Algebra Based Physics PHY210 Midterm Exam Fall 2015

Home Exam Part 1

A ball is dropped from rest from a height H above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released

A ball is dropped from rest from a height H above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the initial speed of the second ball if the two balls are to meet at a height h above the ground.

A mass m1 is at rest on a frictionless horizontal surface and connected to a wall by an ideal spring with a spring constant k

A mass m₁ is at rest on a frictionless horizontal surface and connected to a wall by an ideal spring with a spring constant k as shown in the figure. A second mass m₂ is moving to the left at speed v₀. The two masses then collide elastically. What is the maximum compression of the spring?

 

A block with mass m is being pushed by a constant force F that makes an angle of θ with the horizontal

A block with mass m is being pushed by a constant force F that makes an angle of θ with the horizontal as shown below. The block is moving with constant velocity on a level surface. The coefficient of kinetic friction between the block and the surface is µk. Which one of the following equations is correct for the magnitude of F?
 

Solution

Let |F| is the magnitude of the force F

Given Data: m, F, θ, a=0, µ_k

                      |F| = |F(m, θ, µ_k)|=?

F_NET=0 because a=0

X direction:   |F|·cos θ + f = F_{NET, X} = 0

Y direction: N – mg –  |F|·sin θ = F_{NET, Y} = 0,

where N is the normal force,
             g is the gravity acceleration

f= µ_k N

|F|·cos θ + µ_k (mg- |F|·sin θ) = 0

|F|·cos θ + µ_k mg - µ_k |F|·sin θ = 0

|F|·cos θ  - µ_k |F|·sin θ = µ_k mg

|F|·( cos θ  - µ_k sin θ) = µ_k mg

|F| = µ_k mg/( cos θ  - µ_k sin θ)

Algebra-Oriented Solution

Problem for the homework 3, chapter 3.

You drive your car 5.0 km due East at 35km/hr, and suddenly realize that you forgot your wallet.  So, you return home, driving West at 40 km/hr, and upon arrival spend 10 minutes looking for it, and finally go back on the road for a total of 57.0 km due East.  If your average speed was 40 km/hr for the whole journey, what was your average speed on your last leg?

a)46 km/hr
b)49 km/hr
c)35 km/hr
d) 40 km/hr

2) Algebra-Oriented Solution

To East:
x₁=5km
V₁=35 km/h
t₁=x₁/V₁
Back to home, driving West:
V₂=40 km/h
t₂=x₁/V₂
Looking time t₃=10 minutes = 1/6 h
Distance of the last leg due East x₄ = 57 km
Distance Total (The total length of the total road pass) d = 2x₁+x₄
V_{average total}=V_A = 40km/h
V₄=?

Total time = (Distance Total)/ V_{average total} = d/V_A= (2x₁+x₄)/ V_A
Total time= t1+t2+t3+t4 = x₁/V₁ + x₁/V₂ + t₃ + x₄/V₄
(2x₁+x₄)/ V_A = x₁/V₁ + x₁/V₂ + t₃ + x₄/V₄
x₄/V₄  = (2x₁+x₄)/ V_A - x₁/V₁ - x₁/V₂ - t₃
V₄/ x₄  = 1/(  (2x₁+x₄)/ V_A - x₁/V₁ - x₁/V₂ - t₃   )
V₄  = x₄ /(  (2x₁+x₄)/ V_A - x₁/V₁ - x₁/V₂ - t₃   )

Mathematical Expression for Calculation:

x₁=5km, V₁=35 km/h, V₂=40 km/h, t₃=10 minutes, x₄ = 57 km, V_A = 40km/h

V₄  = x₄ /(  (2x₁+x₄)/ V_A - x₁/V₁ - x₁/V₂ - t₃   )

Expression to the Google calculator:
57km /(  (2* 5km+57km)/ 40 km/h – 5km/35km/h – 5km/40km/h - 10minutes   ) in km/h

Google result: 45.950096 km / hour

Answer: a) 46 km/h

You drive your car 5.0 km due East at 35km/hr ...

Problem for the homework 3, chapter 3.

You drive your car 5.0 km due East at 35km/hr, and suddenly realize that you forgot your wallet.  So, you return home, driving West at 40 km/hr, and upon arrival spend 10 minutes looking for it, and finally go back on the road for a total of 57.0 km due East.  If your average speed was 40 km/hr for the whole journey, what was your average speed on your last leg?

a)46 km/hr

b)49 km/hr

c)35 km/hr

d) 40 km/hr

Solution

To East:

x₁=5km

V₁=35 km/h

t₁=x₁/v₁=(5km)/( 35 km/h)= 1/7 h

Back to home, driving West:

V₂=40 km/h

t₂=x₁/V₂=(5 km)/(40 km/h)=1/8 h

Looking time t₃=10 minutes = 1/6 h

Distance of the last leg due Est = 57 km

Distance Total (The total length of the total pass) = 5km+5km+57km=67km

V_{average total}=40km/h

Total time = (Distance Total)/ V_{average total} = (67km)/(40 km/h)

The time on the last leg = Total time – t₁-t₂-t₃ = (67km)/(40 km/h) – 1/7 h – 1/8 h – 1/6 h

The average speed on the last leg = (Distance of the last leg)/( The time on the last leg) = 57km/ ((67km)/(40 km/h) – 1/7 h – 1/8 h – 1/6 h)

Expression to the Google calculator:
57 km/ ((67 km)/(40 km/hour) – 1/7 hour – 1/8 hour – 1/6 hour) in km/hour

Google result: 45.950096 km / hour

Answer: a) 46 km/h