physics : class11 Range on an Inclined Plane : Projectile Motion : Kinematics MCQ Questions and Answers
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Range on an Inclined Plane
MCQ questions and answers.
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MCQ Questions
1. A projectile is launched with speed u at an angle θ above the horizontal toward an inclined plane that rises at an angle β above the horizontal. If θ > β, what is its range measured along the inclined plane?
A.
2u² cosθ sin(θ − β) / (g cos²β)
B.
2u² sinθ cos(θ − β) / (g cos²β)
C.
2u² cos²θ tan(θ − β) / (g cosβ)
D.
u² sin(2θ − β) / (g cosβ)
Correct Answer:
2u² cosθ sin(θ − β) / (g cos²β)
Explanation:
The projectile trajectory is y = x tanθ − gx²/(2u²cos²θ), while the inclined plane is y = x tanβ. The nonzero intersection gives x = 2u²cosθ sin(θ − β)/(g cosβ). Since the distance along the plane is R = x/cosβ, R = 2u²cosθ sin(θ − β)/(g cos²β).
2. For a projectile launched from the foot of an inclined plane rising at angle β, at what launch angle θ above the horizontal is the range along the plane maximum?
A.
45° − β/2
B.
45° + β/2
C.
90° − β/2
D.
45° + β
Correct Answer:
45° + β/2
Explanation:
The range is proportional to cosθ sin(θ − β) = [sin(2θ − β) − sinβ]/2. This is maximum when sin(2θ − β) = 1, giving 2θ − β = 90°, or θ = 45° + β/2.
3. A projectile is fired with speed 20 m/s onto an inclined plane making 30° with the horizontal. Taking g = 10 m/s², what is the maximum possible range measured along the plane?
A.
20.00 m
B.
23.09 m
C.
26.67 m
D.
30.00 m
Correct Answer:
26.67 m
Explanation:
The maximum range along an incline is Rmax = u²/[g(1 + sinβ)]. Substituting u = 20 m/s and β = 30° gives Rmax = 400/[10(1 + 0.5)] = 26.67 m.
4. For an inclined plane making 30° with the horizontal, two different launch angles can produce the same range along the plane. If one launch angle is 50°, what is the other angle?
A.
60°
B.
65°
C.
70°
D.
80°
Correct Answer:
70°
Explanation:
For a fixed inclined plane and speed, equal ranges occur for angles satisfying θ₁ + θ₂ = 90° + β. Thus θ₂ = 90° + 30° − 50° = 70°.
5. A projectile is launched at speed u and angle θ above the horizontal toward an inclined plane at angle β. What is the horizontal coordinate of the nonzero point where the projectile meets the plane?
A.
2u² cosθ sin(θ − β)/(g cosβ)
B.
2u² cosθ sin(θ − β)/(g cos²β)
C.
2u² sinθ cos(θ − β)/(g cosβ)
D.
u² sin(2θ − β)/(g cos²β)
Correct Answer:
2u² cosθ sin(θ − β)/(g cosβ)
Explanation:
Equating the trajectory y = x tanθ − gx²/(2u²cos²θ) to the plane y = x tanβ gives x = 2u²cos²θ(tanθ − tanβ)/g. Simplifying yields x = 2u²cosθ sin(θ − β)/(g cosβ).
6. A projectile is fired with speed u at angle θ above the horizontal toward an inclined plane making angle β with the horizontal. What is the time taken to strike the plane, excluding the instant of projection?
A.
2u sin(θ − β)/(g cosβ)
B.
2u cos(θ − β)/(g cosβ)
C.
2u sin(θ − β)/(g cos²β)
D.
2u sinθ/(g cosβ)
Correct Answer:
2u sin(θ − β)/(g cosβ)
Explanation:
At time t, x = ucosθ t and y = usinθ t − gt²/2. Using y = x tanβ and taking the nonzero solution gives t = 2u(sinθ cosβ − cosθ sinβ)/(g cosβ) = 2u sin(θ − β)/(g cosβ).
7. A projectile is launched with speed 30 m/s toward a plane inclined at 30° to the horizontal. Taking g = 10 m/s², what is the maximum possible range along the plane?
A.
45.0 m
B.
50.0 m
C.
52.0 m
D.
60.0 m
Correct Answer:
60.0 m
Explanation:
Using Rmax = u²/[g(1 + sinβ)], Rmax = 900/[10(1 + 1/2)] = 900/15 = 60.0 m.
8. A projectile is fired at 60° above the horizontal with speed 10 m/s toward an inclined plane rising at 30°. Taking g = 10 m/s², what is its range measured along the plane?
A.
5.00 m
B.
6.67 m
C.
8.66 m
D.
10.00 m
Correct Answer:
6.67 m
Explanation:
The range along the plane is R = 2u²cosθ sin(θ − β)/(g cos²β). Substituting u = 10, θ = 60°, β = 30°, and g = 10 gives R = 2(100)(1/2)(1/2)/[10(3/4)] = 6.67 m.
9. The maximum range of a projectile along a plane inclined at angle β is measured to be Rmax. What launch speed is required to achieve this maximum range?
A.
√[gRmax(1 + sinβ)]
B.
√[gRmax(1 − sinβ)]
C.
√[gRmax cos²β]
D.
√[gRmax/(1 + sinβ)]
Correct Answer:
√[gRmax(1 + sinβ)]
Explanation:
Since Rmax = u²/[g(1 + sinβ)], rearranging gives u² = gRmax(1 + sinβ), and hence u = √[gRmax(1 + sinβ)].
10. For the same launch speed and the same launch angle θ, a projectile is fired onto two inclined planes having inclinations β₁ and β₂, with θ greater than both inclinations. Which condition makes the ranges along the two planes equal?
Explanation:
For fixed u and θ, the range along an incline is R = [2u²cosθ/g]·[sin(θ − β)/cos²β]. Therefore, equality of the two ranges requires sin(θ − β₁)/cos²β₁ = sin(θ − β₂)/cos²β₂.
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