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~christina~
Gold Member
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1. Homework Statement
A machine designed for quality testing in a handball factory propels the 0.15kg balls toward a wall with a velocity of 9.5m/s at an angle of 60 deg frome the normal to the wall. A typical ball rebounds at 70 deg from the normal at a speed of 9.0m/s.
The impact point of the balls is 0.5m above the floor
a) what impulse does a typical ball deliver to the wall
b) what average force acts on the wall if the ball is in contact with the wall 4.9ms?
c) what is the maximum height that the ball reaches after leaving the wall?
d) How long does it take to reach that height?
e) Find the horizontal distance from the wall to the point where the rebounding
ball lands on the floor
f) How long does it take the ball to travel from the wall to the floor?
g) Find the velocity of the ball (magnitude and direction just before it hits the floor
h) Draw a free body diagram for the ball at the highest point of it’s trajectory
http://img77.imageshack.us/img77/8859/32277524sj2.th.jpg
[tex]I= \Delta P = p_f- p_i= mv_f- mv_i[/tex]
[tex]\Delta P/ \Delta t= Faverage[/tex]
a) what impulse does a typical ball deliver to the wall
[tex]\Delta P= mv_f- mv_i[/tex]
[tex]p_i= m(vf_x- vi_x)= \Delta P_x [/tex]
[tex]m(vf_y- vi_y)= \Delta P_x [/tex]
m= 0.15kg
v1= 9.5m/s
v2= 9.0m/s
theta 1= 60 deg
theta 2= 70 deg
[tex]\Delta P_x= [/tex]0.15kg(9.0cos70 - 9.5cos120)= 1.17[tex]\hat{}i[/tex]
[tex]\Delta P_y= [/tex]0.15kg(9.0sin70 - 9.5sin120)= 0.034[tex]\hat{}j[/tex]
[tex]I= \Delta P= 1.17\hat{}i, 0.034\hat{}j[/tex]
___________________________________________________________________________
b) average force acts on wall if ball is in contact with wall for 4.9 ms?
[tex]Faverage= \Delta P / \Delta t[/tex]
~I'm not sure if this is correct but since I have the x and y component of the impulse to get the change of momentum value to plug into the F average equation I think I need to get the magnitude.
[tex] \Delta P= \sqrt{} P^2 _x + P^2 _y [/tex]
[tex] \Delta P= (1.17)^2 + (0.034)^2 = 1.17kg*m/s [/tex]
t= 4.9ms= 4.9 x 10^-3s
[tex]Faverage= \Delta P / \Delta t= (1.17kg*m/s) / 4.9 x10^-3s = 238.8N [/tex]
c.) Max height the ball reaches after leaving the wall
t= ?
dy= _____?
Voy= 9.0m/s
theta= 70 deg
Vy= 0 ==> at max height
Vy= Voy + at
t= - Voy/ a
[tex]Sy= Soy + Voy(t) + .5 at^2[/tex]
[tex]Sy= Voy(-Voy/ a) + .5 a(-Voy/ a)^2[/tex]
[tex] Sy= -Voy^2 / a + .5 a (-Voy/ a)^2 [/tex]
[tex]Sy= - (9.0 sin 70) ^2 / (-9.81m/s^2) + .5 (-9.81m/s^2) (- 9.0sin70 /(-9.81m/s^2) )^2 [/tex]
Sy= 7.291 - 3.645 = 3.645s
Sy= 3.645s
d) time it takes to reach that height
Vy= Voy + att= -Voy/ a
t= -9.0 sin 70 / (-9.81m/s^2)
[tex]t max= 0.8621s[/tex]
___________________________________________
e) horizontal distance from point where rebounding ball lands on floor
~I'm not quite sure about this since I would think that the max height time would be doubled to see when it reaches the same point again from the rebound on the wall however that wouldn't be when it reaches the floor.
I was thinking that I should see what time is to reach the horizontal distance of 0 since the initial doy= 0.5m so
Soy= 0.5m
Sy= 0 m
t= ?
tmax= 0.8621s
dx= ?
[tex]Sy= Soy + Voy (t) + 0.5 a*t^2[/tex]
0= 0.5 +(9.0sin 70) t + (-4.9) t^2
using quadradic formula to solve for t
[-8.457 +/- [tex]\sqrt{} (8.457)^2 - 4(0.5*-4.9)[/tex]] / 2*-4.9
(-8.457 +/- 9.011)/ -9.8
t = 1.78 s ===> to reach the ground
Sx= Sox + Vx t
Sx= 0 + 9.0 cos 70 (1.78s)
Sx= 5.479m
_________________________________________
f) how long does it take for the ball to travel from the wall to the floor?
found this in last part
t= 1.78s
__________________________________________
g) find velocity of the ball (magnitude and direction just before it hits the floor)
~ how do I determine the magnitude and direction just before it hits the floor?
What time do I use?
I do know that it takes 1.78 sec for the ball to go from the wall to the floor but what is the time right before that?
Is it 1.77s ?
____________________________________________
h) free body diagram of ball at highest point
I think that the ball would only have the Vx as the force on the ball only.
_______________________________________________________________
Could someone please tell me if what I did in the other parts of this long problem are correct?
And also I need help on the last part g) where I have to find the magnitude and velocity of the ball right before it hits the ground.
