## Mini Exam 3

**Golf Ball**

A golf ball with a mass of 47.4 g can be blasted from rest to a speed of 77.80 m/s during impact with a club head. Assume that the impact lasts only about 0.910 ms. Calculate the average force applied.

**4050 N**(in N)

**A**: 0.554

**B**: 1.49

**C**: 3320

**D**: 3360

**E**: 3570

**F**: 4050

**G**: 4780

**H**: 6240

**Approach:**

The force that is impacted on the ball is the change in momentum divided by the impact time. (The word "average" is used since in hitting an object the force applied is not constant. It goes from zero (at instant of impact) to a peak and back to zero.)

**Solution:**

Given quantities:

- mass of ball, m = 47.4 g = 0.0474 kg
- speed of ball, v = 77.8 m/s
- impact time, Δt = 0.91 ms = 0.91×10-3 s

Now, Δp = pf - pi = mv. The initial momentum of the golf ball is zero as the ball is at rest. Putting in the numbers, Fave = 4050 N.

Head-On Collision

Two particles with masses m1 = 2.11 kg and m2 = 3.09 kg collide head on, such that after the collision both are at rest. Initially, m1 had velocity v1 = -2.79 m/s. What was the velocity of m2 before the collision?

**1.91 m/s**(in m/s)

**A**: -4.09

**B**: 0.769

**C**: 1.91

**D**: 4.09

**E**: 4.31

**F**: 4.65

**G**: 5.79

**H**: 5.98

**Approach:**

Whether the collision is elastic or inelastic, momentum is conserved.

**Solution:**

Since both masses are at rest after collision, the final momentum is zero. Thus the initial momentum, . Putting in the numbers for m1, m2 and v1 gives v2 = 1.91 m/s

Conceptual - Rotation For each of the following statements select the correct option.

• Two masses are rotating about a fixed axis of rotation. When the masses are brought closer to the axis, the moment of inertia decreases.

A: increases

B: decreases

C: stays the same

• You are sitting on a rotating chair holding a two-kg mass each on your outstretched arms. When you drop the masses, your angular velocity stays the same.

A: increases

B: decreases

C: stays the same

• Two equal masses, m1 and m2, are on a turning wheel at a distance R from the axis of rotation. When the masses are moved to 2R the angular velocity of masses decreases.

A: increases

B: decreases

C: stays the same

• You are standing at the center of a large freely rotating platform. As you walk toward the edge of the platform, the rate of rotation decreases.

A: increases

B: decreases

C: stays the same

• Two equal masses, m1 and m2, are on a turning wheel at a distance R from the axis of rotation. When the masses are moved to R the linear velocity of masses increases.

A: increases

B: decreases

C: stays the same

**Cyclist**

A cyclist starts from rest and pedals such that the wheels of his bike have a constant angular acceleration. After 17 s, the wheels have made 195 rev. What is the angular acceleration of the wheels?

**8.48 rad/s^2**(in rad/s^2)

**A**: 2.12

**B**: 4.24

**C**: 6.10

**D**: 8.48

**E**: 9.33

**F**: 12.9

**G**: 14.9

**H**: 7130

**Approach**

Since the wheels have a constant angular acceleration, the kinematic equations for rotation can be used.

**Solution**

Given quantities:

- time, t = 17 s
- angular displacement, θ = 195 rev = 2×π ×195 rad

Since the wheels start from rest, ω0 = 0. Putting in the numbers, the angular acceleration, α = 8.48 rad/s2.

If the radius of the wheels is 41 cm, what is the distance traveled after 17 s?

**502 m**(in m)

**A**: 80.0

**B**: 160.

**C**: 251

**D**: 362

**E**: 502

**F**: 593

**G**: 784

**H**: 1450

**Approach**

Since the wheels have a constant angular acceleration, the kinematic equations for rotation can be used.

**Solution**

Given quantities:

- time, t = 17 s
- angular displacement, θ = 195 rev = 2×π ×195 rad

Two-Block Collision

Two blocks are on a straight track without friction. The first one is moving with speed v0 in the positive x direction and collides with the second one which is at rest (see Figure).

• If both blocks get stuck together, the sum of kinetic energies of the blocks after the collision will be less than .

A: greater than

B: less than

C: equal to

D: cannot tell

• If both blocks get stuck together, the sum of momenta of the two blocks after the collision will be equal to m1v0.

A: greater than

B: less than

C: equal to

D: cannot tell

• If the speed of the second block after an elastic collision is v0, the mass of block 1 is equal to the mass of block 2.

A: greater than

B: less than

C: equal to

D: cannot tell

• The speed of the first block after an elastic collision with the second block will be less than v0.

A: greater than

B: less than

C: equal to

D: cannot tell

• In a totally inelastic collision, the speed of the two blocks after the collision is less than v0.

A: greater than

B: less than

C: equal to

D: cannot tell

Car SUV Accident

A 971-kg sports car collides into the rear end of a 1983-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two vehicles skid forward 13.1 m before stopping. The police officer, knowing that the coefficient of kinetic friction between the tires and the road is 0.9, calculates the speed of the sports car just before impact. What was that speed?

**46.2 m/s**(in m/s)

**A**: 7.44

**B**: 23.1

**C**: 31.0

**D**: 36.1

**E**: 46.2

**F**: 66.6

**G**: 80.5

**H**: 86.9

**Approach:**

This problem requires two key concepts, (1) work-energy theorem, and (2) conservation of linear momentum.

**Solution:**

Given quantities:

- Sports car mass, m1 = 971 kg
- SUV mass, m2 = 1983 kg
- Distance, d = 13.1 m
- Coefficient of kinetic friction, μk = 0.9.

we have

Note: The final kinetic energy = 0 since the two vehicles came to a stop. Also, ΔPE = 0 since the road is assumed to be level Rewriting the above expression

This is the speed of the vehicles just after the collision. Now we have to apply the conservation of momentum to find the speed of the sport's car just before the collision.

Putting in the numbers, v1 = 46.2 m/s.

Rolling Down an Incline

A bowling ball of mass 2.5 kg and radius 9.3 cm rolls from rest and without slipping down an inline. What is the speed of the ball at the bottom of the incline. The vertical height of the incline is 21.7 cm.

**1.74 m/s**(in m/s)

**A**: 0.907

**B**: 1.40

**C**: 1.74

**D**: 2.06

**E**: 2.30

**F**: 2.76

**G**: 3.26

**H**: 5.62

**Approach**

The initial potential energy of the ball when it is at the top of the incline is converted into kinetic energy as it rolls down the incline. Since the ball is rolling, it has both translational and rotational kinetic energy. Using the conservation of energy, the total initial mechanical energy is equal to the total final mechanical energy.

**Solution**

Given quantities:

- mass, m = 2.5 kg
- radius, r = 9.3 cm
- height, H = 21.7 cm = 21.7/100 m

Solving gives

Putting in the numbers, v = 1.74 m/s.