Kinematics/mechanics simple question?

Question

hocker players hits ball of mass 0.2kg. The contact time between the stick and the ball is 0.3s and the force exerted on the ball by the stick is 60N. Find the speed in which the ball leaves the stick.

This simple one baffled me for some reason =(

Thanks for the confirmation.

I did it myself but i was really unsure, i did

f=ma
60=0.2xa
60/0.2=a=300ms^-2

at=v
300x0.3=90ms^-1

8D

Answer 1

You are given force, and time
the product of these is impulse
Impulse = Ft = 60 x 0.3 = 18 Ns
impulse is also defined as the change in momentum
so with this we have (let u be initial puck velocity (0) and v be final velocity as it leaves the stick)
Impulse = m(ball)v - 0 (as initial puck velocity is assumed 0)
18 = 0.2v
v = 18/0.2 = 90m/s in the direction of the force applied.
(Wouldnt want to get hit by that)

Answer 2

You can tackle the problem via Newton's second law of motion: law of acceleration
The delivered force through the bat has made the ball to accelerate to a certain speed, Vf (final speed) from intial speed of 0 m/s during the contact period of 0.3s

Here is the procedure:

F= m.a------------------------------------(1)
=m.((vf-vi)/t)
where F= the applied force
m= mass of the body
a= acceleration of body
t= time
vf= final velocity
vi= intial velocity

rearranging equation (1)
you the following expression for final velocity of the ball is determined:

Vf=(F.t/m) + vi---------------------(2)
=(60N*0.3s/0.2kg)+0
= 90m/s
Hence the ball leaves the bat with the velocity of 90m/s

Answer 3

First, let's clarify...impulse is the change in momentum, but it is not the change in momentum over time. This follows from p = mv, the momentum, so that dp/dt = m dv/dt = m a = F force. The change in momentum over time is force.

Thus force times elapsed time (dt) is F dt = m (dv/dt) dt = m dv = I which is impulse by definition. And m dv = dp the change in momentum; while impulse is the change in momentum, it is not the change in momentum over time, which is force.

So we have I = m dv = F dt; so that dv = F dt/m; where F = 60 kg-m/sec^2 (aka Newtons), dt = .3 sec, and m = .2 kg. You can solve for the change in velocity dv upon impact.

Unfortunately, we don't know the velocity of the ball (field hockey?) before impact (v0); where dv = v1 - v0 and v1 is the velocity after impact, the velocity you are looking for. But in general v1 = dv + v0 = F dt/m + v0. If the ball was still before being struck (v0 = 0), then v1 = F dt/m and you can do the math.