EGR232 Dynamics: Homework Set 18
Problem 13:146
Fall 2013
At an intersection car B was traveling south and car A was traveling 30 degrees north of east
when they slammed into each other. Upon investigation it was found that after the crash the two
cars got tuck and skidded off at an angle of 10 degrees north of east. Each driver claimed that he
was going at the speed limit of 50 km/h and that he tried to slow down but couldn’t avoid the
crash because the other driver was going a lot faster. Knowing that the masses of cars A and B
were 1500 kg and 1200 kg, respectively, determine a) which car was going faster, b) the speed
of the faster of the two cars if the slower car was traveling at the speed limit.
solution:
Either car A or car B was traveling at 50 km/hr.
mA = 1500 kg and mB = 1200 kg.
Apply conservation of momentum to
a plastic collision:
 mvi   mv f
Assume Vehicle A is traveling 50 km/hr
then:
vAi  50cos30o iˆ  50sin 30o ˆj  43.3iˆ  25 ˆj
vBi  0iˆ  vB ˆj
v f  vAf  vBf  v f cos10o iˆ  v f sin10o ˆj
then the momentum conservation equation is:
mAvAi  mAvBi  (mA  mB )v f
(1500kg )(43.3iˆ  25 ˆj )km / hr  (1200kg )(0iˆ  vBi ˆj )  (1500  1200)kg (v f cos10o iˆ  v f sin10o ˆj )
(64950iˆ  37500 ˆj )  (1200vBi ˆj)  (2659v f iˆ  469v f ˆj)
Separating out separate i and j equations:
i:
64950  2659v f
v f  24.4km / hr

j:
37500  1200vBi  469v f

vBi 
37500  469v f
1200

37500  469(24.4)
 21.7km / hr
1200
since vehicle B comes in at 21.7 mi/hr < 50 mi/hr…this means that the driver of vehicle A lied.
We need to rerun the problem with vB = 50 mph.
vAi  vAi cos30o iˆ  vAi sin 30o ˆj
vBi  0iˆ  50 ˆj
v f  vAf  vBf  v f cos10o iˆ  v f sin10o ˆj
then the momentum conservation equation is:
mAvAi  mAvBi  (mA  mB )v f
(1500kg )(vAi cos30o iˆ  vAi sin 30o ˆj )km / hr  (1200kg )(0iˆ  50 ˆj )  (1500  1200)kg (v f cos10o iˆ  v f sin10o ˆj )
(1299vAiiˆ  750 Ai ˆj )  (60000 ˆj )  (2659v f iˆ  469v f ˆj )
Separating out separate i and j equations:
i:
v f  0.4885v Ai
1299v Ai  2659v f

j:
750v Ai  60000  469v f

750vAi  60000  469(0.4885vAi )
(750  229)vAi  60000
60000
vAi 
 115.1km / hr
750  229
then
v f  0.4885vAi  0.4885(115.1)  56.3km / hr
EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13.159
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly
different style but of the same weight, has been pushed by dockworkers and hits car B with a
velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5
between A and B, determine the velocity of each car after all collision have taken place.
_______________________________________________________________________
Solution:
_______________________________________________________________________
Solution:
10 ft/s
6 ft/s
1.5 lb
0.9 lb
Apply conservation of momentum and restitution equation:
mAvAi  mB vBi  mAv Af  mB vBf
where
mA = 1.5/g
mB = 0.9/g
vAi = 10
1.5(10)  (0.9)(6)  1.5vAf  0.9vBf
20.4  1.5vAf  0.9vBf
e
vBi = 6
vBf  v Af
v Ai  vBi
e = 0.75
0.75 
vBf  vAf
10  6
3  vBf  v Af
vBf  vAf  3
20.4  1.5vAf  0.9(vAf  3)
20.4  1.5v Af  0.9v Af  2.7
2.4v Af  17.7
and

v Af  7.375 ft / s
vBf  7.375  3  10.375 ft / s
Kinetic Energy:
Before
1
1
KEi  mAvAi 2  mB vBi 2
2
2
After:
1
1
KE f  mAvAf 2  mB vBf 2
2
2
1 1.5lb
1 0.9lb
(10 ft / s) 2 
(6 ft / s)2  2.329  0.503  2.832 ft  lb
2
2
2 32.2 ft / s
2 32.2 ft / s
1 1.5lb
0.9lb
2 1
KEi 
(7.375
ft
/
s
)

