MAE 30B
Additional Review Problems
1. A tennis ball when served horizontally 7.5 ft above the ground strikes the smooth ground at B 20 ft away.
Determine the initial velocity vA of the ball and the velocity vB (and ) of the ball just after it strikes the court at
B. The coefficient of restitution between the ball and the ground is e = 0.7.
v
Step 1) Kinematics to find vB0
2
+  Sy = S0y + v0y t + 0.5 g t
7.5 = 0 + 0 + 0.5 (32.2) t
2
t = 0.6825 s
+  vB0y = v0y + g t = 0 + 32.2 (0.6825) = 21.98 ft/s
As for vB0x , it will be the same as vA
+ → d = vA t  vA = 20 / 0.6825 = 29.3 ft/s
Step 2) Coefficient of restitution in y-direction (line of impact), to find vBy
𝑒=
vB𝑦
vB0𝑦

0.7 =
vB𝑦
21.98
vBy = 15.38 ft/s 
Step 3) conservation of momentum in x-direction (perpendicular to line of impact), to find vBx
Apply conservation of momentum to the system in the x-dir :
+→ m (vB0x) = m (vBx)  vBx = vB0x = vA = 29.3 ft/s →
Velocity and angle of the ball just after it strikes the court :
vB = (29.3 i + 15.4 j) ft/s
-1
 = tan (15.4 / 29.3) = 27.7°
MAE 30B
Additional Review Problems
2- Determine the maximum speed at which the car with mass m can pass over the top point A of the vertical
curved road and still maintain contact with the road. If the car maintains this speed, what is the normal
reaction the road exerts on the car when it passes the lowest point B on the road?
(solution should be done in terms of the known parameters m (mass), g (gravitational acceleration) and r
(road curvature radius).
Step 1) Free-Body Diagram:
The free-body diagram of the car at the top and bottom of the vertical curved road are
shown in Figs. (a) and (b), respectively. Here, an must be directed towards the center of
curvature of the vertical curved road (positive n axis).
Step 2 and 3) Equations of Motion and Kinematics:
When the car is on top of the vertical curved road, “ maximum speed and still maintain
contact with the road” means that car tires are barely in contact with road OR are
about to lose contact with the road surface.
Being on the verge of losing contact with road surface in terms of kinetics means that N
(surface normal force) will go to zero → N=0, (about to lose contact with road surface).
Therefore, at point A with maximum speed, N=0 and we only have the weight force in
normal coordinate.
Can you explain how these situations will influence the experience of the car passengers?
MAE 30B
Additional Review Problems
1
3. A small box of mass m is given a speed of 𝑣 = ට 𝑔𝑟 at the top of the smooth half cylinder. Determine the
4
angle at which the box leaves the cylinder. (g is gravitational acceleration).
Step 1 & 2 ) FBD & Work – Energy to find the velocity at angle of departure (θ)
Assuming that the box leaves the circular path at the unknown angle θ, we have the following FBD
at angle θ:
Step 3) Equation of Motion and N = 0 condition for box to leave the circular track.
MAE 30B
Additional Review Problems
4- A 60-kg skateboarder in coasts down the circular track. If she starts from rest when θ=0°, determine the
magnitude of the normal reaction the track exerts on her when θ=60°. Neglect his size for the calculation (i.e.
approach the problem with a particle kinetics).
Solution
Step 1) Free-Body Diagram. The free-body diagram of the skateboarder when he is at an arbitrary position θ is drawn
below.
At θ=60° there are three unknowns, Ns, at, and an (or v , as an and v are connected)
Step 2) equations of motion (2nd Law)
Step 3) Finding v2,
using work- energy principle T1 + U1-2 = T2,
Knowing: T1 = 0 (starting from rest) and T2 = 0.5mv2
U1-2 = mg . h (where h = 4m sin 60 0 downward)
MAE 30B
We will get v2 = 69.97 (m/s)2
Additional Review Problems
We can also solve for v2 just by using kinematics:
Since at is expressed in terms of θ (and therefore is NOT constant), the general equation vdv=atds must be used to
determine the speed of the skateboarder when θ=60°. Using the geometric relation s=rθ, where ds=rdθ=(4m)dθ, and
remembering the initial condition v=0 at θ=0°, we have,
Step 4) Finding the Normal fore on the skater
Substituting v2 into equation of motion (and θ = 600 ) we will get
Ns=1529.23 N=1.53kN
MAE 30B
Additional Review Problems
5- The pilot of the airplane performs a vertical loop which in part follows the path of a cardioid, r = 200(1 +
cosθ) m, where θ is in radians. If the speed at A is a constant vp = 85 m/s, determine the vertical reaction the
seat of the plane exerts on the pilot when the plane is at A. The pilot has a mass of 80 kg.
Solution:
Step 1) Free-Body Diagram
(remember that the speed of the plane at A is a constant, vp = 85 m/s, which is in θ direction)
Step 2) Equations of Motion (Second Law)
Referring to FBD of the pilot in step 1 we will have:
80(9.81) – N = 80 (ar)
Step 3) Kinematics
To determine the time derivatives necessary to calculate the acceleration ar, we have to take the first and second timederivatives of r = 200(1 + cosθ). Remember that we have to use chain-rule here to find the time derivatives
MAE 30B
Additional Review Problems
Substituting ar in equation of motion we will get
(Note: if we get a negative value for N (vertical reaction of the seat on the pilot), it means that the pilot will lose contact
with the seat and the determined force N is the vertical force that the seat belt exerts on the pilot to hold him/her to the
seat.)