Q. A particle moves in a circular path with a radius r and a constant speed v. If the speed is doubled, what happens to the angular momentum of the particle?
A.It remains the same
B.It doubles
C.It quadruples
D.It halves
Solution
Angular momentum L = mvr; if v is doubled, L also doubles.
Q. A particle moves in a straight line under the influence of a constant force. If the initial kinetic energy is 100 J and the work done by the force is 50 J, what is the final kinetic energy?
A.50 J
B.100 J
C.150 J
D.200 J
Solution
Final kinetic energy = Initial kinetic energy + Work done = 100 J + 50 J = 150 J.
Q. A particle moves in a straight line with a velocity v. What is its angular momentum about a point P located at a distance d from the line of motion?
A.mv
B.mvd
C.mdv
D.0
Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.
Q. A particle of mass m is moving in a circular path of radius r with a constant speed v. What is the angular momentum of the particle about the center of the circle?
A.mv
B.mvr
C.mr^2
D.mv^2
Solution
Angular momentum L = mvr, where v is the linear speed and r is the radius.
Q. A particle with charge q moves with velocity v in a magnetic field B. What is the expression for the magnetic force acting on the particle?
A.F = qvB
B.F = qvB sin(θ)
C.F = qB
D.F = qvB cos(θ)
Solution
The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity vector and the magnetic field vector.
Q. A pendulum of length 2 m swings from a height of 1 m. What is the speed at the lowest point of the swing? (g = 9.8 m/s²)
A.4.4 m/s
B.3.1 m/s
C.2.8 m/s
D.5.0 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.
Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
A.2 m/s
B.4 m/s
C.6 m/s
D.8 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.
Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.
Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased to four times its original length, what will be the new period?
A.1 s
B.2 s
C.4 s
D.√4 s
Solution
The period of a pendulum is given by T = 2π√(L/g). If L is increased to 4L, T becomes 2π√(4L/g) = 2T = 2 seconds.