(10.375 ft / s) 2  1.267  1.505  2.771 ft  lb
2
2
2 32.2 ft / s
2 32.2 ft / s
KEi 
∆KE = 2.832 - 2.771 = 0.0607 ft-lb
EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13.165
A 600 g ball is moving with a velocity of magnitude 6 m/s when it is hit as shown by a 1 kg ball B which
has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and assuming no
friction, determine the velocity of each ball after impact.
------------------------------------------------------Solution:
Apply conservation of momentum
and restitution equation:
mAvAi  mB vBi  mAv Af  mB vBf
e
and
where
vBf _ n  v Af _ n
v Ai _ n  vBi _ n
mA = 0.6kg
vAi = 10 m/s
mB =1kg
vBi =4m/s
e = 0.8
so:
mAvAi  mB vBi  mAv Af  mB vBf
(0.6)(6cos40o eˆn  6sin 40o eˆt )  (1.0)(4eˆn  0eˆt )  (0.6)(vA f _ neˆn  vA f _ t eˆt )  (1.0)(vB f _ neˆn  vB f _ t eˆt )
Along Normal Direction:
Along tangent direction:
2.76  4  0.6vA f _ n  1.0vB f _ n
for body A:
3.6sin 40o  0.6vA f _ t

vA f _ t  3.857
for body B:
vB f _ t  0
Restitution equation:
e
vBf _ n  v Af _ n
v Ai _ n  vBi _ n

0.8 
vBf _ n  vAf _ n
6cos 40o  4

0.5962  vBf _ n  vAf _ n
vBf _ n  vAf _ n  0.5962
Solving:
2.76  4  0.6vA f _ n  1.0vB f _ n
and
vBf _ n  vAf _ n  0.5962
1.24  0.6vA f _ n  1.0(vAf _ n  0.5962)
1.836  1.6vA f _ n
vA f _ n  1.148

and
vBf _ n  1.148  0.5962  0.5518
vAf  vA f _ n eˆn  vA f _ t eˆt
 1.148eˆn  3.857eˆt m / s
vBf  vB f _ n eˆn  vB f _ t eˆt
 0.5518eˆn  0eˆt m / s
EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13:170
A girl throws a ball at an inclined wall from a height of 1.2 m, hitting the wall at A with a horizontal
velocity v0 of magnitude 15 m/s. Knowing that the coefficient of restitution between the ball and the wall
is 0.9 and neglecting friction, determine the distance d from the foot of the wall to the point B where the
ball will hit the ground after bouncing off the wall.
----------------------------------------------------------------------------------------------------------Solution:
vo = 15 m/s
e = 0.9
For wall collision:
vo_n= 15 cos 30o = 12.99
vo_t =15sin 30o = 7.5
Restitution applied at wall:
e
vBf _ n  v Af _ n
v Ai _ n  vBi _ n
0.9 
0  vo '_ n
vo _ n  0

vo '_ n  0.9vo _ n  (0.9)12.99  11.69
Momentum is conserved along the tangent direction:
vo’_t = vo_t =15sin 30o = 7.5
vo '  11.692  7.52  13.89
and angle
vo’_n =11.69
θ
vo’_t =7.5
o
30
7.5
)  32.68o
11.69
  tan 1 (
1.2 m
therefore the velocity off the wall is 13.89 m/s up at 62.68
from the horizontal.
Projectile motion:
x0 = 0.693 y0 = 1.2 vox = -13.89cos(62.68o) voy = 13.89 sin 62.68o
= -6.375
= 12.34
Horizontal direction:
Vertical direction:
x f  xo  vx t
d  0.693  6.37t
then
0.693m
ay = -9.81
1
y f  yo  v yot  a y t 2
2
0  1.2  12.34t  0.5(9.81)t 2
0  1.2  12.34t  4.9t 2  t  0.0938 or 2.612
d  6.37(2.612)  0.693  15.95m