Worksheet
7: Linear Momentum
IMPULSE
Problem 1: In a game of softball,
a 0.200-kg softball crossed the plate at 15.0 m/s at an angle of 45.00
below the horizontal. The ball was hit at 40.0 m/s, 30.00 above
the horizontal.
(a) Determine the impulse applied to the ball.
Solution:
Initial momentum:
pix = (0.200 kg)(15.0 m/s)(cos -45) = 2.12 kg m/s
piy = (0.200 kg)(15.0 m/s)(sin -45) = -2.12 kg m/s
Final momentum:
pfx = -(0.200 kg)(40.0 m/s)(cos 30) = -6.93 kg m/s
pfy = (0.200 kg)(40.0 m/s)(sin 30) = 4.00 kg m/s
Ix = Dpx
= (-6.93 -2.12) kg m/s = -9.05 kg m/s
Iy = Dpy
= (4.00 + 2.12) kg m/s = 6.12 kg m/s
I = (-9.05 kg m/s)i +(6.12 kg m/s) j
(b) If the force on the ball increased linearly
for 4.00 ms, held constant for 20.0 ms, then decreased linearly to 0 in
another 4.00 ms, find the maximum force on the ball.
Solution:
The graph of the force looks like the graph below (see next problem) the
only difference is that F is unknown and the base of each triangular section
is 4 ms long and the length of the rectangle is 20 ms long.
I = F(area of triangle) +F(area of rectangle) +F(area
of triangle)
= F (0.004 s) + F(0.020s) + F(0.004 s)
= F(0.028 s)
F = (1/0.028s)[(-9.05 kg m/s)i +(6.12 kg m/s)
j]
F = -323 N i + 219 N j
Problem 2: The force Fx
acting on a 2.0-kg particle varies in time as shown in the figure.
a) Find the impulse of the force.
Solution:
Recall: the Impulse of a force, F, is the area under the curve.
The curve above could be broken up into three geometric figures:
a triangle, a rectangle, and a triangle. The area of a triangle is
base
times height and the area of a rectangle is length times width.
Triangle 1 has base = 2 s and a height of 4 N
Rectangle has a length = 4 N and a width = 1 s
Triangle 2 has base = 2 s and a height of 4 N
I = (1/2)(2 s)(4 N) + (4 N)(1 s) + (1/2)(2 s)(4 N)
= 4 N-s + 4 N-s + 4 N-s
I = 12 N-s
b) Find the final velocity of the particle
if it is initially at rest.
Solution:
I = pf - pi
12 N-s = mvf - mvi where vi = 0
12 N-s = mvf
vf = (12 N-s)/(2.0 kg) = 6.0 m/s
c) Find its final velocity if it is
initially moving along the x-axis with a velocity of -2.0 m/s
Solution:
I = pf - pi
12 N-s = mvf - mvi where vi = -2.0
m/s
12 N-s = (2.0 kg)vf + 4.0 kg m/s
vf = (12 N-s)/(2.0 kg) - 2.0 m/s
= (6.0 - 2.0) m/s
= 4.0 m/s
d) Find the average force exerted on
the particle for the time interval ti = 0 to tf =
5 s
Solution:
I = FDt
F = I/Dt =
(12 N-s)/(5 s) = 2.40 N
COLLISIONS: 1-DIMENSIONAL
Problem 1: A solid metal ball of mass
3.0-kg is moving to the right with a speed of 5.0 m/s. It collides with
another solid metal ball of mass 6.0-kg that is at rest. Find the final
speed of each ball after the collision assuming the collision is elastic.
Solution:
INITIAL SITUATION:
pi = (3.0 kg)(5.0 m/s) = 15 kg m/s
Ki= (1/2)(3.0 kg)(5.0 m/s)2 = 37.5 J
AFTER COLLISION:
pf= (3.0 kg)v1 + (6.0 kg)v2
Kf = (1/2)(3.0 kg)v12 + (1/2)(6.0 kg)v22
Assuming the collision is elastic, then both momentum and kinetic energy
is conserved.
(3.0 kg)v1 + (6.0 kg)v2 = 15 kg m/s
v1 = 5 - 2v2
(1/2)(3.0 kg)v12 + (1/2)(6.0
kg)v22 = 37.5 J
1.50(5 - 2v2)2
+ 3.0v22 = 37.5
37.5 -30v2 + 6.0v22 + 3.0v22
= 37.5
-30v2 + 9.0v22 = 0
v2=0 or v2= 3.33 m/s
v1 = 5 or v1
=
-1.66 m/s
Problem 2: A 75 kg ice skater moving
at 10 m/s crashes into a stationary skater of equal mass. After the
collision, both skaters move as one with a speed of 5.0 m/s. The
average force a human can endure without breaking a bone is 4500 N.
