Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?
A.10 m/s
B.5 m/s
C.20 m/s
D.15 m/s
Solution
Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).
Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?
A.10 m/s
B.15 m/s
C.5 m/s
D.20 m/s
Solution
Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?
A.10 m/s
B.15 m/s
C.20 m/s
D.5 m/s
Solution
Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?
A.10 m/s
B.15 m/s
C.40 m/s
D.25 m/s
Solution
Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.
Q. A cyclist is moving at a speed of 15 km/h. If a car is moving in the same direction at 30 km/h, what is the relative speed of the car with respect to the cyclist?
A.15 km/h
B.30 km/h
C.45 km/h
D.0 km/h
Solution
Relative speed = speed of car - speed of cyclist = 30 - 15 = 15 km/h.
Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?
A.180 N
B.120 N
C.90 N
D.60 N
Solution
Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.
Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?
Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?
A.50 N
B.75 N
C.100 N
D.125 N
Solution
Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.
Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?
A.0.25
B.0.5
C.0.75
D.1
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.
Q. A cyclist is pedaling at a constant speed and exerts a power of 200 W. If the cyclist increases their power output to 400 W, what happens to their speed assuming no other forces act?
A.Speed remains the same
B.Speed doubles
C.Speed increases by 41%
D.Speed increases by 100%
Solution
Power is proportional to the cube of the speed in cycling. If power doubles, speed increases by a factor of (2)^(1/3) which is approximately 1.26, or about a 41% increase.
Q. A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
A.√(2gh)
B.√(3gh)
C.√(4gh)
D.√(5gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = (1/2)mR^2 and ω = v/R. Solving gives v = √(3gh).
Q. A cylinder rolls down a hill. If it has a radius R and rolls without slipping, what is the relationship between its linear velocity v and its angular velocity ω?
A.v = Rω
B.v = 2Rω
C.v = ω/R
D.v = R^2ω
Solution
For rolling without slipping, the relationship is v = Rω.
Q. A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
A.√(gh)
B.√(2gh)
C.√(3gh)
D.√(4gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = 1/2 mr^2, leading to v = √(2gh).
Q. A cylindrical conductor of radius R carries a uniform charge per unit length λ. What is the electric field at a distance r from the axis of the cylinder (r > R)?
A.0
B.λ/(2πε₀r)
C.λ/(2πε₀R)
D.λ/(4πε₀r²)
Solution
For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).
