You are not logged in.
Pages: 1
A baseball has a mass of 0.14 kg. It is moving with a velocity of 38 m/s LEFTWARDS towards the bat just before impact. The bat applies an average force of 17,000 N RIGHTWARDS for 0.0007 s until they lose contact.
Calculate the ball's velocity when it leaves the bat in the following scenario.
P of ball = 0.14(38)
=5.32kg.m/s LEFTWARDS
***
J=Ft
J=17,000(0.0007)
=11.9N RIGHTWARDS
J=delta P
∴ Delta P also = 11.9N RIGHTWARDS
***
Delta P=Pf-Pi
Delta P=Pf-5.32 RIGHTWARDS
Delta P=0.14v-5.32 RIGHTWARDS
I’ll now consider rightwards to be positive, leftwards to be negative
11.9N=0.14v-(-5.32kg.m/s)
11.9N=0.14v+5.32kg.m/s
0.14v=11.9N-5.32kg.m/s
I’m not sure if this feels right, subtracting kg.m/s from N, but I’ll try it for now.
0.14v=6.58
V=6.58/0.14
V=47m/s?
"The secret of getting ahead is getting started."
Mark Twain
Offline
hi paulb203
Good to hear from you.
Your answer is correct. The working obviouly works but there are one thing not quite right about what you've written.
Back to basics:
So 11.9 isn't in Newtons as it's an impulse not a force. What are the units for impulse? Ns ie Newton seconds. So you are not subtracting Newtons from Ns. The equation underlying this calculation is
I had
That gives V = 47 m/s
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Ah, Ns, of course.
Thanks, Bob.
"The secret of getting ahead is getting started."
Mark Twain
Offline
Pages: 1