If the impact time was 0.10 s, what was the force of impact on each skater
and did a bone break?
Solution:
Looking just at one skater before and after the collision:
pi = (75 kg)(10 m/s) = 750 kg m/s
pf = (75 kg)(5 m/s) = 375 kg m/s
|F| =| Dp/Dt|
= 3,750 N is the magnitude of the force on each skater as they collide,
it is less than 4500 N therefore there will be NO broken bones.
COLLISIONS: 2-DIMENSIONAL
Problem 1: A 0.30 kg puck, initially
at rest on a horizontal frictionless surface is struck by a 0.20 kg puck
moving initially along the x- axis with a speed of 2.0 m/s. After the collision,
the lighter puck has a speed of 1.0 m/s at an angle of 53.00
with respect to the positive x-axis.
(a) Determine the velocity of the other puck
after the collision?
Solution:
pix = (0.20 kg)(2.0 m/s) = 0.40 kg m/s
piy = 0
After collision:
pfx= (0.20 kg)(1.0 m/s)(cos 53) + (0.30 kg)(vx)
= 0.12 + 0.30 vx
pfy= (0.20 kg)(1.0 m/s)(sin 53) + (0.30 kg)(vy)
= 0.16 + 0.30 vy
Conservation of momentum:
pix = pfx
0.40 kg m/s = 0.12 + 0.30 vx
vx= (0.40 - 0.12)/0.30 = 0.93 m/s
piy = pfy
0 = 0.16 + 0.30 vy
vy= -0.53 m/s
v = [(vx)2 + (vy)2]1/2
= 1.07 m/s
q = tan-1
(vy /vx ) = -30.0o
(b) Find the fraction of kinetic energy lost
in the collision.
Solution:
Ki = (1/2)mv2 = (0.5)(0.20 kg)(2.0 m/s)2
= 0.40 J
Kf = (1/2)mv12 + (1/2)mv22
=
0.10 J + 0.17 J = 0.27 J
Ki - Kf = 0.13
Fraction = (0.13)/(0.40) = 0.325 lost.
CENTER OF MASS:
[Problem 1] A uniform piece
of sheet steel is shaped as in the figure. Compute the x and y coordinates
of the center of mass of the piece.
Solution:
We break up the sheet into three areas: area 1 (A1)= green,
area 2 (A2) =yellow, and area 3 (A3) = blue.
Therefore A = A1+A2 +A3 and the total
mass is M = M1+M2 +M3 . Therefore:
A1 = (10 cm)(30 cm) = 300 cm2
A2 = (20 cm)(10 cm) = 200 cm2
A3 = (10 cm)(10 cm) = 100 cm2
A = 600 cm2
M1= M(A1/A) = M(300 cm2/600 cm2)
= M/2
M2= M(A2/A) = M(200 cm2/600 cm2)
= M/3
M3= M(A3/A) = M(100 cm2/600 cm2)
= M/6
xcm =(1/M)(x1M1 + x2M2
+ x3M3 ) =(1/M)[15 cm(M/2) + 5 cm(M/3) + 15 cm(M/6)]
= 11.7 cm
ycm =(1/M)(y1M1 + y2M2
+ y3M3 ) =(1/M)[5 cm(M/2) + 20 cm(M/3) + 25 cm(M/6)
= 13.3 cm
Problem 2: The table below
lists three masses, their positions, and their velocities.
| Mass (kg) |
Position (m) |
Velocity (m/s) |
| M1
= 3.0 |
R1
= (0.0, 1.0) |
V1
= (0, 2.5) |
| M2
= 2.5 |
R2
= (3.0, 1.0) |
V2
=
(1.5, 0) |
| M3 = 1.5 |
R3 =
(-1.0, -1.0) |
V3 = (2.0,
-2.5) |
a) Calculate the coordinates of the center of
mass, Rcm
Solution:
Rcm= (3.0 kg)(0.0, 1.0) +(2.5 kg)(3.0, 1.0)+ (1.5
kg)(-1.0, -1.0)
3.0 kg + 2.5 kg + 1.5 kg
= (1/7.0 kg)[(0,3.0) + (7.5, 2.5) + (-1.5, -1.5)] kg-m
= (1/7.0)(6.0,4.0)m
= (0.86, 0.57) m
b)Calculate the momentum of the center of
mass, Pcm
Solution:
Pcm=MVcm = (7.0 kg)[(3.0 kg)(0,
2.5) +(2.5 kg)(1.5, 0)+ (1.5 kg)(2.0, -2.5)] m/s
7.0 kg
= [(0,7.5)+(3.8,0)+(3.0,-3.8)]kg m/s
=(6.8, 3.7) kg m/s
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