Q. A ball is thrown horizontally from the top of a cliff 45 m high. How far from the base of the cliff will it land if it is thrown with a speed of 10 m/s?
A.20 m
B.30 m
C.40 m
D.50 m
Solution
Time to fall = sqrt(2h/g) = sqrt(2*45/9.8) ≈ 3.03 s. Horizontal distance = speed * time = 10 * 3.03 ≈ 30.3 m, approximately 30 m.
Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and initial horizontal speed = 20 m/s)
A.40 m
B.60 m
C.80 m
D.100 m
Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 * 4 = 80 m.
Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and horizontal speed = 20 m/s)
A.20 m
B.40 m
C.60 m
D.80 m
Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 m/s * 4 s = 80 m.
Q. A ball is thrown horizontally from the top of a cliff with a speed of 15 m/s. If the cliff is 45 m high, how far from the base of the cliff will the ball land?
A.30 m
B.45 m
C.60 m
D.75 m
Solution
Time to fall = √(2h/g) = √(2*45/10) = 3 s. Horizontal distance = speed * time = 15 m/s * 3 s = 45 m.
Q. A ball is thrown vertically upwards with a speed of 30 m/s. How high will it rise before coming to rest momentarily?
A.45 m
B.30 m
C.60 m
D.75 m
Solution
Using the equation v² = u² + 2as, where v = 0, u = 30 m/s, and a = -9.8 m/s² (acceleration due to gravity), we have 0 = (30)² + 2*(-9.8)*s. Solving gives s = 45.92 m, approximately 45 m.
Q. A ball is thrown vertically upwards with a speed of 30 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.45.9 m
B.46.0 m
C.46.1 m
D.46.2 m
Solution
Using conservation of energy, initial kinetic energy = potential energy at maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (30)²/(2 * 9.8) = 45.9 m.
Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to remain in circular motion?
A.Tension must be zero
B.Tension must be maximum
C.Weight must be zero
D.Centripetal force must be zero
Solution
At the highest point, the tension can be zero if the centripetal force is provided entirely by the weight.
Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain circular motion?
A.Tension = 0
B.Tension = mg
C.Tension > mg
D.Tension < mg
Solution
At the highest point, the centripetal force is provided by the weight, so T + mg = mv²/r, T = 0.
Q. A ball is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 2 N and the weight of the ball is 3 N. What is the speed of the ball at the highest point if the radius of the circle is 1 m?
A.1 m/s
B.2 m/s
C.3 m/s
D.4 m/s
Solution
At the highest point, T + mg = mv²/r. 2 N + 3 N = mv²/1. v² = 5, v = √5 ≈ 2.24 m/s.
Q. A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
A.(2/5)m(10^2)
B.(1/2)m(10^2)
C.(1/3)m(10^2)
D.(5/2)m(10^2)
Solution
Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).
Q. A ball rolls without slipping on a flat surface. If the ball's radius is doubled while keeping its mass constant, how does its moment of inertia change?
A.Increases by a factor of 2
B.Increases by a factor of 4
C.Increases by a factor of 8
D.Remains the same
Solution
The moment of inertia of a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.
Q. A beam of light enters a prism with an angle of incidence of 45 degrees. If the refractive index of the prism is 1.5, what is the angle of refraction inside the prism?
Q. A beam of light in glass (n=1.5) strikes the glass-air interface at an angle of 60°. Will total internal reflection occur?
A.Yes
B.No
C.Only if the angle is increased
D.Only if the angle is decreased
Solution
To determine if total internal reflection occurs, we first find the critical angle using sin(θc) = 1/n = 1/1.5, which gives θc ≈ 41.8°. Since 60° > 41.8°, total internal reflection will not occur.