THANK YOU
A machine designed for quality testing in a handball factory propels the 0.15kg balls toward a wall with a velocity of 9.5m/s at an angle of 60 deg frome the normal to the wall. A typical ball rebounds at 70 deg from the normal at a speed of 9.0m/s.
The impact point of the balls is 0.5m above the floor
a) what impulse does a typical ball deliver to the wall
b) what average force acts on the wall if the ball is in contact with the wall 4.9ms?
c) what is the maximum height that the ball reaches after leaving the wall?
d) How long does it take to reach that height?
e) Find the horizontal distance from the wall to the point where the rebounding
ball lands on the floor
f) How long does it take the ball to travel from the wall to the floor?
g) Find the velocity of the ball (magnitude and direction just before it hits the floor
h) Draw a free body diagram for the ball at the highest point of it’s trajectory
http://img77.imageshack.us/img77/8859/32277524sj2.th.jpg
Homework Equations
[tex]I= \Delta P = p_f- p_i= mv_f- mv_i[/tex]
[tex]\Delta P/ \Delta t= Faverage[/tex]
The Attempt at a Solution
a) what impulse does a typical ball deliver to the wall
[tex]\Delta P= mv_f- mv_i[/tex]
[tex]p_i= m(vf_x- vi_x)= \Delta P_x [/tex]
[tex]m(vf_y- vi_y)= \Delta P_x [/tex]
m= 0.15kg
v1= 9.5m/s
v2= 9.0m/s
theta 1= 60 deg
theta 2= 70 deg
[tex]\Delta P_x= [/tex]0.15kg(9.0cos70 - 9.5cos120)= 1.17[tex]\hat{}i[/tex]
[tex]\Delta P_y= [/tex]0.15kg(9.0sin70 - 9.5sin120)= 0.034[tex]\hat{}j[/tex]
[tex]I= \Delta P= 1.17\hat{}i, 0.034\hat{}j[/tex]
___________________________________________________________________________
b) average force acts on wall if ball is in contact with wall for 4.9 ms?
[tex]Faverage= \Delta P / \Delta t[/tex]
~I'm not sure if this is correct but since I have the x and y component of the impulse to get the change of momentum value to plug into the F average equation I think I need to get the magnitude.
[tex] \Delta P= \sqrt{} P^2 _x + P^2 _y [/tex]
[tex] \Delta P= (1.17)^2 + (0.034)^2 = 1.17kg*m/s [/tex]
t= 4.9ms= 4.9 x 10^-3s
[tex]Faverage= \Delta P / \Delta t= (1.17kg*m/s) / 4.9 x10^-3s = 238.8N [/tex]
c.) Max height the ball reaches after leaving the wall
t= ?
dy= _____?
Voy= 9.0m/s
theta= 70 deg
Vy= 0 ==> at max height
Vy= Voy + at
t= - Voy/ a
[tex]Sy= Soy + Voy(t) + .5 at^2[/tex]
[tex]Sy= Voy(-Voy/ a) + .5 a(-Voy/ a)^2[/tex]
[tex] Sy= -Voy^2 / a + .5 a (-Voy/ a)^2 [/tex]
[tex]Sy= - (9.0 sin 70) ^2 / (-9.81m/s^2) + .5 (-9.81m/s^2) (- 9.0sin70 /(-9.81m/s^2) )^2 [/tex]
Sy= 7.291 - 3.645 = 3.645s
Sy= 3.645s
d) time it takes to reach that height
Vy= Voy + att= -Voy/ a
t= -9.0 sin 70 / (-9.81m/s^2)
[tex]t max= 0.8621s[/tex]
___________________________________________
e) horizontal distance from point where rebounding ball lands on floor
~I'm not quite sure about this since I would think that the max height time would be doubled to see when it reaches the same point again from the rebound on the wall however that wouldn't be when it reaches the floor.
I was thinking that I should see what time is to reach the horizontal distance of 0 since the initial doy= 0.5m so
Soy= 0.5m
Sy= 0 m
t= ?
tmax= 0.8621s
dx= ?
[tex]Sy= Soy + Voy (t) + 0.5 a*t^2[/tex]
0= 0.5 +(9.0sin 70) t + (-4.9) t^2
using quadradic formula to solve for t
[-8.457 +/- [tex]\sqrt{} (8.457)^2 - 4(0.5*-4.9)[/tex]] / 2*-4.9
(-8.457 +/- 9.011)/ -9.8
t = 1.78 s ===> to reach the ground
Sx= Sox + Vx t
Sx= 0 + 9.0 cos 70 (1.78s)
Sx= 5.479m
_________________________________________
f) how long does it take for the ball to travel from the wall to the floor?
found this in last part
t= 1.78s
__________________________________________
g) find velocity of the ball (magnitude and direction just before it hits the floor)
~ how do I determine the magnitude and direction just before it hits the floor?
What time do I use?
I do know that it takes 1.78 sec for the ball to go from the wall to the floor but what is the time right before that?
Is it 1.77s ?
____________________________________________
h) free body diagram of ball at highest point
I think that the ball would only have the Vx as the force on the ball only.
_______________________________________________________________
Could someone please tell me if what I did in the other parts of this long problem are correct?
And also I need help on the last part g) where I have to find the magnitude and velocity of the ball right before it hits the ground.
THANK YOU
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