +x direction. It hits a wall and reverses direction to the -x direction at 20m/s. The total change in velocity is -30m/s.

Answer 21

Yeah, it breaks down into the same formula, but mine doesn't require an understanding of momentum or impulse :P.

Answer 22

Isn't the entirety of physics like that?

Answer 23

right thanks a lot @Xishem and @Shane_B

@Xishem can u please give a medal to shane_B since m not able to give medal to both

Answer 24

My crystal ball says his next chapter will cover impulse and momentum :)

Answer 25

Well I had studied momentum : p = m v and the :

momentum before collision = momentum after collision
like this ... but just forgot about "impulse"

I got this now

thanks a lot again

Answer 26

Perhaps. Maybe it will be circular motion, though :P. Or energy. Or... etc.

Answer 27

:)

Answer 28

Or maybe... relativity :D

Answer 29

the answer is "270 N " why not -270?

Answer 30

It depends on your frame of reference. Either answer should be acceptable. Make sure you draw a diagram of the situation which indicates the direction of the x-axis to clarify. As long as the force is directed in the opposite direction of the initial velocity, you're okay.

Answer 31

oh k got it thanks again :)

Answer 32

The bottom line is the magnitude of the force = |F|=|-270N|=270N

Answer 33

right ~~