Impulse produced by the ball??

Question

A ball of mass 0.25 kg is thrown 3 metres vertically into the air and is caught
by an athlete 1 metre from the ground as it falls. The acceleration of the ball
due to gravity is 9.81 m s-1 and the ball is in contact with the athlete’s hand
for 0.2 s before it has stopped moving. Calculate:
(i) the force exerted by the ball on the athlete’s hand
(ii) the impulse produced by the ball on the athlete’s hand
(iii) the velocity of the ball immediately before impacting on the athlete’s
hand.

Answer 1

Using conservation of energy techniques, we can find the initial / final (depending on how you want to look at it) speed of the ball just before it is caught 1 meter off the ground.
The ball was originally at rest at 3 meters off the ground and posses some amount of gravitational Potential Energy.

PE = mgh
PE = (.25 kg) * (9.81 m/s) * (3 meters)
PE = 7.36 Joules

As the ball fell, this PE was converted into Kinetic Energy. When the ball was 1 meter off the ground, it had converted 2 meters worth of PE into its KE.
ΔKE = ΔPE = mg(Δh) = (.25 kg) * (9.81 m/s) * (2 meters)
KE = 4.91 Joules

We also know that KE = ½ mv^2
So,
KE = (4.91 Joules) = ½ (.25 kg) * v^2,
Solving for v,
V^2 = 2 * (4.91 Joules) / (.25 kg)
V = sqrt (2 * (4.91 Joules) / (.25 kg))
V = 6.26 meters per second.


The ball is initially traveling at 6.26 m/s as it enters the catcher’s glove. The catcher of the ball exerts some (we assume constant) force on the ball and brings it to a stop in .2 second.

Δv = acceleration * time
Acceleration = Δv / time = (6.26 m/s) / (.2 seconds)
Acceleration = 31.3 m/s^2

Force = mass * acceleration
Force = (.25 kg) * (31.3 m/s^2)
Force = 7.83 Newtons

The impulse the ball exerts on the athlete’s hand can be found as the product of the average force exerted on the ball multiplied by the time.

Impulse = Force * time
Impulse = (7.83 Newtons) * (.2 seconds)
Impulse = 1.57 Newtons seconds

The impulse delivered to the ball is the same as the impulse delivered to the glove.

Answer 2

I'll assume the ball was thrown upwards at 0 meters. When the ball reaches the top it has zero velocity so then it's just 2 meters of free fall.
vf^2=vo^2+2ad
=0+2*9.81*2
vf^2=39.24
vf=6.26 m/s

fΔt=mΔv
f*.2=.25*(6.26-0)
f=7.83 N
impluse=f*Δt
=7.83*.2
fΔt=1.57 Ns

Your answers are
i)7.83 N
ii)1.57 Ns
iii)6.26 m/s

The impluse provided by the hand equals the impluse provided by the ball, Newton's Third Law.

Answer 3

Momentum = mass * velocity = impulse = tension * time Plug in what you be attentive to and clean up for what you like Conservation of Momentum entire momentum in the previous = entire momentum after pay interest to signs and indicators. Momentum is a vector, so course is substantial a million) impulse = tension * time = momentum = mass * velocity Im = Mo = 0.4 kg * 20 m/s = 8.0 kg*m/s 2) Im = tension * time = Mo = mass * velocity F* T = M*v 0.1s *F = 44kg * 15 m/s clean up for F attempt the different 2

Answer 4

(i) F = -7.83 N
(ii) J = 1.57 N*s
(iii) v = 6.26 m/s

F = change in mv / t of impact. So

v when it hits the hand = sqrt(2g(d-do)) = 6.26 m/s. The ball comes to rest in 0.2 s; F = - (0.25 kg)(6.26m/s) / (0.2s) = - 7.83 N. Negative because the force is in the opposite diection of the motion of the ball. Impulse, or J, = Impact force (7.83N) * impact time (0.2 s) = 1.57 N*s.

Answer 5

first we have to find the velocity of the ball when it is caught

Vf^2 = 2ad + Vi^2
Vf^2 = 2(-9.8)(1-3) + 0^2
Vf = -6.26m/s

impluse is change in momentum

J = m * v
J = .25 ( -6.26)
J = -1.565 kgm/s

F*x = m*v
F (.2) = (.25)(0 - (-6.62) )
F = 7.825N

Answer 6

F = ma
F = (0.25 kg)(9.80 m/s^2)

p = Ft
p = (F calculated above)(0.2 s)

Vf^2 = Vi^2 + 2ad
On the downward path of the ball:
Vi = 0.00 m/s
a = 9.80 m/s^2
d = 2.00 m